Wikibuku idwikibooks https://id.wikibooks.org/wiki/Halaman_Utama MediaWiki 1.46.0-wmf.24 first-letter Media Istimewa Pembicaraan Pengguna Pembicaraan Pengguna Wikibuku Pembicaraan Wikibuku Berkas Pembicaraan Berkas MediaWiki Pembicaraan MediaWiki Templat Pembicaraan Templat Bantuan Pembicaraan Bantuan Kategori Pembicaraan Kategori Resep Pembicaraan Resep Wisata Pembicaraan Wisata TimedText TimedText talk Modul Pembicaraan Modul Acara Pembicaraan Acara 0 9946 115017 101366 2026-04-26T23:31:29Z ~2026-25672-55 43045 115017 wikitext text/x-wiki <div style="text-align: center; font-size: 400%;">'''S''' untuk '''S'''api</div> [[File:Cow_female_black_white.jpg|450px|center]] {{Wikijunior:Alfabet}} <!-- <br/> {{listen | filename = S is for sun.ogg | title = S is for sun }} --> [[az:Vikiuşaq:Əlifba/S]] [[fr:Wikijunior:Alphabet/S]] el9vjp1s7sfx3t5wfis0lllgz57ou2c 0 9951 115018 101372 2026-04-26T23:33:33Z ~2026-25672-55 43045 115018 wikitext text/x-wiki <div style="text-align: center; font-size: 400%;">'''Y''' untuk '''Y'''ak</div> [[File:Himalayan_Yak_(40848200765).jpg|500px|center]] {{Wikijunior:Alfabet}} <!-- <br/> {{listen | filename = Y is for yoyo.ogg | title = Y is for yo-yo }} --> [[az:Vikiuşaq:Əlifba/Y]] [[fr:Wikijunior:Alphabet/Y]] 0bj2vfr3x55y249jraokme76ivl9ktv 1 20816 115019 101376 2026-04-26T23:36:04Z ~2026-25672-55 43045 115019 wikitext text/x-wiki <div style="text-align: center; font-size: 400%;">'''A''' untuk '''A'''pel</div> [[File:Fuji_apple.jpg|center|500px]] <!-- <br/> {{listen | filename = A is for apple.ogg | title = Bahasa Inggris |description = A is for apple }} --> [[az:Vikiuşaq:Əlifba/A]] [[en:Wikijunior:Alphabet/A]] [[fr:Wikijunior:Alphabet/A]] [[tr:Vikiçocuk:Alfabe/A]] <div style="text-align: center; font-size: 400%;">'''B''' untuk '''B'''ola</div> [[File:Generic_football.png|400px|center]] [[az:Vikiuşaq:Əlifba/B]] [[fr:Wikijunior:Alphabet/B]] <div style="text-align: center; font-size: 400%;">'''C''' untuk '''C'''abai</div> [[File:Colours_of_Capsicum_-_chili_-_chilli_-_piment.jpg|400px|center]] [[az:Vikiuşaq:Əlifba/C]] [[fr:Wikijunior:Alphabet/C]] <div style="text-align: center; font-size: 400%;">'''D''' untuk '''D'''anau</div> [[File:Emerald_Bay.jpg|500px|center]] <br/> [[fr:Wikijunior:Alphabet/D]] <div style="text-align: center; font-size: 400%;">'''E''' untuk '''E'''mber</div> [[File:Bucket.agr.jpg|320px|center]] <!-- <br/> {{listen | filename = E is for egg.ogg | title = E is for egg }} --> [[az:Vikiuşaq:Əlifba/E]] [[fr:Wikijunior:Alphabet/E]] <div style="text-align: center; font-size: 400%;">'''F''' untuk '''F'''osil</div> [[File:Tarbosaurus_museum_Muenster.jpg|500px|center]] <!-- <br/> {{listen | filename = F is for fish.ogg | title = F is for fish }} --> [[az:Vikiuşaq:Əlifba/F]] [[fr:Wikijunior:Alphabet/F]] <div style="text-align: center; font-size: 400%;">'''G''' untuk '''G'''erbang</div> [[File:Hargimont 051030 (5).JPG|500px|center]] <!-- <br/> {{listen | filename = G is for gate.ogg | title = G is for gate }} --> [[az:Vikiuşaq:Əlifba/G]] [[fr:Wikijunior:Alphabet/G]] <div style="text-align: center; font-size: 400%;">'''H''' untuk '''H'''anduk</div> [[File:Zusammengelegte Handtücher.jpg|500px|center]] <!-- <br/> {{listen | filename = H is for horse.ogg | title = H is for horse }} --> [[az:Vikiuşaq:Əlifba/H]] [[fr:Wikijunior:Alphabet/H]] <div style="text-align: center; font-size: 400%;">'''I''' untuk '''I'''stana</div> [[File:Budapest_four_seasons.jpg|450px|center]] <!-- <br/> {{listen | filename = I is for igloo.ogg | title = I is for igloo }} --> [[az:Vikiuşaq:Əlifba/I]] [[fr:Wikijunior:Alphabet/I]] <div style="text-align: center; font-size: 400%;">'''J''' untuk '''J'''amur</div> [[File:Amanita_rubescens_I.jpg|450px|center]] <!-- <br/> {{listen | filename = J is for jam.ogg | title = J is for jam }} --> [[az:Vikiuşaq:Əlifba/J]] [[fr:Wikijunior:Alphabet/J]] <div style="text-align: center; font-size: 400%;">'''K''' untuk '''K'''asur</div> [[Berkas:Shifman_Mattress_Set.JPG|450px|pus]] <!-- <br/> {{listen | filename = K is for kangaroo.ogg | title = K is for kangaroo }} --> [[az:Vikiuşaq:Əlifba/K]] [[fr:Wikijunior:Alphabet/K]] <div style="text-align: center; font-size: 400%;">'''L''' untuk '''L'''ampu</div> [[File:Energiesparlampe_01.jpg|400px|center]] <!-- <br/> {{listen | filename = L is for lion.ogg | title = L is for lion }} --> [[az:Vikiuşaq:Əlifba/L]] [[fr:Wikijunior:Alphabet/L]] <div style="text-align: center; font-size: 400%;">'''M''' untuk '''M'''eja</div> [[Berkas:Teapoy, ചെറിയ മേശ.JPG|450px|pus]] <!-- <br/> {{listen | filename = M is for mouse.ogg | title = M is for mouse }} --> [[az:Vikiuşaq:Əlifba/M]] [[fr:Wikijunior:Alphabet/M]] <div style="text-align: center; font-size: 400%;">'''N''' untuk '''N'''anas</div> [[File:Ananas~May 2008-1.jpg|300px|center]] <!-- <br/> {{listen | filename = N is for nest.ogg | title = N is for nest }} --> [[az:Vikiuşaq:Əlifba/N]] [[fr:Wikijunior:Alphabet/N]] <div style="text-align: center; font-size: 400%;">'''O''' untuk '''O'''bor</div> [[File:Burning_torches.JPG|500px|center]] <!-- <br/> {{listen | filename = O is for octopus.ogg | title = O is for octopus }} --> [[az:Vikiuşaq:Əlifba/O]] [[fr:Wikijunior:Alphabet/O]] <div style="text-align: center; font-size: 400%;">'''P''' untuk '''P'''intu</div> [[File:Bressanone,_santi_gottardo_ed_erardo_03_porta.JPG|300px|center]] <!-- <br/> {{listen | filename = P is for pig.ogg | title = P is for pig }} --> [[az:Vikiuşaq:Əlifba/P]] [[fr:Wikijunior:Alphabet/P]] <div style="text-align: center; font-size: 400%;">'''Q''' untuk '''Q'''uran</div> [[File:Opened_Qur'an.jpg|400px|center]] <!-- <br/> {{listen | filename = Q is for queen.ogg | title = Q is for queen }} --> [[az:Vikiuşaq:Əlifba/Q]] [[fr:Wikijunior:Alphabet/Q]] <div style="text-align: center; font-size: 400%;">'''R''' untuk '''R'''adio</div> [[File:Радиоприемник Верас РП-225.JPG|400px|border|center]] <!-- <br/> {{listen | filename = R is for road.ogg | title = R is for road }} --> [[az:Vikiuşaq:Əlifba/R]] [[fr:Wikijunior:Alphabet/R]] <div style="text-align: center; font-size: 400%;">'''S''' untuk '''S'''api</div> [[File:Cow_female_black_white.jpg|450px|center]] <!-- <br/> {{listen | filename = S is for sun.ogg | title = S is for sun }} --> [[az:Vikiuşaq:Əlifba/S]] [[fr:Wikijunior:Alphabet/S]] <div style="text-align: center; font-size: 400%;">'''T''' untuk '''T'''as</div> [[File:Rucksack_Schweizer_Armee_1960er_a.jpg|300px|center]] <!-- <br/> {{listen | filename = T is for turtle.ogg | title = T is for turtle }} --> [[az:Vikiuşaq:Əlifba/T]] [[fr:Wikijunior:Alphabet/T]] <div style="text-align: center; font-size: 400%;">'''U''' untuk '''U'''ang</div> [[File:Indonesian_Rupiah_(IDR)_banknotes2009.jpg|350px|center]] <!-- <br/> {{listen | filename = U is for umbrella.ogg | title = U is for umbrella }} --> [[az:Vikiuşaq:Əlifba/U]] [[fr:Wikijunior:Alphabet/U]] <div style="text-align: center; font-size: 400%;">'''V''' untuk '''V'''as</div> [[File:State Gifts White Vase.JPG|300px|center]] <!-- <br/> {{listen | filename = V is for volcano.ogg | title = V is for volcano }} --> [[az:Vikiuşaq:Əlifba/V]] [[fr:Wikijunior:Alphabet/V]] <div style="text-align: center; font-size: 400%;">'''W''' untuk '''W'''ajan</div> [[File:Stickhandle-pekingpan.jpg|500px|center]] <!-- <br/> {{listen | filename = W is for window.ogg | title = W is for window }} --> [[fr:Wikijunior:Alphabet/W]] <div style="text-align: center; font-size: 400%;">'''X''' untuk '''X'''ysticus</div> [[File:Xysticus.spec.6891.jpg|center|500px]] [[en:Wikijunior:Alphabet/X]] <div style="text-align: center; font-size: 400%;">'''Y''' untuk '''Y'''ak</div> [[File:Himalayan_Yak_(40848200765).jpg|500px|center]] <!-- <br/> {{listen | filename = Y is for yoyo.ogg | title = Y is for yo-yo }} --> [[az:Vikiuşaq:Əlifba/Y]] [[fr:Wikijunior:Alphabet/Y]] <div style="text-align: center; font-size: 400%;">'''Z''' untuk '''Z'''ebra</div> [[File:Zebra_rownikowa_Equus_burchelli_boehmi_RB3.jpg|500px|center]] <!-- <br/> {{listen | filename = Z is for zebra.ogg | title = Z is for zebra }} --> [[az:Vikiuşaq:Əlifba/Z]] [[fr:Wikijunior:Alphabet/Z]] 96iauktia7i6x6l8qarwx31nqm6uun9 0 23123 115049 114142 2026-04-27T11:02:37Z ~2026-25546-23 43043 115049 wikitext text/x-wiki Banyaknya pemetaan yang mungkin dari A ke B yaitu n(B)^n(A) untuk seluruh fungsinya. Banyaknya pemetaan yang mungkin dari A ke B dalam fungsi bijektif yaitu n(n-1). Sifat fungsi yaitu: # Fungsi surjektif (fungsi onto atau fungsi kepada). rumusnya: dari A ke B maka hasilnya B^A. contoh: A = {1, 2, 3, 4} dan B = {2, 4, 6} maka HP = {(1,2), (2,4), (3,6), (4,6)} ## Buktikan f(x)=x²-4 dengan batas-batas -3<x<3 merupakan fungsi surjektif! Misalkan x = -2 maka y = 0, x = -1 maka y = -3 jadi sebagai berikut: HP = {(-2,0), (-1,-3), (0,-4), (1,-3), (2,0)} Dalam tersebut adalah fungsi surjektif. # Fungsi injektif (fungsi into atau fungsi ke dalam) contoh: A = {1, 2, 3} dan B = {2, 4, 6, 8} maka HP = {(1,2), (2,4), (3,6)} # Fungsi bijektif (fungsi satu-satu atau korespondensi satu-satu) rumusnya: Bamyaknya n untuk A dan B: 2ⁿ. contoh: A = {1, 2, 3, 4} dan B = {2, 4, 6, 8} maka HP = {(1,2), (2,4), (3,6), (4,8)} ## Buktikan f(x)=1/2x dengan batas-batas 1<x<7 merupakan fungsi bijektif! Misalkan x = 2 maka y = 1/4, x = 3 maka y = 1/6 jadi sebagai berikut: HP = {(2,1/4), (3,1/6), (4,1/8), (5,1/10), (6,1/12)} Dalam tersebut adalah fungsi bijektif. # Bukan Fungsi surjektif dan injektif contoh: A = {1, 2, 3} dan B = {2, 4, 6, 8} maka HP = {(1,2), (2,4), (3,4)} Dalam pembuktian sebagai berikut: f:A -> B : x -> y Kalau membuktikan bahwa f(x) bersifat surjektif maka ∀y ∈ B, ∃x ∈ A, y = f(x). Kalau membuktikan bahwa f(x) bersifat injektif maka ∀p,q ∈ A, f(p) = f(q) ⟹ p = q. Kalau membuktikan bahwa f(x) bersifat bijektif maka memiliki fungsi surjektif dan injektif sekaligus. contoh # Apakah f:R -> R : x -> x³ merupakan fungsi bijektif? : Membuktikan f(x) merupakan fungsi surjektif :: Ambil sembarangan y ∈ B :: y = x³ :: x = <math>\sqrt[3]{y}</math> ∈ R maka terbukti f(x) tersebut adalah fungsi surjektif. : Membuktikan f(x) merupakan fungsi injektif :: Misalkan f(p) = f(q) dimana p,q ∈ R :: p³ = q³ :: <math>\sqrt[3]{(p)^3}</math> = <math>\sqrt[3]{(q)^3}</math> :: p = q maka terbukti f(x) tersebut adalah fungsi injektif. karena f(x) terbukti merupakan fungsi surjektif dan injektif jadi f(x) merupakan fungsi bijektif. == operasi aritmetika fungsi == * f(x)+g(x) = (f+g) x * f(x)-g(x) = (f-g) x * f(x).g(x) = (f.g) x * <math>\frac{f(x)}{g(x)} = (\frac{f}{g}) x</math> == bentuk fungsi lain == Domain disebut daerah asal sedangkan kodomain atau range disebut daerah hasil Beberapa fungsi-fungsi sebagai berikut: # Fungsi linear contoh soal 1 tentukan daerah asal serta hasil dari y = 5x+2! daerah asal: <math>HP = {x|-\infty<x<\infty, x \in R}</math><br> daerah hasil: <math>HP = {y|-\infty<y<\infty, y \in R}</math> # Fungsi kuadrat ; Sumbu simetri dan harga ekstrem/titik balik fungsi kuadrat <math>y=ax^2+bx+c</math> persamaan fungsi kuadrat sebagai berikut: : y = a(x-x₁)(x-x₂) : y = a(x-x_p)²+y_p sumbu simetri adalah x = <math>-\frac{b}{2a}</math>, nilai maksimum/minimum adalah <math>-\frac{D}{4a}</math> serta titik puncak/titik balik adalah <math>(-\frac{b}{2a}, -\frac{D}{4a})</math>. {| class="wikitable" |+ Kriteria akar-akar |- ! !! colspan=3| Pernyataan |- ! !! D>0 !! D=0 !! D<0 |- | a>0 (terbuka ke atas; nilai minimum) || rowspan=2| memotong || rowspan=2| menyinggung || rowspan=2| tidak memotong dan tidak menyinggung |- | a<0 (terbuka ke bawah; nilai maksimum) |- |} contoh soal 1 tentukan daerah asal serta hasil dari y = x²+5x+4! cari sumbu simetri x yaitu -b/2a adalah -5/2(1) = -5/2 serta hasil y yaitu (-5/2)²+5(-5/2)+4 = -9/4. daerah asal: <math>HP = {x|-\infty<x<\infty, x \in R}</math><br> daerah hasil: <math>HP = {y|y \ge -\frac{9}{4}, y \in R}</math> 2 tentukan sumbu simetri dan titik balik dari persamaan y = x²-4x+3! ; cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} y &= x^2-4x+3 \\ x &= -\frac{-4}{2 \cdot 1} = 2 \\ y &= -\frac{(-4)^2-4 \cdot 1 \cdot 3}{4 \cdot 1} \\ &= -\frac{4}{4} = -1 \\ \end{align} </math> </div></div> ; cara 2 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} y &= x^2-4x+3 \\ x &= -\frac{-4}{2 \cdot 1} = 2 \\ y &= 2^2-4(2)+3 = -1 \\ \end{align} </math> </div></div> ; cara 3 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} y &= x^2-4x+3 \\ y' &= 0 \\ 2x-4 &= 0 \\ x &= 2 \\ y &= 2^2-4(2)+3 = -1 \\ \end{align} </math> </div></div> sumbu simetri x=2 dan titik baliknya (2,-1). # agar garis lurus y=x-10 memotong parabola y=x²-(m-1)x+6 maka berapa batasan nilai m serta titik potongnya? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Pembuktian</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} y &= x-10 \\ y &= x^2-(m-1)x+6 \\ x-10 &= x^2-(m-1)x+6 \\ x-10 &= x^2-mx+x+6 \\ x^2-mx+16 &= 0 \\ \text{syarat memotong yaitu } D > 0 \\ (-m)^2-4(1)(16) &> 0 \\ m^2-64 &> 0 \\ \text{harga nol } \\ m^2-64 &= 0 \\ (m-8)(m+8) &= 0 \\ m_1 = 8 &\text{ atau } m_2 = -8 \\ \text{batasan melalui irisan menjadi } m<-8 \text{ atau } m>8 \\ \text{untuk } m_1=8 \\ x^2-8x+16 &= 0 \\ (x-4)^2 &= 0 \\ x &= 4 \\ y &= 4-10 \\ &= -6 \\ \text{titik potong adalah (4,6) } \\ \text{untuk } m_2=-8 \\ x^2+8x+16 &= 0 \\ (x+4)^2 &= 0 \\ x &= -4 \\ y &= -4-10 \\ &= -14 \\ \text{titik potong adalah (-4,-16) } \\ \end{align} </math> </div></div> # agar garis lurus y=2x-1 menyinggung parabola y=mx²+(m-5)x+8 maka berapa nilai m serta titik singgungnya? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Pembuktian</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} y &= 2x-1 \\ y &= mx^2+(m-5)x+8 \\ 2x-1 &= mx^2+(m-5)x+8 \\ 2x-1 &= mx^2+mx-5x+8 \\ mx^2+mx-7x+9 &= 0 \\ mx^2+(m-7)x+9 &= 0 \\ \text{syarat menyinggung yaitu } D = 0 \\ (m-7)^2-4(m)(9) &= 0 \\ m^2-14m+49-36m &= 0 \\ m^2-50m+49 &= 0 \\ (m-1)(m-49) &= 0 \\ m_1 = 1 &\text{ atau } m_2 = 49 \\ \text{nilainya } m=1 \text{ atau } m=49 \\ \text{untuk } m_1=1 \\ x^2-6x+9 &= 0 \\ (x-3)^2 &= 0 \\ x &= 3 \\ y &= 2(3)-1 \\ &= 5 \\ \text{titik singgung adalah (3,5) } \\ \text{untuk } m_2=49 \\ 49x^2+42x+9 &= 0 \\ (7x+3)^2 &= 0 \\ x &= -\frac{3}{7} \\ y &= 2(-\frac{3}{7})-1 \\ &= -\frac{13}{7} \\ \text{titik singgung adalah } (-\frac{3}{7},-\frac{13}{7}) \\ \end{align} </math> </div></div> # agar garis lurus y=2mx-1 tidak memotong dan tidak menyinggung parabola y=x²-x+3 maka berapa batasan nilai m? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Pembuktian</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} y &= 2mx-1 \\ y &= x^2-x+3 \\ 2mx-1 &= x^2-x+3 \\ x^2-2mx-x+4 &= 0 \\ x^2+(-2m-1)x+4 &= 0 \\ \text{syarat tidak memotong dan tidak menyinggung yaitu } D < 0 \\ (-2m-1)^2-4(1)(4) &< 0 \\ 4m^2+4m+1-16 &< 0 \\ 4m^2+4m-15 &< 0 \\ \text{harga nol } \\ 4m^2+4m-15 &= 0 \\ (2m-3)(2m+5) &= 0 \\ m_1 = \frac{3}{2} &\text{ atau } m_2 = -\frac{5}{2} \\ \text{batasan melalui irisan menjadi } -\frac{5}{2} < m < \frac{3}{2} \\ \end{align} </math> </div></div> # tentukan nilai k dan titik potongnya jika garis lurus y=x+k memotong parabola y=x²-4x+3 yang berabsis 6! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Pembuktian</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} y &= x^2-4x+3 \\ &= 6^2-4(6)+3 \\ &= 15 \\ y &= x+k \\ 15 &= 6+k \\ k &= 9 \\ y &= x^2-4x+3 \\ y &= x+9 \\ x+9 &= x^2-4x+3 \\ x^2-5x-6 &= 0 \\ (x-6)(x+1) &= 0 \\ x_1 = 6 &\text{ atau } x_2 = -1 \\ \text{untuk } x_1 = 6 \\ y &= 6 + 9 \\ &= 15 \\ \text{untuk } x_2 = -1 \\ y &= -1 + 9 \\ &= 8 \\ \text{titik potongnya adalah (-1,8) dan (6,15) } \\ \end{align} </math> </div></div> ; gambar fungsi {| class="wikitable" |+ |- ! !! D > 0<br> Terdapat 2 titik terbuka !! D = 0<br> Terdapat 1 titik terbuka || D < 0<br> Tidak terdapat titik terbuka |- | a > 0 || colspan=3| Melengkung ke atas dari titik balik |- | a < 0 || colspan=3| Melengkung ke bawah dari titik balik |} # Fungsi akar contoh soal 1 tentukan daerah asal serta hasil dari <math>y = \sqrt{2x + 1}</math>! cari batasan sumbu x yaitu 2x + 1 ≥ 0 adalah x ≥ -1/2. daerah asal: <math>HP = {x|x \ge -\frac{1}{2}, x \in R}</math><br> daerah hasil: <math>HP = {y|y \ge 0, y \in R}</math> # Fungsi pecahan contoh soal 1 tentukan daerah asal serta hasil dari <math>y = \frac{2x+1}{3x-1}</math>! daerah asal: <math>HP = {x|x \neq \frac{1}{3}, x \in R}</math><br> daerah hasil: <math>HP = {y|y \neq \frac{2}{3}, y \in R}</math> 2 tentukan daerah asal serta hasil dari <math>y = \frac{x^2-4x-5}{x^2-x-6}</math>! daerah asal: <math>HP = {x|x \neq {-2, 3} , x \in R}</math><br> daerah hasil: <math>HP = {y|-\infty<y<1 \text{ v } <1<y<\infty , y \in R}</math> 3 tentukan daerah asal serta hasil dari <math>y = \frac{x^2+x-20}{2x+3}</math>! daerah asal: <math>HP = {x|x \neq -\frac{3}{2}, x \in R}</math><br> daerah hasil: <math>HP = {y|-\infty<y<\infty, y \in R}</math> ; Pembuatan grafik ;fungsi pecahan linear <math>\frac{ax+b}{px+q}</math> dengan p≠0. Langkah-langkahnya sebagai berikut: # titik potong sumbu x ((-b/a,0)) # titik potong sumbu y ((0,b/q)) # asymtot tegak (x=-q/p) # asymtot datar (y=a/p) # titik-titik lainnya ;fungsi pecahan kuadrat <math>\frac{ax^2+bx+c}{px^2+qx+r}</math> dengan {a,p}≠0. Langkah-langkahnya sebagai berikut: # titik potong sumbu x (ax²+bx+c=0) # titik potong sumbu y ((0,c/r)) # asymtot tegak (px²+qx+r=0) # asymtot datar (y=a/p) # harga ekstrem/titik balik (cari diskriminan dari persamaan yang bernilai y dimana y sebagai misalnya serta syarat D ≥ 0. kalau y didapat tinggal cari x nya) # titik potong tegak dengan asymtot datar (cari nilai x dimana y adalah asymtot datar) # titik-titik lainnya ;fungsi pecahan kuadrat linear <math>\frac{ax^2+bx+c}{px+q}</math> dengan {a,p}≠0. Langkah-langkahnya sebagai berikut: # titik potong sumbu x (ax²+bx+c=0) # titik potong sumbu y ((0,c/q)) # asymtot tegak (x=-q/p) # asymtot miring (hasil bagi dari pembilang dengan penyebut) # harga ekstrem/titik balik (cari diskriminan dari persamaan yang bernilai y dimana y sebagai misalnya serta syarat D ≥ 0. kalau y didapat tinggal cari x nya) # titik-titik lainnya ;fungsi pecahan linear kuadrat <math>\frac{ax+b}{px^2+qx+r}</math> dengan p≠0. Langkah-langkahnya sebagai berikut: # titik potong sumbu x ((-b/a,0)) # titik potong sumbu y ((0,b/r)) # asymtot tegak (px²+qx+r=0) # asymtot datar (y=0) # harga ekstrem/titik balik (cari diskriminan dari persamaan yang bernilai y dimana y sebagai misalnya serta syarat D ≥ 0. kalau y didapat tinggal cari x nya) # titik potong tegak dengan asymtot datar (cari nilai x dimana y adalah asymtot datar) # titik-titik lainnya # Fungsi mutlak Fungsi mutlak adalah setiap fungsi memiliki aturan nilainya mutlak dan selalu bernilai positif. rumus: <math>|f(x)| = \begin{cases} x & \text{ untuk } x \ge 0 \\ -x & \text{ untuk } x < 0 \end{cases} </math> contoh soal 1 tentukan daerah asal serta hasil dari y = |2x + 1]! cari batasan sumbu x yaitu 2x + 1 = 0 adalah 0. daerah asal: <math>HP = {x|-\infty<x<\infty, x \in R}</math><br> daerah hasil: <math>HP = {y|y \ge 0, y \in R}</math> # Fungsi sepenggal (piecewise) Fungsi sepenggal adalah setiap fungsi dari daerah hasil memiliki beberapa fungsi dari daerah asal yang berbeda. contoh: <math>f(x) = \begin{cases} 2x-3 & \text{ untuk } x \ge \frac{3}{2} \\ x & \text{ untuk } x < \frac{3}{2} \\ \end{cases} </math>, <math>f(x) = \begin{cases} cos x & \text{ untuk } x \ge 0 \\ sin x & \text{ untuk } x < 0 \end{cases} </math> # Fungsi tangga Fungsi tangga adalah fungsi yang bernilai konstan pada interval-interval dimana ia didefinisikan. Fungsi tangga terbagi tiga jenis yaitu ## Fungsi atap Fungsi atap disebut juga fungsi bilangan bulat terkecil. Dilambangkan sebagai berikut: <math>f(x) = \lceil x \rceil = \text{atap}(x)</math> dan rumus <math>\lceil x \rceil = min{m \in Z, m \ge x}</math>. Contoh: <math>\lceil 0,\!8 \rceil = 1</math>, <math>\lceil 2,\!4 \rceil = 3</math> dan <math>\lceil -3,\!5 \rceil = -3</math> sifat-sifat: * <math>\lceil x \rceil = x</math> jika x bulat * <math>x \le \lceil x \rceil < x+1</math> * <math>\lceil x \rceil = n \text{ maka } n-1 < x \le n</math> * <math>n < \lceil x \rceil \text{ maka } n < x</math> * <math>\lceil x \rceil \le n \text{ maka } x \le n</math> * <math>\lceil x + n \rceil = \lceil x \rceil + n</math> * <math>\lceil x \rceil + \lceil y \rceil \ge \lceil x+y \rceil</math> * <math>\lceil xy \rceil \ge \lceil x \rceil \cdot \lceil y \rceil</math> ## Fungsi lantai Fungsi lantai disebut juga fungsi bilangan bulat terbesar. Dilambangkan sebagai berikut: <math>f(x) = \lfloor x \rfloor = \text{lantai}(x)</math> dan rumus <math>\lfloor x \rfloor = max{m \in Z, m \le x}</math>. Contoh: <math>\lfloor 4,\!27 \rfloor = 4</math>, <math>\lceil 5,\!69 \rceil = 5</math> dan <math>\lfloor -7,\!5 \rfloor = -8</math> sifat-sifat: * <math>\lfloor x \rfloor = x</math> jika x bulat * <math>x-1 < \lfloor x \rfloor \le x</math> * <math>\lfloor x \rfloor = n \text{ maka } n < x \le n+1</math> * <math>\lfloor x \rfloor < n \text{ maka } n < x</math> * <math>n \le \lfloor x \rfloor \text{ maka } x \le n</math> * <math>\lfloor x + n \rfloor = \lfloor x \rfloor + n</math> * <math>\lfloor x \rfloor + \lfloor y \rfloor \le \lfloor x+y \rfloor</math> * <math>\lfloor xy \rfloor \le \lfloor x \rfloor \cdot \lfloor y \rfloor</math> ## Fungsi pembulatan Dilambangkan sebagai berikut: <math>f(x) = [x] = \text{int}(x)</math>. Contoh: [8,2] = 8, [9,7] = 10 dan [10,5] = 11 :: tambahan; * x = <math>\lfloor x \rfloor</math>+{x} :: hubungan fungsi atap dan fungsi lantai * <math>x-1 < \lfloor x \rfloor \le \lceil x \rceil < x+1</math> * <math>-\lfloor x \rfloor = \lceil -x \rceil</math> * <math>\lfloor -x \rfloor = -\lceil x \rceil</math> * <math>\lceil x \rceil - \lfloor x \rfloor</math> = 0 (?) jika x bulat * <math>\lceil x \rceil - \lfloor x \rfloor</math> = 1 jika x takbulat # Fungsi eksponensial rumus: f(x) = a^x # Fungsi logaritma rumus: f(x) = ^alog x, a>0, x>0, <math>a \neq 1</math> ; Tambahan soal * Tentukan hasil nilai dibawah ini! # f(1) jika <math>f(\frac{26}{3^x-1}) = \frac{16}{x^2-1}</math> # f(2) jika <math>f(\sqrt{3^x-5}) = log 5x</math> # f(10) jika f(6) = 61 serta <math>f(6) \cdot f(\frac{5}{6}) = f(6)+25 f(\frac{5}{6})</math> # f(2025) jika <math>x^3f(x)+f(2-x) = 3x+3x^4</math> 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(\frac{26}{3^x-1}) &= \frac{16}{x^2-1} \\ \text{ anggap bahwa } f(1) &= f(\frac{26}{3^x-1}) \\ 1 &= \frac{26}{3^x-1} \\ 3^x-1 &= 26 \\ 3^x &= 27 \\ 3^x &= 3^3 \\ x &= 3 \\ \text { nah x adalah 3 } \\ \frac{16}{x^2-1} &= \frac{16}{3^2-1} \\ &= \frac{16}{8} \\ &= 2 \\ f(1) &= 2 \\ \end{align} </math> </div></div> 2 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(\sqrt{3^x-5}) &= log 5x \\ \text{ anggap bahwa } f(2) &= f(\sqrt{3^x-5}) \\ 2 &= \sqrt{3^x-5} \\ 3^x-5 &= 2^2 \\ 3^x-5 &= 4 \\ 3^x &= 9 \\ 3^x &= 3^2 \\ x &= 2 \\ \text { nah x adalah 2 } \\ log 5x &= log 5(2) \\ &= log 10 \\ &= 1 \\ f(2) &= 1 \\ \end{align} </math> </div></div> 3 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(6) \cdot f(\frac{5}{6}) &= f(6) + 25 f(\frac{5}{6}) \\ 61 f(\frac{5}{6}) &= 61 + 25 f(\frac{5}{6}) \\ 36 f(\frac{5}{6}) &= 61 \\ f(\frac{5}{6}) &= \frac{61}{36} \\ f(\frac{5}{6}) &= 1+\frac{25}{36} \\ f(\frac{5}{6}) &= 1+\frac{5^2}{6^2} \\ f(\frac{5}{6}) &= 1+(\frac{5}{6})^2 \\ f(x) &= 1+x^2 \\ f(x) &= 1+x^2 \\ f(10) &= 1+10^2 \\ &= 101 \\ \end{align} </math> </div></div> 4 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^3f(x)+f(2-x) &= 3x+3x^4 \\ \text{jika f(x) adalah f(2025) maka } \\ x^3f(2025)+f(-2023) &= 3x+3x^4 \\ \text{jika f(2-x) adalah f(2025) maka } \\ x^3f(-2023)+f(2025) &= 3x+3x^4 \\ \text{eliminasi untuk kedua persamaan menjadi } \\ x^6f(2025)-f(2025) &= (x^3-1)(3x+3x^4) \\ f(2025)(x^6-1) &= (x^3-1)3x(x^3+1) \\ f(2025)(x^6-1) &= 3x(x^6-1) \\ f(2025) &= 3x \\ f(2025) &= 3(2025) \\ f(2025) &= 6075 \\ \end{align} </math> </div></div> == Fungsi ganjil dan genap == fungsi ganjil apabila f(x) = -f(-x) serta fungsi genap apabila f(x) = f(-x). contoh soal Apakah fungsi ganjil atau genap sebagai berikut: * f(x) = x² * f(x) = x⁷ * f(x) = x² sin x * f(x) = x² cos 3x ;jawaban * f(x) = x² -> f(-x) = (-x)² : f(-x) = x² = f(x) -> fungsi genap * f(x) = x⁷ -> f(-x) = (-x)⁷ : f(-x) = -x⁷ = f(x) -> fungsi ganjil * f(x) = x² sin x -> f(-x) = (-x)² sin (-x) : f(-x) = x² (- sin x) = -x² sin x = -f(x) -> fungsi ganjil * f(x) = x² cos x -> f(-x) = (-x)² cos (-3x) : f(-x) = x² cos x = f(x) -> fungsi genap [[Kategori:Soal-Soal Matematika]] l3exjw3v8zg04s6ixqunn1we519nbtr 0 23140 115015 115010 2026-04-26T23:23:10Z ~2026-25546-23 43043 /* integral pecahan parsial */ 115015 wikitext text/x-wiki == Kaidah umum == :# <math>\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ konstan)}\,\!</math> :# <math>\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx</math> :# <math>\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx</math> :# <math>\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx</math> :# <math>\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\! </math> :# <math>\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C </math> :# <math>\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C </math> :# <math>\int f(x) \, dx = F(x) + C</math> :# <math>\int_a^b f(x) \, dx = F(b) - F(a)</math> :# <math>\int f(x) \circ g(x) dx</math> = f’(g(x)) g’(x) :# <math>\int f^{-1}(x) dx = \frac{1}{f'(f'(x))}</math> == rumus sederhana == :<math>\int \, dx = x + C</math> :<math>\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int {dx \over x} = \ln{\left|x\right|} + C</math> <math>\int {dx \over \sqrt{a^2-b^2 x^2}} = \frac{1}{b} \arcsin \, {bx \over a} + C</math> :<math>\int {dx \over {a^2+b^2x^2}} = {1 \over ab}\arctan \, {bx \over a} + C</math> <math>\int {dx \over x\sqrt{b^2x^2-a^2}} = {1 \over a}\arcsec \, {bx \over a} + C</math> ; Eksponen dan logaritma :<math>\int e^x\,dx = e^x + C</math> :<math>\int a^x\,dx = \frac{a^x}{\ln{a}} + C</math> :<math>\int \ln \, x\,dx = x \cdot \, \ln {x} - x + C</math> :<math>\int \, ^b\!\log \, x\,dx = x \cdot \, ^b\!\log \, x - x \cdot \,^b\!\log e + C</math> ; Trigonometri :<math>\int \sin \, x \, dx = -\cos \, x + C</math> :<math>\int \cos \, x \, dx = \sin \, x + C</math> :<math>\int \tan \, x \, dx = -\ln{\left| \cos \, x \right|} + C</math> :<math>\int \cot \, x \, dx = \ln{\left| \sin \, x \right|} + C</math> :<math>\int \sec \, x \, dx = \ln{\left| \sec \, x + \tan \, x \right|} + C</math> :<math>\int \csc \, x \, dx = -\ln{\left| \csc \, x + \cot \, x \right|} + C</math> ; Hiperbolik :<math>\int \sinh \, x \, dx = \cosh \, x + C</math> :<math>\int \cosh \, x \, dx = \sinh \, x + C</math> :<math>\int \tanh \, x \, dx = \ln{\left| \cosh \, x \right|} + C</math> :<math>\int \coth \, x \, dx = \ln{\left| \sinh \, x \right|} + C</math> :<math>\int \mbox{sech} \, x \, dx = \arctan \,(\sinh \, x) + C</math> :<math>\int \mbox{csch} \, x \, dx = \ln{\left| \tanh \, {x \over 2} \right|} + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : 1. <math>\int x^3 - cos 3x + e^{5x+2} - \frac{1}{2x+5}\,dx</math> : <math>= \frac{1}{3+1}x^{3+1} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : <math>= \frac{x^4}{4} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : 2. <math>\int x(x-5)^4\,dx</math> : <math>= \int x(x^4-4x^3(5)+6x^2(25)-4x(125)+625)\,dx</math> : <math>= \int x(x^4-20x^3+150x^2-500x+625)\,dx</math> : <math>= \int x^5-20x^4+150x^3-500x^2+625x\,dx</math> : <math>= \frac{x^6}{6}-4x^5+\frac{150x^4}{4}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> : <math>= \frac{x^6}{6}-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : 1. <math>\int \frac{\ln \, x}{x}\,dx</math> : <math>u = \ln \, x\, du = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{\ln(x)}{x} dx \\ \int u du \\ \frac{1}{2} u^2 + C \\ \frac{1}{2} ln^2 x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x-5,\, du = dx,\, x = u+5</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4\,dx \\ \int (u+5)u^4\,du \\ \int (u^5+5u^4)\,du \\ \frac{1}{6} u^6+u^5 + C \\ \frac{1}{6} (x-5)^6+(x-5)^5 + C \\ \frac{1}{6} (x^6-6x^5(5)+15x^4(25)-20x^3(125)+15x^2(625)-6x(3125)+15625)+(x^5-5x^4(5)+10x^3(25)-10x^2(125)+5x(625)+3125) + C \\ \frac{1}{6}x^6-5x^5+\frac{125x^4}{2}-\frac{1250x^3}{3}+\frac{3125x^2}{2}-3125x+\frac{15625}{6}+x^5-25x^4+250x^3-1250x^2+3125x+3125 + C \\ \frac{1}{6}x^6-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C \\ \end{aligned} </math> </div></div> === integral parsial === ;Cara 1: Rumus Integral parsial diselesaikan dengan rumus berikut. : <math>\int u\,dv = uv - \int v\,du </math> Berikut contoh penyelesaian cara parsial dengan rumus. : 1. <math>\int x \sin \, x \, dx</math> : <math>u = x, du = 1 \, dx, dv = \sin \, x \, dx, v = -\cos \, x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \,dx \\ (x)(-\cos \, x) - \int (-\cos \, x)(1 dx) \\ -x \cos \, x + \int \cos \, x dx \\ -x \cos \, x + \sin \, x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x,\, du = 1\,dx,\, dv = (x-5)^4\,dx,\, v = \frac{1}{5}(x-5)^5</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - \int (\frac{1}{5}(x-5)^5)(1\,dx) \\ \frac{x}{5}(x-5)^5 - \int \frac{1}{5}(x-5)^5\,dx \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6\,dx \\ \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x\,dx</math> : <math>u = \ln \,x,\, du = \frac{1}{x}\,dx,\, dv = x\,dx,\, v = \frac{x^2}{2}</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x\, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{x^2}{2})(\frac{1}{x}\,dx) \\ \frac{x^2}{2} \ln \,x - \int \frac{x^2}{2}(\frac{1}{x})\,dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \,dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x\,dx</math> ini terjadi dua kali integral parsial : <math>u = \sin \,x,\, du = \cos \,x\,dx,\, dv = e^x\,dx,\, v = e^x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x\, dx \\ (\sin \,x)(e^x) - \int (e^x)(\cos \,x\,dx) \\ e^x \sin \,x - \int e^x \cos \,x\,dx \\ u = \cos \,x,\, du = -\sin \,x\,dx,\, dv = e^x\,dx,\, v = e^x \\ (\cos \,x)(e^x) - \int (e^x)(-\sin \,x\,dx) \\ e^x \cos \,x + \int e^x \sin \,x\,dx \\ e^x \sin \,x - (e^x \cos \,x + \int e^x \sin \,x\,dx) \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x\,dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> ;Cara 2: Tabel Untuk <math display="inline">\int u\,dv</math>, berlaku ketentuan sebagai berikut. {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>u</math> || <math>dv</math> |- | - || <math>\frac{du}{dx}</math> || <math>v</math> |- | + || <math>\frac{d^2u}{dx^2}</math> || <math>\int v\,dx</math> |} Berikut contoh penyelesaian cara parsial dengan tabel. : 1. <math>\int x \sin \, x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>\sin \, x</math> |- | - || <math>1</math> || <math>-\cos \, x</math> |- | + || <math>0</math> || <math>-\sin \, x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \, dx \\ (x)(-\cos \, x) - (1)(-\sin \, x) + C \\ -x \cos \, x + \sin \, x + C \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4 \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>(x-5)^4</math> |- | - || <math>1</math> || <math>\frac{1}{5}(x-5)^5</math> |- | + || <math>0</math> || <math>\frac{1}{30}(x-5)^6</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - (1)(\frac{1}{30}(x-5)^6) + C \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6 + C \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\ln \,x</math> || <math>x</math> |- | - || <math>\frac{1}{x}</math> || <math>\frac{x^2}{2}</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x \, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{1}{x})(\frac{x^2}{2}) \, dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \, dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x \, dx</math> ini terjadi dua kali integral parsial {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\sin \,x</math> || <math>e^x</math> |- | - || <math>\cos \, x</math> || <math>e^x</math> |- | + || <math>-\sin \,x</math> || <math>e^x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x \, dx \\ (\sin \,x)(e^x) - (\cos \,x)(e^x) + \int (-\sin \,x)(e^x) \, dx \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x) \, dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{1}{x^2 - 4} dx</math> Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{1}{x^2 - 4} \\ =&\; \frac{1}{(x + 2)(x - 2)} \\ =&\; \frac{A}{x + 2} + \frac{B}{x - 2} \\ =&\; \frac{A(x - 2) + B(x + 2)}{x^2 - 4} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4} dx \\ =&\; \int -\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)}\,dx \\ =&\; \frac{1}{4} \int \frac{1}{x - 2} - \frac{1}{x + 2}\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> : <math>\int \frac{x^2+5x+4}{(x-3)^3}dx</math> : cara 1 Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^2} \\ ; \text{untuk mencari C yaitu } (3)^2+5(3)+4=28 \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{28}{(x - 3)^2} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{dx}{x^2 - 4} \\ =&\; \int -\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)}\,dx \\ =&\; \frac{1}{4} \int \frac{1}{x - 2} - \frac{1}{x + 2}\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> === integral substitusi trigonometri === {| class="wikitable" |- ! Bentuk !! Substitusi Trigonometri |- | <math>\sqrt{a^2 - b^2 x^2}</math> || <math>x = \frac{a}{b} sin \, \alpha </math> |- | <math>\sqrt{a^2 + b^2 x^2}</math> || <math>x = \frac{a}{b} tan \, \alpha</math> |- | <math>\sqrt{b^2 x^2 - a^2}</math> || <math>x = \frac{a}{b} sec \, \alpha</math> |} Berikut contoh penyelesaian cara substitusi trigonometri. : <math>\int \frac{dx}{x^2 \sqrt{x^2 + 4}}</math> : <math>x = 2 \tan \, A, dx = 2 \sec^2 \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \int \frac{dx}{x^2\sqrt{x^2+4}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(2 \tan \, A)^2 \sqrt{4 + (2 \tan \, A)^2}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 + 4 \tan^2 \, A}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4(1 + \tan^2 \, A)}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 \sec^2 \, A }} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(4 \tan^2 \, A)(2 \sec \, A)} \\ =&\; \int \frac{\sec \, A \,dA}{4 \tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\sec \, A \,dA}{\tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A} \,dA \end{aligned} </math> Substitusi berikut dapat dibuat. : <math>\int \frac{\cos \, A}{\sin^2 \, A} \,dA</math> : <math>t = \sin \, A, dt = \cos \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A}\,dA \\ =&\; \frac{1}{4} \int \frac{dt}{t^2} \\ =&\; \frac{1}{4} \left(-\frac{1}{t}\right) + C \\ =&\; -\frac{1}{4 t} + C \\ =&\; -\frac{1}{4 \sin \, A} + C \end{aligned} </math> Ingat bahwa <math display="inline">\sin \, A = \frac{x}{\sqrt{x^2 + 4}}</math> berlaku. : <math> \begin{aligned} =&\; -\frac{1}{4 \sin \, A} + C \\ =&\; -\frac{\sqrt{x^2 + 4}}{4x} + C \end{aligned} </math> === integral mutlak === : <math>\int |f(x)| \,dA</math> buatlah <math>f(x) \ge 0</math> jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x) : <math>\int_a^c |f(x)| \, dx = \int_a^b f(x) \, dx + \int_b^c -f(x) \, dx = F(b) - F(a) - F(c) + F(b)</math> jika <math>|f(x)| = \begin{cases} f(x), & \mbox{jika } f(x) \ge b, \\ -f(x), & \mbox{jika } f(x) < b. \end{cases}</math>. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif. === integral fungsi ganjil dan integral fungsi genap === : <math>\int_{-a}^a f(x) \,dx</math> dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral: : integral fungsi ganjil : <math>\int_{-a}^a f(x) \,dx</math> = 0 : integral fungsi genap : <math>\int_{-a}^a f(x) \,dx</math> = <math>2 \int_{0}^a f(x) \,dx</math> === integral notasi === : <math>\int_a^b f(x) \,dx = \int_{a-p}^{b-p} f(x) \,dx = \int_{a+p}^{b+p} f(x) \,dx</math> : <math>\int_a^c f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx</math> : <math>\int_a^b f(x \pm k) \,dx = \int_{a \pm k}^{b \pm k} f(x) \,dx</math> === integral terbalik === : <math>\int_a^b f(x) \,dx = -\int_b^a f(x) \,dx</math> == Jenis integral lainnya == === panjang busur === ; Sumbu ''x'' : <math>S = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' : <math>S = \int_{y_1}^{y_2} \sqrt{1 + (f'(y))^2}\,dy</math> === luas daerah === ; Satu kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} f(x)\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} f(y)\,dy</math> ; Dua kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} (f(x_2) - f(x_1))\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} (f(y_2) - f(y_1))\,dy</math> : atau juga <math>L = \frac {D \sqrt{D}}{6 a^2}</math> === luas permukaan benda putar === ; Sumbu ''x'' sebagai poros : <math>L_p = 2 \pi \int_{x_1}^{x_2} f(x)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' sebagai poros : <math>L_p = 2 \pi \int_{y_1}^{y_2} f(y)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(y))^2}\,dy</math> === volume benda putar === ; Satu kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} (f(x))^2\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} (f(y))^2\,dy</math> ; Dua kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} ((f(x_2))^2 - (f(x_1))^2)\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} ((f(y_2))^2 - (f(y_1))^2)\,dy</math> : atau juga <math>V = |\frac {D^2 \sqrt{D}}{30 a^3}|</math> == Integral lipat == Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga. ;contoh # Tentukan hasil dari: * <math>\int (5x-1)(5x^2-2x+7)^4 dx</math> * <math>\int x^2 cos 3x dx</math> * <math>\int sin^2 x cos x dx</math> * <math>\int cos^2 x sin x dx</math> * <math>\int sec x dx</math> * <math>\int csc x dx</math> * <math>\int \frac{1}{2x^2-5x+3} dx</math> * <math>\int \frac{1}{\sqrt{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> * <math>\int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx</math> * <math>\int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx</math> * <math>\int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx</math> * <math>\int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx</math> * <math>\int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx \\ a=1, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-sin^2 x} \\ &= \frac{1}{cos^2 x} \\ &= sec^2 x \\ \int sec^2 x dx &= tan x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx \\ a=1, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-cos^2 x} \\ &= \frac{1}{sin^2 x} \\ &= csc^2 x \\ \int csc^2 x dx &= -cot x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx \\ a=sin x, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{sin x}{1-sin^2 x} \\ &= \frac{sin x}{cos^2 x} \\ &= sec x \cdot tan x \\ \int sec x \cdot tan x dx &= sec x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx \\ a=cos x, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{cos x}{1-cos^2 x} \\ &= \frac{cos x}{sin^2 x} \\ &= csc x \cdot cot x \\ \int csc x \cdot cot x dx &= -csc x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx \\ \text{misalkan } y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y &= \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y^2 &= \frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}} \\ &= \frac{x}{y} \\ y^3 &= x \\ 3y^2 dy &= dx \\ \int y \cdot 3y^2 dy \\ \int 3y^3 dy \\ \frac{3y^4}{4}+C \\ \end{aligned} </math> </div></div> # Tentukan hasil dari: * <math>\int_{1}^{5} 2x-3 dx</math> * <math>\int_{0}^{4} |x-2| dx</math> * <math>\int_{2}^{4} |3-x| dx</math> * <math>\int_{1}^{5} x^2-6x+8 dx</math> * <math>\int_{-3}^{5} |x^2-2x-8| dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} 2x-3 dx &= \left. x^2 - 3x \right|_{1}^{5} \\ &= 10 - (-2) \\ &= 12 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{0}^{4} |x-2| dx \\ \text{ tentukan nilai harga nol } \\ x-2 &= 0 \\ x &= 2 \\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti } \\ |x-2| = \begin{cases} x-2, & \mbox{jika } x \ge 2, \\ -(x-2), & \mbox{jika } x < 2. \end{cases} \\ \int_{0}^{4} |x-2| dx &= \int_{2}^{4} x-2 dx + \int_{0}^{2} -(x-2) dx \\ &= \left. (\frac{x^2}{2}-2x) \right|_{2}^{4} + \left. (\frac{-x^2}{2}+2x) \right|_{0}^{2} \\ &= 0 - (-2) + 2 - 0 \\ &= 4 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{2}^{4} |3-x| dx \\ \text{ tentukan nilai syarat-syarat } \\ 3-x &\ge 0 \\ \text{ harga nol } \\ x &= 3 \\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti } \\ |3-x| = \begin{cases} 3-x, & \mbox{jika } x \le 3, \\ -(3-x), & \mbox{jika } x > 3. \end{cases} \\ \int_{2}^{4} |3-x| dx &= \int_{2}^{3} 3-x dx + \int_{3}^{4} -(3-x) dx \\ &= \left. (3x-\frac{x^2}{2}) \right|_{2}^{3} + \left. (-3x+\frac{x^2}{2}) \right|_{3}^{4} \\ &= 4.5 - 4 - 4 - (-4.5) \\ &= 1 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} x^2-6x+8 dx &= \left. \frac{x^3}{3}-3x^2+8x \right|_{1}^{5} \\ &= \frac{20}{3} - \frac{16}{3} \\ &= \frac{4}{3} \\ &= 1\frac{1}{3} \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{-3}^{5} |x^2-2x-8| dx \\ \text{ tentukan nilai syarat-syarat } \\ x^2-2x-8 &\ge 0 \\ \text{ harga nol } \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x = 4 \\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti } \\ |x^2-2x-8| = \begin{cases} x^2-2x-8, & \mbox{jika } x \le -2, \\ -(x^2-2x-8), & \mbox{jika } -2 < x < 4, \\ x^2-2x-8, & \mbox{jika } x \ge 4 \end{cases} \\ \int_{-3}^{5} |x^2-2x-8| dx &= \int_{-3}^{-2} x^2-2x-8 dx + \int_{-2}^{4} -(x^2-2x-8) dx + \int_{4}^{5} x^2-2x-8 dx \\ &= \left. (\frac{x^3}{3}-x^2-8x) \right|_{-3}^{-2} + \left. (\frac{-x^3}{3}+x^2+8x) \right|_{-2}^{4} + \left. (\frac{x^3}{3}-x^2-8x) \right|_{4}^{5} \\ &= \frac{28}{3} - 6 + \frac{80}{3} - (-\frac{28}{3}) - \frac{70}{3} - (-\frac{80}{3}) \\ &= \frac{128}{3} \\ &= 42\frac{2}{3} \\ \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x\,dx \\ L &= \frac{1}{2} x^2 \\ L &= \frac{1}{2} xy \quad (y = x) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x^2</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x^2\,dx \\ L &= \frac{1}{3} x^3 \\ L &= \frac{1}{3} xy \quad (y = x^2) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = \sqrt{x}</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int \sqrt{x}\,dx \\ L &= \frac{2}{3} x^{\frac{3}{2}} \\ L &= \frac{2}{3} xy \quad (y = \sqrt{x}) \end{aligned} </math> </div></div> # Buktikan luas persegi <math>L = s^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = s \text{ dan titik (''s'', ''s''), } \\ L &= \int_{0}^{s} s\,dx \\ L &= sx |_{0}^{s} \\ L &= ss - 0 \\ L &= s^2 \end{aligned} </math> </div></div> # Buktikan luas persegi panjang <math>L = pl</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = l \text{dan titik (''p'', ''l''), } \\ L &= \int_{0}^{p} l\,dx \\ L &= lx |_{0}^{p} \\ L &= pl - 0 \\ L &= pl \end{aligned} </math> </div></div> * Buktikan luas segitiga <math>L = \frac{at}{2}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{-tx}{a} + t \text { dan titik (''a'', ''t''), } \\ L &= \int_{0}^{a} \left (\frac{-tx}{a} + t \right) \,dx \\ L &= \left. \frac{-tx^2}{2a} + tx \right|_{0}^{a} \\ L &= \frac{-ta^2}{2a} + ta - 0 + 0 \\ L &= \frac{-ta}{2} + ta \\ L &= \frac{at}{2} \end{aligned} </math> </div></div> # Buktikan volume tabung <math>V = \pi r^2t</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ Dengan posisi } y = r \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} r^2\,dx \\ V &= \pi \left. r^2x \right|_{0}^{t} \\ V &= \pi r^2t - 0 \\ V &= \pi r^2t \end{aligned} </math> </div></div> # Buktikan volume kerucut <math>V = \frac{\pi r^2t}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{rx}{t} \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} \left (\frac{rx}{t} \right)^2 \,dx \\ V &= \pi \left. \frac{r^2 x^3}{3t^2} \right|_{0}^{t} \\ V &= \pi \frac{r^2 t^3}{3t^2} - 0 \\ V &= \frac{\pi r^2t}{3} \end{aligned} </math> </div></div> # Buktikan volume bola <math>V = \frac{4 \pi r^3}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), } \\ V &= \pi \int_{-r}^{r} \left(\sqrt {r^2 - x^2}\right)^2\,dx \\ V &= \pi \int_{-r}^{r} r^2 - x^2 dx \\ V &= \pi \left. r^2x - \frac{x^3}{3} \right|_{-r}^{r} \\ V &= \pi \left (r^3 - \frac{r^3}{3} - \left (-r^3 + \frac{r^3}{3} \right) \right) \\ V &= \frac{4 \pi r^3}{3} \end{aligned} </math> </div></div> # Buktikan luas permukaan bola <math>L = 4 \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{selanjutnya } \\ ds &= \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ ds &= \sqrt{1 + \frac{x^2}{r^2 - x^2}}\,dx \\ ds &= \sqrt{\frac{r^2}{r^2 - x^2}}\,dx \\ ds &= \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ \text{sehingga } \\ L &= 2 \pi \int_{-r}^{r} \sqrt{r^2 - x^2} \cdot \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ L &= 2 \pi \int_{-r}^{r} r\,dx \\ L &= 2 \pi rx|_{-r}^{r} \\ L &= 2 \pi (r r - r (-r)) \\ L &= 2 \pi (r^2 + r^2) \\ L &= 4 \pi r^2 \end{aligned} </math> </div></div> # Buktikan keliling lingkaran <math>K = 2 \pi r</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0) } \\ \text{kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{sehingga } \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{x^2}{r^2 - x^2})}\,dx \\ K &= 4 \int_{0}^{r} \sqrt\frac{r^2}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \sqrt\frac{1}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \frac{1}{\sqrt{r^2 - x^2}}\,dx \\ K &= 4r \left. \arcsin \left (\frac{x}{r} \right) \right|_{0}^{r} \\ K &= 4r \left (\arcsin \left (\frac{r}{r}\right) - \arcsin \left (\frac{0}{r} \right) \right) \\ K &= 4r \left (\arcsin\left (1 \right) - \arcsin \left(0 \right) \right)) \\ K &= 4r \left (\frac{\pi}{2} \right) \\ K &= 2 \pi r \end{aligned} </math> </div></div> # Buktikan luas lingkaran <math>L = \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. } \\ \sin(\theta) &= \frac{x}{r} \\ x &= r \sin(\theta) \\ dx &= r \cos(\theta)\,d\theta \\ \text{dengan turunan di atas } \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - x^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - (r \sin(\theta))^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - r^2 \sin^2(\theta)}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 (1 - \sin^2(\theta))}\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta)\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta) (r \cos(\theta)\,d\theta) \\ L &= 4 \int_{0}^{r} r^2 \cos^2(\theta)\,d\theta \\ L &= 4r^2 \int_{0}^{r} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta \\ L &= 2r^2 \int_{0}^{r} (1 + \cos(2\theta))\,d\theta \\ L &= 2r^2 \left (\theta + \frac{1}{2} \sin(2\theta) \right)|_{0}^{r} \\ L &= 2r^2 (\theta + \sin(\theta) \cos(\theta))|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left (\frac{x}{r} \right) + \left (\frac{x}{r} \right) \left (\frac{r^2 - x^2}{r} \right) \right)|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left(\frac{r}{r} \right) + \left (\frac{r}{r} \right) \left (\frac {r^2 - r^2}{r} \right) \right) - \left(\arcsin \left (\frac{0}{r} \right) + \left (\frac{0}{r}\right) \left (\frac{r^2 - 0^2}{r} \right) \right) \\ L &= 2r^2 (\arcsin(1) + 0 - (\arcsin(0) + 0)) \\ L &= 2r^2 \left (\frac{\pi}{2} \right) \\ L &= \pi r^2 \end{aligned} </math> </div></div> # Buktikan luas elips <math>L = \pi ab</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \frac{b \sqrt{a^2 - x^2}}{a} \text{ serta (-''a'', 0) dan (''a'', 0), } \\ L &= 4 \int_{0}^{r} \frac{b \sqrt{a^2 - x^2}}{a}\,dx \\ L &= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx \\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, } \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{\frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx}{4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx} \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{b}{a} \\ L_\text{elips} &= \frac{b}{a} L_\text{ling} \\ L_\text{elips} &= \frac{b}{a} \pi a^2 \\ L_\text{elips} &= \pi ab \end{aligned} </math> </div></div> # Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ x^-2x &= 4x+7 \\ x^2-6x-7 &= 0 \\ (x+1)(x-7) &= 0 \\ x=-1 &\text{ atau } x=7 \\ y &= 4(-1)+7 = 3 \\ y &= 4(7)+7 = 35 \\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut } \\ L &= \int_{-1}^{7} 4x+7-(x^2-2x)\,dx \\ &= \int_{-1}^{7} -x^2+6x+7\,dx \\ &= -\frac{x^3}{3}+3x^2+7x|_{-1}^{7} \\ &= -\frac{7^3}{3}+3(7)^2+7(7)-(-\frac{(-1)^3}{3}+3(-1)^2+7(-1)) \\ &= \frac{245}{3}-(-\frac{13}{3}) \\ &= \frac{258}{3} \\ &= 86 \\ \end{aligned} </math> </div></div> # Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ 6-1,5x &= x-4 \\ 2,5x &= 10 \\ x &= 4 \\ y &= 4-4 = 0 \\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut } \\ \text{untuk x=0 } \\ y &= 6-1,5(0) = 6 \\ y &= 0-4 = -4 \\ L &= \pi \int_{0}^{4} ((6-1,5x)^2-(x-4)^2)\,dx \\ &= \pi \int_{0}^{4} (36-18x+2,25x^2-(x^2-8x+16))\,dx \\ &= \pi \int_{0}^{4} (1,25x^2-10x+20)\,dx \\ &= \pi (\frac{1,25x^3}{3}-5x^2+20x)|_{0}^{4} \\ &= \pi (\frac{1,25(4)^3}{3}-5(4)^2+20(4)-(\frac{1,25(0)^3}{3}-5(0)^2+20(0))) \\ &= \pi (\frac{30}{3}-0) \\ &= 10\pi \\ \end{aligned} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 1by0vzs2jzrqy7cat7a0fjiw6ti3ox6 115016 115015 2026-04-26T23:30:59Z ~2026-25546-23 43043 /* integral pecahan parsial */ 115016 wikitext text/x-wiki == Kaidah umum == :# <math>\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ konstan)}\,\!</math> :# <math>\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx</math> :# <math>\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx</math> :# <math>\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx</math> :# <math>\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\! </math> :# <math>\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C </math> :# <math>\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C </math> :# <math>\int f(x) \, dx = F(x) + C</math> :# <math>\int_a^b f(x) \, dx = F(b) - F(a)</math> :# <math>\int f(x) \circ g(x) dx</math> = f’(g(x)) g’(x) :# <math>\int f^{-1}(x) dx = \frac{1}{f'(f'(x))}</math> == rumus sederhana == :<math>\int \, dx = x + C</math> :<math>\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int {dx \over x} = \ln{\left|x\right|} + C</math> <math>\int {dx \over \sqrt{a^2-b^2 x^2}} = \frac{1}{b} \arcsin \, {bx \over a} + C</math> :<math>\int {dx \over {a^2+b^2x^2}} = {1 \over ab}\arctan \, {bx \over a} + C</math> <math>\int {dx \over x\sqrt{b^2x^2-a^2}} = {1 \over a}\arcsec \, {bx \over a} + C</math> ; Eksponen dan logaritma :<math>\int e^x\,dx = e^x + C</math> :<math>\int a^x\,dx = \frac{a^x}{\ln{a}} + C</math> :<math>\int \ln \, x\,dx = x \cdot \, \ln {x} - x + C</math> :<math>\int \, ^b\!\log \, x\,dx = x \cdot \, ^b\!\log \, x - x \cdot \,^b\!\log e + C</math> ; Trigonometri :<math>\int \sin \, x \, dx = -\cos \, x + C</math> :<math>\int \cos \, x \, dx = \sin \, x + C</math> :<math>\int \tan \, x \, dx = -\ln{\left| \cos \, x \right|} + C</math> :<math>\int \cot \, x \, dx = \ln{\left| \sin \, x \right|} + C</math> :<math>\int \sec \, x \, dx = \ln{\left| \sec \, x + \tan \, x \right|} + C</math> :<math>\int \csc \, x \, dx = -\ln{\left| \csc \, x + \cot \, x \right|} + C</math> ; Hiperbolik :<math>\int \sinh \, x \, dx = \cosh \, x + C</math> :<math>\int \cosh \, x \, dx = \sinh \, x + C</math> :<math>\int \tanh \, x \, dx = \ln{\left| \cosh \, x \right|} + C</math> :<math>\int \coth \, x \, dx = \ln{\left| \sinh \, x \right|} + C</math> :<math>\int \mbox{sech} \, x \, dx = \arctan \,(\sinh \, x) + C</math> :<math>\int \mbox{csch} \, x \, dx = \ln{\left| \tanh \, {x \over 2} \right|} + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : 1. <math>\int x^3 - cos 3x + e^{5x+2} - \frac{1}{2x+5}\,dx</math> : <math>= \frac{1}{3+1}x^{3+1} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : <math>= \frac{x^4}{4} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : 2. <math>\int x(x-5)^4\,dx</math> : <math>= \int x(x^4-4x^3(5)+6x^2(25)-4x(125)+625)\,dx</math> : <math>= \int x(x^4-20x^3+150x^2-500x+625)\,dx</math> : <math>= \int x^5-20x^4+150x^3-500x^2+625x\,dx</math> : <math>= \frac{x^6}{6}-4x^5+\frac{150x^4}{4}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> : <math>= \frac{x^6}{6}-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : 1. <math>\int \frac{\ln \, x}{x}\,dx</math> : <math>u = \ln \, x\, du = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{\ln(x)}{x} dx \\ \int u du \\ \frac{1}{2} u^2 + C \\ \frac{1}{2} ln^2 x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x-5,\, du = dx,\, x = u+5</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4\,dx \\ \int (u+5)u^4\,du \\ \int (u^5+5u^4)\,du \\ \frac{1}{6} u^6+u^5 + C \\ \frac{1}{6} (x-5)^6+(x-5)^5 + C \\ \frac{1}{6} (x^6-6x^5(5)+15x^4(25)-20x^3(125)+15x^2(625)-6x(3125)+15625)+(x^5-5x^4(5)+10x^3(25)-10x^2(125)+5x(625)+3125) + C \\ \frac{1}{6}x^6-5x^5+\frac{125x^4}{2}-\frac{1250x^3}{3}+\frac{3125x^2}{2}-3125x+\frac{15625}{6}+x^5-25x^4+250x^3-1250x^2+3125x+3125 + C \\ \frac{1}{6}x^6-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C \\ \end{aligned} </math> </div></div> === integral parsial === ;Cara 1: Rumus Integral parsial diselesaikan dengan rumus berikut. : <math>\int u\,dv = uv - \int v\,du </math> Berikut contoh penyelesaian cara parsial dengan rumus. : 1. <math>\int x \sin \, x \, dx</math> : <math>u = x, du = 1 \, dx, dv = \sin \, x \, dx, v = -\cos \, x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \,dx \\ (x)(-\cos \, x) - \int (-\cos \, x)(1 dx) \\ -x \cos \, x + \int \cos \, x dx \\ -x \cos \, x + \sin \, x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x,\, du = 1\,dx,\, dv = (x-5)^4\,dx,\, v = \frac{1}{5}(x-5)^5</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - \int (\frac{1}{5}(x-5)^5)(1\,dx) \\ \frac{x}{5}(x-5)^5 - \int \frac{1}{5}(x-5)^5\,dx \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6\,dx \\ \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x\,dx</math> : <math>u = \ln \,x,\, du = \frac{1}{x}\,dx,\, dv = x\,dx,\, v = \frac{x^2}{2}</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x\, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{x^2}{2})(\frac{1}{x}\,dx) \\ \frac{x^2}{2} \ln \,x - \int \frac{x^2}{2}(\frac{1}{x})\,dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \,dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x\,dx</math> ini terjadi dua kali integral parsial : <math>u = \sin \,x,\, du = \cos \,x\,dx,\, dv = e^x\,dx,\, v = e^x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x\, dx \\ (\sin \,x)(e^x) - \int (e^x)(\cos \,x\,dx) \\ e^x \sin \,x - \int e^x \cos \,x\,dx \\ u = \cos \,x,\, du = -\sin \,x\,dx,\, dv = e^x\,dx,\, v = e^x \\ (\cos \,x)(e^x) - \int (e^x)(-\sin \,x\,dx) \\ e^x \cos \,x + \int e^x \sin \,x\,dx \\ e^x \sin \,x - (e^x \cos \,x + \int e^x \sin \,x\,dx) \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x\,dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> ;Cara 2: Tabel Untuk <math display="inline">\int u\,dv</math>, berlaku ketentuan sebagai berikut. {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>u</math> || <math>dv</math> |- | - || <math>\frac{du}{dx}</math> || <math>v</math> |- | + || <math>\frac{d^2u}{dx^2}</math> || <math>\int v\,dx</math> |} Berikut contoh penyelesaian cara parsial dengan tabel. : 1. <math>\int x \sin \, x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>\sin \, x</math> |- | - || <math>1</math> || <math>-\cos \, x</math> |- | + || <math>0</math> || <math>-\sin \, x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \, dx \\ (x)(-\cos \, x) - (1)(-\sin \, x) + C \\ -x \cos \, x + \sin \, x + C \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4 \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>(x-5)^4</math> |- | - || <math>1</math> || <math>\frac{1}{5}(x-5)^5</math> |- | + || <math>0</math> || <math>\frac{1}{30}(x-5)^6</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - (1)(\frac{1}{30}(x-5)^6) + C \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6 + C \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\ln \,x</math> || <math>x</math> |- | - || <math>\frac{1}{x}</math> || <math>\frac{x^2}{2}</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x \, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{1}{x})(\frac{x^2}{2}) \, dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \, dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x \, dx</math> ini terjadi dua kali integral parsial {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\sin \,x</math> || <math>e^x</math> |- | - || <math>\cos \, x</math> || <math>e^x</math> |- | + || <math>-\sin \,x</math> || <math>e^x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x \, dx \\ (\sin \,x)(e^x) - (\cos \,x)(e^x) + \int (-\sin \,x)(e^x) \, dx \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x) \, dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{1}{x^2 - 4} dx</math> Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{1}{x^2 - 4} \\ =&\; \frac{1}{(x + 2)(x - 2)} \\ =&\; \frac{A}{x + 2} + \frac{B}{x - 2} \\ =&\; \frac{A(x - 2) + B(x + 2)}{x^2 - 4} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4} dx \\ =&\; \int -\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)}\,dx \\ =&\; \frac{1}{4} \int \frac{1}{x - 2} - \frac{1}{x + 2}\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> : <math>\int \frac{x^2+5x+4}{(x-3)^3}dx</math> : cara 1 Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^3} \\ : \text{untuk mencari C yaitu } (3)^2+5(3)+4=28 \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x - 3)^2 + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x^2 - 6x + 9) + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 - 6Ax + 9A + Bx - 3B}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{dx}{x^2 - 4} \\ =&\; \int -\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)}\,dx \\ =&\; \frac{1}{4} \int \frac{1}{x - 2} - \frac{1}{x + 2}\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> === integral substitusi trigonometri === {| class="wikitable" |- ! Bentuk !! Substitusi Trigonometri |- | <math>\sqrt{a^2 - b^2 x^2}</math> || <math>x = \frac{a}{b} sin \, \alpha </math> |- | <math>\sqrt{a^2 + b^2 x^2}</math> || <math>x = \frac{a}{b} tan \, \alpha</math> |- | <math>\sqrt{b^2 x^2 - a^2}</math> || <math>x = \frac{a}{b} sec \, \alpha</math> |} Berikut contoh penyelesaian cara substitusi trigonometri. : <math>\int \frac{dx}{x^2 \sqrt{x^2 + 4}}</math> : <math>x = 2 \tan \, A, dx = 2 \sec^2 \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \int \frac{dx}{x^2\sqrt{x^2+4}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(2 \tan \, A)^2 \sqrt{4 + (2 \tan \, A)^2}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 + 4 \tan^2 \, A}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4(1 + \tan^2 \, A)}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 \sec^2 \, A }} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(4 \tan^2 \, A)(2 \sec \, A)} \\ =&\; \int \frac{\sec \, A \,dA}{4 \tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\sec \, A \,dA}{\tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A} \,dA \end{aligned} </math> Substitusi berikut dapat dibuat. : <math>\int \frac{\cos \, A}{\sin^2 \, A} \,dA</math> : <math>t = \sin \, A, dt = \cos \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A}\,dA \\ =&\; \frac{1}{4} \int \frac{dt}{t^2} \\ =&\; \frac{1}{4} \left(-\frac{1}{t}\right) + C \\ =&\; -\frac{1}{4 t} + C \\ =&\; -\frac{1}{4 \sin \, A} + C \end{aligned} </math> Ingat bahwa <math display="inline">\sin \, A = \frac{x}{\sqrt{x^2 + 4}}</math> berlaku. : <math> \begin{aligned} =&\; -\frac{1}{4 \sin \, A} + C \\ =&\; -\frac{\sqrt{x^2 + 4}}{4x} + C \end{aligned} </math> === integral mutlak === : <math>\int |f(x)| \,dA</math> buatlah <math>f(x) \ge 0</math> jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x) : <math>\int_a^c |f(x)| \, dx = \int_a^b f(x) \, dx + \int_b^c -f(x) \, dx = F(b) - F(a) - F(c) + F(b)</math> jika <math>|f(x)| = \begin{cases} f(x), & \mbox{jika } f(x) \ge b, \\ -f(x), & \mbox{jika } f(x) < b. \end{cases}</math>. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif. === integral fungsi ganjil dan integral fungsi genap === : <math>\int_{-a}^a f(x) \,dx</math> dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral: : integral fungsi ganjil : <math>\int_{-a}^a f(x) \,dx</math> = 0 : integral fungsi genap : <math>\int_{-a}^a f(x) \,dx</math> = <math>2 \int_{0}^a f(x) \,dx</math> === integral notasi === : <math>\int_a^b f(x) \,dx = \int_{a-p}^{b-p} f(x) \,dx = \int_{a+p}^{b+p} f(x) \,dx</math> : <math>\int_a^c f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx</math> : <math>\int_a^b f(x \pm k) \,dx = \int_{a \pm k}^{b \pm k} f(x) \,dx</math> === integral terbalik === : <math>\int_a^b f(x) \,dx = -\int_b^a f(x) \,dx</math> == Jenis integral lainnya == === panjang busur === ; Sumbu ''x'' : <math>S = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' : <math>S = \int_{y_1}^{y_2} \sqrt{1 + (f'(y))^2}\,dy</math> === luas daerah === ; Satu kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} f(x)\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} f(y)\,dy</math> ; Dua kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} (f(x_2) - f(x_1))\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} (f(y_2) - f(y_1))\,dy</math> : atau juga <math>L = \frac {D \sqrt{D}}{6 a^2}</math> === luas permukaan benda putar === ; Sumbu ''x'' sebagai poros : <math>L_p = 2 \pi \int_{x_1}^{x_2} f(x)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' sebagai poros : <math>L_p = 2 \pi \int_{y_1}^{y_2} f(y)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(y))^2}\,dy</math> === volume benda putar === ; Satu kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} (f(x))^2\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} (f(y))^2\,dy</math> ; Dua kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} ((f(x_2))^2 - (f(x_1))^2)\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} ((f(y_2))^2 - (f(y_1))^2)\,dy</math> : atau juga <math>V = |\frac {D^2 \sqrt{D}}{30 a^3}|</math> == Integral lipat == Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga. ;contoh # Tentukan hasil dari: * <math>\int (5x-1)(5x^2-2x+7)^4 dx</math> * <math>\int x^2 cos 3x dx</math> * <math>\int sin^2 x cos x dx</math> * <math>\int cos^2 x sin x dx</math> * <math>\int sec x dx</math> * <math>\int csc x dx</math> * <math>\int \frac{1}{2x^2-5x+3} dx</math> * <math>\int \frac{1}{\sqrt{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> * <math>\int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx</math> * <math>\int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx</math> * <math>\int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx</math> * <math>\int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx</math> * <math>\int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx \\ a=1, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-sin^2 x} \\ &= \frac{1}{cos^2 x} \\ &= sec^2 x \\ \int sec^2 x dx &= tan x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx \\ a=1, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-cos^2 x} \\ &= \frac{1}{sin^2 x} \\ &= csc^2 x \\ \int csc^2 x dx &= -cot x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx \\ a=sin x, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{sin x}{1-sin^2 x} \\ &= \frac{sin x}{cos^2 x} \\ &= sec x \cdot tan x \\ \int sec x \cdot tan x dx &= sec x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx \\ a=cos x, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{cos x}{1-cos^2 x} \\ &= \frac{cos x}{sin^2 x} \\ &= csc x \cdot cot x \\ \int csc x \cdot cot x dx &= -csc x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx \\ \text{misalkan } y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y &= \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y^2 &= \frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}} \\ &= \frac{x}{y} \\ y^3 &= x \\ 3y^2 dy &= dx \\ \int y \cdot 3y^2 dy \\ \int 3y^3 dy \\ \frac{3y^4}{4}+C \\ \end{aligned} </math> </div></div> # Tentukan hasil dari: * <math>\int_{1}^{5} 2x-3 dx</math> * <math>\int_{0}^{4} |x-2| dx</math> * <math>\int_{2}^{4} |3-x| dx</math> * <math>\int_{1}^{5} x^2-6x+8 dx</math> * <math>\int_{-3}^{5} |x^2-2x-8| dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} 2x-3 dx &= \left. x^2 - 3x \right|_{1}^{5} \\ &= 10 - (-2) \\ &= 12 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{0}^{4} |x-2| dx \\ \text{ tentukan nilai harga nol } \\ x-2 &= 0 \\ x &= 2 \\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti } \\ |x-2| = \begin{cases} x-2, & \mbox{jika } x \ge 2, \\ -(x-2), & \mbox{jika } x < 2. \end{cases} \\ \int_{0}^{4} |x-2| dx &= \int_{2}^{4} x-2 dx + \int_{0}^{2} -(x-2) dx \\ &= \left. (\frac{x^2}{2}-2x) \right|_{2}^{4} + \left. (\frac{-x^2}{2}+2x) \right|_{0}^{2} \\ &= 0 - (-2) + 2 - 0 \\ &= 4 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{2}^{4} |3-x| dx \\ \text{ tentukan nilai syarat-syarat } \\ 3-x &\ge 0 \\ \text{ harga nol } \\ x &= 3 \\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti } \\ |3-x| = \begin{cases} 3-x, & \mbox{jika } x \le 3, \\ -(3-x), & \mbox{jika } x > 3. \end{cases} \\ \int_{2}^{4} |3-x| dx &= \int_{2}^{3} 3-x dx + \int_{3}^{4} -(3-x) dx \\ &= \left. (3x-\frac{x^2}{2}) \right|_{2}^{3} + \left. (-3x+\frac{x^2}{2}) \right|_{3}^{4} \\ &= 4.5 - 4 - 4 - (-4.5) \\ &= 1 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} x^2-6x+8 dx &= \left. \frac{x^3}{3}-3x^2+8x \right|_{1}^{5} \\ &= \frac{20}{3} - \frac{16}{3} \\ &= \frac{4}{3} \\ &= 1\frac{1}{3} \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{-3}^{5} |x^2-2x-8| dx \\ \text{ tentukan nilai syarat-syarat } \\ x^2-2x-8 &\ge 0 \\ \text{ harga nol } \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x = 4 \\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti } \\ |x^2-2x-8| = \begin{cases} x^2-2x-8, & \mbox{jika } x \le -2, \\ -(x^2-2x-8), & \mbox{jika } -2 < x < 4, \\ x^2-2x-8, & \mbox{jika } x \ge 4 \end{cases} \\ \int_{-3}^{5} |x^2-2x-8| dx &= \int_{-3}^{-2} x^2-2x-8 dx + \int_{-2}^{4} -(x^2-2x-8) dx + \int_{4}^{5} x^2-2x-8 dx \\ &= \left. (\frac{x^3}{3}-x^2-8x) \right|_{-3}^{-2} + \left. (\frac{-x^3}{3}+x^2+8x) \right|_{-2}^{4} + \left. (\frac{x^3}{3}-x^2-8x) \right|_{4}^{5} \\ &= \frac{28}{3} - 6 + \frac{80}{3} - (-\frac{28}{3}) - \frac{70}{3} - (-\frac{80}{3}) \\ &= \frac{128}{3} \\ &= 42\frac{2}{3} \\ \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x\,dx \\ L &= \frac{1}{2} x^2 \\ L &= \frac{1}{2} xy \quad (y = x) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x^2</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x^2\,dx \\ L &= \frac{1}{3} x^3 \\ L &= \frac{1}{3} xy \quad (y = x^2) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = \sqrt{x}</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int \sqrt{x}\,dx \\ L &= \frac{2}{3} x^{\frac{3}{2}} \\ L &= \frac{2}{3} xy \quad (y = \sqrt{x}) \end{aligned} </math> </div></div> # Buktikan luas persegi <math>L = s^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = s \text{ dan titik (''s'', ''s''), } \\ L &= \int_{0}^{s} s\,dx \\ L &= sx |_{0}^{s} \\ L &= ss - 0 \\ L &= s^2 \end{aligned} </math> </div></div> # Buktikan luas persegi panjang <math>L = pl</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = l \text{dan titik (''p'', ''l''), } \\ L &= \int_{0}^{p} l\,dx \\ L &= lx |_{0}^{p} \\ L &= pl - 0 \\ L &= pl \end{aligned} </math> </div></div> * Buktikan luas segitiga <math>L = \frac{at}{2}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{-tx}{a} + t \text { dan titik (''a'', ''t''), } \\ L &= \int_{0}^{a} \left (\frac{-tx}{a} + t \right) \,dx \\ L &= \left. \frac{-tx^2}{2a} + tx \right|_{0}^{a} \\ L &= \frac{-ta^2}{2a} + ta - 0 + 0 \\ L &= \frac{-ta}{2} + ta \\ L &= \frac{at}{2} \end{aligned} </math> </div></div> # Buktikan volume tabung <math>V = \pi r^2t</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ Dengan posisi } y = r \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} r^2\,dx \\ V &= \pi \left. r^2x \right|_{0}^{t} \\ V &= \pi r^2t - 0 \\ V &= \pi r^2t \end{aligned} </math> </div></div> # Buktikan volume kerucut <math>V = \frac{\pi r^2t}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{rx}{t} \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} \left (\frac{rx}{t} \right)^2 \,dx \\ V &= \pi \left. \frac{r^2 x^3}{3t^2} \right|_{0}^{t} \\ V &= \pi \frac{r^2 t^3}{3t^2} - 0 \\ V &= \frac{\pi r^2t}{3} \end{aligned} </math> </div></div> # Buktikan volume bola <math>V = \frac{4 \pi r^3}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), } \\ V &= \pi \int_{-r}^{r} \left(\sqrt {r^2 - x^2}\right)^2\,dx \\ V &= \pi \int_{-r}^{r} r^2 - x^2 dx \\ V &= \pi \left. r^2x - \frac{x^3}{3} \right|_{-r}^{r} \\ V &= \pi \left (r^3 - \frac{r^3}{3} - \left (-r^3 + \frac{r^3}{3} \right) \right) \\ V &= \frac{4 \pi r^3}{3} \end{aligned} </math> </div></div> # Buktikan luas permukaan bola <math>L = 4 \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{selanjutnya } \\ ds &= \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ ds &= \sqrt{1 + \frac{x^2}{r^2 - x^2}}\,dx \\ ds &= \sqrt{\frac{r^2}{r^2 - x^2}}\,dx \\ ds &= \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ \text{sehingga } \\ L &= 2 \pi \int_{-r}^{r} \sqrt{r^2 - x^2} \cdot \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ L &= 2 \pi \int_{-r}^{r} r\,dx \\ L &= 2 \pi rx|_{-r}^{r} \\ L &= 2 \pi (r r - r (-r)) \\ L &= 2 \pi (r^2 + r^2) \\ L &= 4 \pi r^2 \end{aligned} </math> </div></div> # Buktikan keliling lingkaran <math>K = 2 \pi r</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0) } \\ \text{kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{sehingga } \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{x^2}{r^2 - x^2})}\,dx \\ K &= 4 \int_{0}^{r} \sqrt\frac{r^2}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \sqrt\frac{1}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \frac{1}{\sqrt{r^2 - x^2}}\,dx \\ K &= 4r \left. \arcsin \left (\frac{x}{r} \right) \right|_{0}^{r} \\ K &= 4r \left (\arcsin \left (\frac{r}{r}\right) - \arcsin \left (\frac{0}{r} \right) \right) \\ K &= 4r \left (\arcsin\left (1 \right) - \arcsin \left(0 \right) \right)) \\ K &= 4r \left (\frac{\pi}{2} \right) \\ K &= 2 \pi r \end{aligned} </math> </div></div> # Buktikan luas lingkaran <math>L = \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. } \\ \sin(\theta) &= \frac{x}{r} \\ x &= r \sin(\theta) \\ dx &= r \cos(\theta)\,d\theta \\ \text{dengan turunan di atas } \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - x^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - (r \sin(\theta))^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - r^2 \sin^2(\theta)}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 (1 - \sin^2(\theta))}\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta)\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta) (r \cos(\theta)\,d\theta) \\ L &= 4 \int_{0}^{r} r^2 \cos^2(\theta)\,d\theta \\ L &= 4r^2 \int_{0}^{r} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta \\ L &= 2r^2 \int_{0}^{r} (1 + \cos(2\theta))\,d\theta \\ L &= 2r^2 \left (\theta + \frac{1}{2} \sin(2\theta) \right)|_{0}^{r} \\ L &= 2r^2 (\theta + \sin(\theta) \cos(\theta))|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left (\frac{x}{r} \right) + \left (\frac{x}{r} \right) \left (\frac{r^2 - x^2}{r} \right) \right)|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left(\frac{r}{r} \right) + \left (\frac{r}{r} \right) \left (\frac {r^2 - r^2}{r} \right) \right) - \left(\arcsin \left (\frac{0}{r} \right) + \left (\frac{0}{r}\right) \left (\frac{r^2 - 0^2}{r} \right) \right) \\ L &= 2r^2 (\arcsin(1) + 0 - (\arcsin(0) + 0)) \\ L &= 2r^2 \left (\frac{\pi}{2} \right) \\ L &= \pi r^2 \end{aligned} </math> </div></div> # Buktikan luas elips <math>L = \pi ab</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \frac{b \sqrt{a^2 - x^2}}{a} \text{ serta (-''a'', 0) dan (''a'', 0), } \\ L &= 4 \int_{0}^{r} \frac{b \sqrt{a^2 - x^2}}{a}\,dx \\ L &= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx \\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, } \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{\frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx}{4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx} \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{b}{a} \\ L_\text{elips} &= \frac{b}{a} L_\text{ling} \\ L_\text{elips} &= \frac{b}{a} \pi a^2 \\ L_\text{elips} &= \pi ab \end{aligned} </math> </div></div> # Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ x^-2x &= 4x+7 \\ x^2-6x-7 &= 0 \\ (x+1)(x-7) &= 0 \\ x=-1 &\text{ atau } x=7 \\ y &= 4(-1)+7 = 3 \\ y &= 4(7)+7 = 35 \\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut } \\ L &= \int_{-1}^{7} 4x+7-(x^2-2x)\,dx \\ &= \int_{-1}^{7} -x^2+6x+7\,dx \\ &= -\frac{x^3}{3}+3x^2+7x|_{-1}^{7} \\ &= -\frac{7^3}{3}+3(7)^2+7(7)-(-\frac{(-1)^3}{3}+3(-1)^2+7(-1)) \\ &= \frac{245}{3}-(-\frac{13}{3}) \\ &= \frac{258}{3} \\ &= 86 \\ \end{aligned} </math> </div></div> # Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ 6-1,5x &= x-4 \\ 2,5x &= 10 \\ x &= 4 \\ y &= 4-4 = 0 \\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut } \\ \text{untuk x=0 } \\ y &= 6-1,5(0) = 6 \\ y &= 0-4 = -4 \\ L &= \pi \int_{0}^{4} ((6-1,5x)^2-(x-4)^2)\,dx \\ &= \pi \int_{0}^{4} (36-18x+2,25x^2-(x^2-8x+16))\,dx \\ &= \pi \int_{0}^{4} (1,25x^2-10x+20)\,dx \\ &= \pi (\frac{1,25x^3}{3}-5x^2+20x)|_{0}^{4} \\ &= \pi (\frac{1,25(4)^3}{3}-5(4)^2+20(4)-(\frac{1,25(0)^3}{3}-5(0)^2+20(0))) \\ &= \pi (\frac{30}{3}-0) \\ &= 10\pi \\ \end{aligned} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 9590ezj8d8tg7oyr3dggdgkr34il28k 115020 115016 2026-04-26T23:42:31Z ~2026-25546-23 43043 /* integral pecahan parsial */ 115020 wikitext text/x-wiki == Kaidah umum == :# <math>\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ konstan)}\,\!</math> :# <math>\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx</math> :# <math>\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx</math> :# <math>\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx</math> :# <math>\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\! </math> :# <math>\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C </math> :# <math>\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C </math> :# <math>\int f(x) \, dx = F(x) + C</math> :# <math>\int_a^b f(x) \, dx = F(b) - F(a)</math> :# <math>\int f(x) \circ g(x) dx</math> = f’(g(x)) g’(x) :# <math>\int f^{-1}(x) dx = \frac{1}{f'(f'(x))}</math> == rumus sederhana == :<math>\int \, dx = x + C</math> :<math>\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int {dx \over x} = \ln{\left|x\right|} + C</math> <math>\int {dx \over \sqrt{a^2-b^2 x^2}} = \frac{1}{b} \arcsin \, {bx \over a} + C</math> :<math>\int {dx \over {a^2+b^2x^2}} = {1 \over ab}\arctan \, {bx \over a} + C</math> <math>\int {dx \over x\sqrt{b^2x^2-a^2}} = {1 \over a}\arcsec \, {bx \over a} + C</math> ; Eksponen dan logaritma :<math>\int e^x\,dx = e^x + C</math> :<math>\int a^x\,dx = \frac{a^x}{\ln{a}} + C</math> :<math>\int \ln \, x\,dx = x \cdot \, \ln {x} - x + C</math> :<math>\int \, ^b\!\log \, x\,dx = x \cdot \, ^b\!\log \, x - x \cdot \,^b\!\log e + C</math> ; Trigonometri :<math>\int \sin \, x \, dx = -\cos \, x + C</math> :<math>\int \cos \, x \, dx = \sin \, x + C</math> :<math>\int \tan \, x \, dx = -\ln{\left| \cos \, x \right|} + C</math> :<math>\int \cot \, x \, dx = \ln{\left| \sin \, x \right|} + C</math> :<math>\int \sec \, x \, dx = \ln{\left| \sec \, x + \tan \, x \right|} + C</math> :<math>\int \csc \, x \, dx = -\ln{\left| \csc \, x + \cot \, x \right|} + C</math> ; Hiperbolik :<math>\int \sinh \, x \, dx = \cosh \, x + C</math> :<math>\int \cosh \, x \, dx = \sinh \, x + C</math> :<math>\int \tanh \, x \, dx = \ln{\left| \cosh \, x \right|} + C</math> :<math>\int \coth \, x \, dx = \ln{\left| \sinh \, x \right|} + C</math> :<math>\int \mbox{sech} \, x \, dx = \arctan \,(\sinh \, x) + C</math> :<math>\int \mbox{csch} \, x \, dx = \ln{\left| \tanh \, {x \over 2} \right|} + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : 1. <math>\int x^3 - cos 3x + e^{5x+2} - \frac{1}{2x+5}\,dx</math> : <math>= \frac{1}{3+1}x^{3+1} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : <math>= \frac{x^4}{4} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : 2. <math>\int x(x-5)^4\,dx</math> : <math>= \int x(x^4-4x^3(5)+6x^2(25)-4x(125)+625)\,dx</math> : <math>= \int x(x^4-20x^3+150x^2-500x+625)\,dx</math> : <math>= \int x^5-20x^4+150x^3-500x^2+625x\,dx</math> : <math>= \frac{x^6}{6}-4x^5+\frac{150x^4}{4}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> : <math>= \frac{x^6}{6}-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : 1. <math>\int \frac{\ln \, x}{x}\,dx</math> : <math>u = \ln \, x\, du = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{\ln(x)}{x} dx \\ \int u du \\ \frac{1}{2} u^2 + C \\ \frac{1}{2} ln^2 x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x-5,\, du = dx,\, x = u+5</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4\,dx \\ \int (u+5)u^4\,du \\ \int (u^5+5u^4)\,du \\ \frac{1}{6} u^6+u^5 + C \\ \frac{1}{6} (x-5)^6+(x-5)^5 + C \\ \frac{1}{6} (x^6-6x^5(5)+15x^4(25)-20x^3(125)+15x^2(625)-6x(3125)+15625)+(x^5-5x^4(5)+10x^3(25)-10x^2(125)+5x(625)+3125) + C \\ \frac{1}{6}x^6-5x^5+\frac{125x^4}{2}-\frac{1250x^3}{3}+\frac{3125x^2}{2}-3125x+\frac{15625}{6}+x^5-25x^4+250x^3-1250x^2+3125x+3125 + C \\ \frac{1}{6}x^6-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C \\ \end{aligned} </math> </div></div> === integral parsial === ;Cara 1: Rumus Integral parsial diselesaikan dengan rumus berikut. : <math>\int u\,dv = uv - \int v\,du </math> Berikut contoh penyelesaian cara parsial dengan rumus. : 1. <math>\int x \sin \, x \, dx</math> : <math>u = x, du = 1 \, dx, dv = \sin \, x \, dx, v = -\cos \, x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \,dx \\ (x)(-\cos \, x) - \int (-\cos \, x)(1 dx) \\ -x \cos \, x + \int \cos \, x dx \\ -x \cos \, x + \sin \, x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x,\, du = 1\,dx,\, dv = (x-5)^4\,dx,\, v = \frac{1}{5}(x-5)^5</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - \int (\frac{1}{5}(x-5)^5)(1\,dx) \\ \frac{x}{5}(x-5)^5 - \int \frac{1}{5}(x-5)^5\,dx \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6\,dx \\ \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x\,dx</math> : <math>u = \ln \,x,\, du = \frac{1}{x}\,dx,\, dv = x\,dx,\, v = \frac{x^2}{2}</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x\, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{x^2}{2})(\frac{1}{x}\,dx) \\ \frac{x^2}{2} \ln \,x - \int \frac{x^2}{2}(\frac{1}{x})\,dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \,dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x\,dx</math> ini terjadi dua kali integral parsial : <math>u = \sin \,x,\, du = \cos \,x\,dx,\, dv = e^x\,dx,\, v = e^x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x\, dx \\ (\sin \,x)(e^x) - \int (e^x)(\cos \,x\,dx) \\ e^x \sin \,x - \int e^x \cos \,x\,dx \\ u = \cos \,x,\, du = -\sin \,x\,dx,\, dv = e^x\,dx,\, v = e^x \\ (\cos \,x)(e^x) - \int (e^x)(-\sin \,x\,dx) \\ e^x \cos \,x + \int e^x \sin \,x\,dx \\ e^x \sin \,x - (e^x \cos \,x + \int e^x \sin \,x\,dx) \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x\,dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> ;Cara 2: Tabel Untuk <math display="inline">\int u\,dv</math>, berlaku ketentuan sebagai berikut. {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>u</math> || <math>dv</math> |- | - || <math>\frac{du}{dx}</math> || <math>v</math> |- | + || <math>\frac{d^2u}{dx^2}</math> || <math>\int v\,dx</math> |} Berikut contoh penyelesaian cara parsial dengan tabel. : 1. <math>\int x \sin \, x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>\sin \, x</math> |- | - || <math>1</math> || <math>-\cos \, x</math> |- | + || <math>0</math> || <math>-\sin \, x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \, dx \\ (x)(-\cos \, x) - (1)(-\sin \, x) + C \\ -x \cos \, x + \sin \, x + C \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4 \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>(x-5)^4</math> |- | - || <math>1</math> || <math>\frac{1}{5}(x-5)^5</math> |- | + || <math>0</math> || <math>\frac{1}{30}(x-5)^6</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - (1)(\frac{1}{30}(x-5)^6) + C \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6 + C \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\ln \,x</math> || <math>x</math> |- | - || <math>\frac{1}{x}</math> || <math>\frac{x^2}{2}</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x \, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{1}{x})(\frac{x^2}{2}) \, dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \, dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x \, dx</math> ini terjadi dua kali integral parsial {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\sin \,x</math> || <math>e^x</math> |- | - || <math>\cos \, x</math> || <math>e^x</math> |- | + || <math>-\sin \,x</math> || <math>e^x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x \, dx \\ (\sin \,x)(e^x) - (\cos \,x)(e^x) + \int (-\sin \,x)(e^x) \, dx \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x) \, dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{1}{x^2 - 4}\,dx</math> Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{1}{x^2 - 4} \\ =&\; \frac{1}{(x + 2)(x - 2)} \\ =&\; \frac{A}{x + 2} + \frac{B}{x - 2} \\ =&\; \frac{A(x - 2) + B(x + 2)}{x^2 - 4} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int -\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)}\,dx \\ =&\; \frac{1}{4} \int \frac{1}{x - 2} - \frac{1}{x + 2}\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> : <math>\int \frac{x^2+5x+4}{(x-3)^3}\,dx</math> : cara 1 Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^3} \\ : \text{untuk mencari C yaitu } (3)^2+5(3)+4=28 \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x - 3)^2 + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x^2 - 6x + 9) + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 - 6Ax + 9A + Bx - 3B}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 + (- 6A + B)x +(9 - 3B)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ \end{aligned} </math> Kita tahu bahwa <math display="inline">A = 1</math> dan <math display="inline">-6A + B = 5</math> dapat diselesaikan, yaitu <math display="inline">A = 1</math> dan {{nowrap|<math display="inline">B = 11</math>.}} : <math> \begin{aligned} &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int \frac{1}{x - 3} + \frac{11}{(x - 3)^2} + \frac{28}{(x - 3)^3}\,dx \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> === integral substitusi trigonometri === {| class="wikitable" |- ! Bentuk !! Substitusi Trigonometri |- | <math>\sqrt{a^2 - b^2 x^2}</math> || <math>x = \frac{a}{b} sin \, \alpha </math> |- | <math>\sqrt{a^2 + b^2 x^2}</math> || <math>x = \frac{a}{b} tan \, \alpha</math> |- | <math>\sqrt{b^2 x^2 - a^2}</math> || <math>x = \frac{a}{b} sec \, \alpha</math> |} Berikut contoh penyelesaian cara substitusi trigonometri. : <math>\int \frac{dx}{x^2 \sqrt{x^2 + 4}}</math> : <math>x = 2 \tan \, A, dx = 2 \sec^2 \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \int \frac{dx}{x^2\sqrt{x^2+4}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(2 \tan \, A)^2 \sqrt{4 + (2 \tan \, A)^2}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 + 4 \tan^2 \, A}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4(1 + \tan^2 \, A)}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 \sec^2 \, A }} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(4 \tan^2 \, A)(2 \sec \, A)} \\ =&\; \int \frac{\sec \, A \,dA}{4 \tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\sec \, A \,dA}{\tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A} \,dA \end{aligned} </math> Substitusi berikut dapat dibuat. : <math>\int \frac{\cos \, A}{\sin^2 \, A} \,dA</math> : <math>t = \sin \, A, dt = \cos \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A}\,dA \\ =&\; \frac{1}{4} \int \frac{dt}{t^2} \\ =&\; \frac{1}{4} \left(-\frac{1}{t}\right) + C \\ =&\; -\frac{1}{4 t} + C \\ =&\; -\frac{1}{4 \sin \, A} + C \end{aligned} </math> Ingat bahwa <math display="inline">\sin \, A = \frac{x}{\sqrt{x^2 + 4}}</math> berlaku. : <math> \begin{aligned} =&\; -\frac{1}{4 \sin \, A} + C \\ =&\; -\frac{\sqrt{x^2 + 4}}{4x} + C \end{aligned} </math> === integral mutlak === : <math>\int |f(x)| \,dA</math> buatlah <math>f(x) \ge 0</math> jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x) : <math>\int_a^c |f(x)| \, dx = \int_a^b f(x) \, dx + \int_b^c -f(x) \, dx = F(b) - F(a) - F(c) + F(b)</math> jika <math>|f(x)| = \begin{cases} f(x), & \mbox{jika } f(x) \ge b, \\ -f(x), & \mbox{jika } f(x) < b. \end{cases}</math>. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif. === integral fungsi ganjil dan integral fungsi genap === : <math>\int_{-a}^a f(x) \,dx</math> dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral: : integral fungsi ganjil : <math>\int_{-a}^a f(x) \,dx</math> = 0 : integral fungsi genap : <math>\int_{-a}^a f(x) \,dx</math> = <math>2 \int_{0}^a f(x) \,dx</math> === integral notasi === : <math>\int_a^b f(x) \,dx = \int_{a-p}^{b-p} f(x) \,dx = \int_{a+p}^{b+p} f(x) \,dx</math> : <math>\int_a^c f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx</math> : <math>\int_a^b f(x \pm k) \,dx = \int_{a \pm k}^{b \pm k} f(x) \,dx</math> === integral terbalik === : <math>\int_a^b f(x) \,dx = -\int_b^a f(x) \,dx</math> == Jenis integral lainnya == === panjang busur === ; Sumbu ''x'' : <math>S = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' : <math>S = \int_{y_1}^{y_2} \sqrt{1 + (f'(y))^2}\,dy</math> === luas daerah === ; Satu kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} f(x)\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} f(y)\,dy</math> ; Dua kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} (f(x_2) - f(x_1))\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} (f(y_2) - f(y_1))\,dy</math> : atau juga <math>L = \frac {D \sqrt{D}}{6 a^2}</math> === luas permukaan benda putar === ; Sumbu ''x'' sebagai poros : <math>L_p = 2 \pi \int_{x_1}^{x_2} f(x)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' sebagai poros : <math>L_p = 2 \pi \int_{y_1}^{y_2} f(y)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(y))^2}\,dy</math> === volume benda putar === ; Satu kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} (f(x))^2\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} (f(y))^2\,dy</math> ; Dua kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} ((f(x_2))^2 - (f(x_1))^2)\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} ((f(y_2))^2 - (f(y_1))^2)\,dy</math> : atau juga <math>V = |\frac {D^2 \sqrt{D}}{30 a^3}|</math> == Integral lipat == Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga. ;contoh # Tentukan hasil dari: * <math>\int (5x-1)(5x^2-2x+7)^4 dx</math> * <math>\int x^2 cos 3x dx</math> * <math>\int sin^2 x cos x dx</math> * <math>\int cos^2 x sin x dx</math> * <math>\int sec x dx</math> * <math>\int csc x dx</math> * <math>\int \frac{1}{2x^2-5x+3} dx</math> * <math>\int \frac{1}{\sqrt{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> * <math>\int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx</math> * <math>\int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx</math> * <math>\int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx</math> * <math>\int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx</math> * <math>\int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx \\ a=1, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-sin^2 x} \\ &= \frac{1}{cos^2 x} \\ &= sec^2 x \\ \int sec^2 x dx &= tan x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx \\ a=1, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-cos^2 x} \\ &= \frac{1}{sin^2 x} \\ &= csc^2 x \\ \int csc^2 x dx &= -cot x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx \\ a=sin x, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{sin x}{1-sin^2 x} \\ &= \frac{sin x}{cos^2 x} \\ &= sec x \cdot tan x \\ \int sec x \cdot tan x dx &= sec x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx \\ a=cos x, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{cos x}{1-cos^2 x} \\ &= \frac{cos x}{sin^2 x} \\ &= csc x \cdot cot x \\ \int csc x \cdot cot x dx &= -csc x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx \\ \text{misalkan } y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y &= \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y^2 &= \frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}} \\ &= \frac{x}{y} \\ y^3 &= x \\ 3y^2 dy &= dx \\ \int y \cdot 3y^2 dy \\ \int 3y^3 dy \\ \frac{3y^4}{4}+C \\ \end{aligned} </math> </div></div> # Tentukan hasil dari: * <math>\int_{1}^{5} 2x-3 dx</math> * <math>\int_{0}^{4} |x-2| dx</math> * <math>\int_{2}^{4} |3-x| dx</math> * <math>\int_{1}^{5} x^2-6x+8 dx</math> * <math>\int_{-3}^{5} |x^2-2x-8| dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} 2x-3 dx &= \left. x^2 - 3x \right|_{1}^{5} \\ &= 10 - (-2) \\ &= 12 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{0}^{4} |x-2| dx \\ \text{ tentukan nilai harga nol } \\ x-2 &= 0 \\ x &= 2 \\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti } \\ |x-2| = \begin{cases} x-2, & \mbox{jika } x \ge 2, \\ -(x-2), & \mbox{jika } x < 2. \end{cases} \\ \int_{0}^{4} |x-2| dx &= \int_{2}^{4} x-2 dx + \int_{0}^{2} -(x-2) dx \\ &= \left. (\frac{x^2}{2}-2x) \right|_{2}^{4} + \left. (\frac{-x^2}{2}+2x) \right|_{0}^{2} \\ &= 0 - (-2) + 2 - 0 \\ &= 4 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{2}^{4} |3-x| dx \\ \text{ tentukan nilai syarat-syarat } \\ 3-x &\ge 0 \\ \text{ harga nol } \\ x &= 3 \\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti } \\ |3-x| = \begin{cases} 3-x, & \mbox{jika } x \le 3, \\ -(3-x), & \mbox{jika } x > 3. \end{cases} \\ \int_{2}^{4} |3-x| dx &= \int_{2}^{3} 3-x dx + \int_{3}^{4} -(3-x) dx \\ &= \left. (3x-\frac{x^2}{2}) \right|_{2}^{3} + \left. (-3x+\frac{x^2}{2}) \right|_{3}^{4} \\ &= 4.5 - 4 - 4 - (-4.5) \\ &= 1 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} x^2-6x+8 dx &= \left. \frac{x^3}{3}-3x^2+8x \right|_{1}^{5} \\ &= \frac{20}{3} - \frac{16}{3} \\ &= \frac{4}{3} \\ &= 1\frac{1}{3} \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{-3}^{5} |x^2-2x-8| dx \\ \text{ tentukan nilai syarat-syarat } \\ x^2-2x-8 &\ge 0 \\ \text{ harga nol } \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x = 4 \\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti } \\ |x^2-2x-8| = \begin{cases} x^2-2x-8, & \mbox{jika } x \le -2, \\ -(x^2-2x-8), & \mbox{jika } -2 < x < 4, \\ x^2-2x-8, & \mbox{jika } x \ge 4 \end{cases} \\ \int_{-3}^{5} |x^2-2x-8| dx &= \int_{-3}^{-2} x^2-2x-8 dx + \int_{-2}^{4} -(x^2-2x-8) dx + \int_{4}^{5} x^2-2x-8 dx \\ &= \left. (\frac{x^3}{3}-x^2-8x) \right|_{-3}^{-2} + \left. (\frac{-x^3}{3}+x^2+8x) \right|_{-2}^{4} + \left. (\frac{x^3}{3}-x^2-8x) \right|_{4}^{5} \\ &= \frac{28}{3} - 6 + \frac{80}{3} - (-\frac{28}{3}) - \frac{70}{3} - (-\frac{80}{3}) \\ &= \frac{128}{3} \\ &= 42\frac{2}{3} \\ \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x\,dx \\ L &= \frac{1}{2} x^2 \\ L &= \frac{1}{2} xy \quad (y = x) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x^2</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x^2\,dx \\ L &= \frac{1}{3} x^3 \\ L &= \frac{1}{3} xy \quad (y = x^2) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = \sqrt{x}</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int \sqrt{x}\,dx \\ L &= \frac{2}{3} x^{\frac{3}{2}} \\ L &= \frac{2}{3} xy \quad (y = \sqrt{x}) \end{aligned} </math> </div></div> # Buktikan luas persegi <math>L = s^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = s \text{ dan titik (''s'', ''s''), } \\ L &= \int_{0}^{s} s\,dx \\ L &= sx |_{0}^{s} \\ L &= ss - 0 \\ L &= s^2 \end{aligned} </math> </div></div> # Buktikan luas persegi panjang <math>L = pl</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = l \text{dan titik (''p'', ''l''), } \\ L &= \int_{0}^{p} l\,dx \\ L &= lx |_{0}^{p} \\ L &= pl - 0 \\ L &= pl \end{aligned} </math> </div></div> * Buktikan luas segitiga <math>L = \frac{at}{2}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{-tx}{a} + t \text { dan titik (''a'', ''t''), } \\ L &= \int_{0}^{a} \left (\frac{-tx}{a} + t \right) \,dx \\ L &= \left. \frac{-tx^2}{2a} + tx \right|_{0}^{a} \\ L &= \frac{-ta^2}{2a} + ta - 0 + 0 \\ L &= \frac{-ta}{2} + ta \\ L &= \frac{at}{2} \end{aligned} </math> </div></div> # Buktikan volume tabung <math>V = \pi r^2t</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ Dengan posisi } y = r \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} r^2\,dx \\ V &= \pi \left. r^2x \right|_{0}^{t} \\ V &= \pi r^2t - 0 \\ V &= \pi r^2t \end{aligned} </math> </div></div> # Buktikan volume kerucut <math>V = \frac{\pi r^2t}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{rx}{t} \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} \left (\frac{rx}{t} \right)^2 \,dx \\ V &= \pi \left. \frac{r^2 x^3}{3t^2} \right|_{0}^{t} \\ V &= \pi \frac{r^2 t^3}{3t^2} - 0 \\ V &= \frac{\pi r^2t}{3} \end{aligned} </math> </div></div> # Buktikan volume bola <math>V = \frac{4 \pi r^3}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), } \\ V &= \pi \int_{-r}^{r} \left(\sqrt {r^2 - x^2}\right)^2\,dx \\ V &= \pi \int_{-r}^{r} r^2 - x^2 dx \\ V &= \pi \left. r^2x - \frac{x^3}{3} \right|_{-r}^{r} \\ V &= \pi \left (r^3 - \frac{r^3}{3} - \left (-r^3 + \frac{r^3}{3} \right) \right) \\ V &= \frac{4 \pi r^3}{3} \end{aligned} </math> </div></div> # Buktikan luas permukaan bola <math>L = 4 \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{selanjutnya } \\ ds &= \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ ds &= \sqrt{1 + \frac{x^2}{r^2 - x^2}}\,dx \\ ds &= \sqrt{\frac{r^2}{r^2 - x^2}}\,dx \\ ds &= \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ \text{sehingga } \\ L &= 2 \pi \int_{-r}^{r} \sqrt{r^2 - x^2} \cdot \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ L &= 2 \pi \int_{-r}^{r} r\,dx \\ L &= 2 \pi rx|_{-r}^{r} \\ L &= 2 \pi (r r - r (-r)) \\ L &= 2 \pi (r^2 + r^2) \\ L &= 4 \pi r^2 \end{aligned} </math> </div></div> # Buktikan keliling lingkaran <math>K = 2 \pi r</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0) } \\ \text{kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{sehingga } \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{x^2}{r^2 - x^2})}\,dx \\ K &= 4 \int_{0}^{r} \sqrt\frac{r^2}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \sqrt\frac{1}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \frac{1}{\sqrt{r^2 - x^2}}\,dx \\ K &= 4r \left. \arcsin \left (\frac{x}{r} \right) \right|_{0}^{r} \\ K &= 4r \left (\arcsin \left (\frac{r}{r}\right) - \arcsin \left (\frac{0}{r} \right) \right) \\ K &= 4r \left (\arcsin\left (1 \right) - \arcsin \left(0 \right) \right)) \\ K &= 4r \left (\frac{\pi}{2} \right) \\ K &= 2 \pi r \end{aligned} </math> </div></div> # Buktikan luas lingkaran <math>L = \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. } \\ \sin(\theta) &= \frac{x}{r} \\ x &= r \sin(\theta) \\ dx &= r \cos(\theta)\,d\theta \\ \text{dengan turunan di atas } \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - x^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - (r \sin(\theta))^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - r^2 \sin^2(\theta)}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 (1 - \sin^2(\theta))}\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta)\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta) (r \cos(\theta)\,d\theta) \\ L &= 4 \int_{0}^{r} r^2 \cos^2(\theta)\,d\theta \\ L &= 4r^2 \int_{0}^{r} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta \\ L &= 2r^2 \int_{0}^{r} (1 + \cos(2\theta))\,d\theta \\ L &= 2r^2 \left (\theta + \frac{1}{2} \sin(2\theta) \right)|_{0}^{r} \\ L &= 2r^2 (\theta + \sin(\theta) \cos(\theta))|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left (\frac{x}{r} \right) + \left (\frac{x}{r} \right) \left (\frac{r^2 - x^2}{r} \right) \right)|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left(\frac{r}{r} \right) + \left (\frac{r}{r} \right) \left (\frac {r^2 - r^2}{r} \right) \right) - \left(\arcsin \left (\frac{0}{r} \right) + \left (\frac{0}{r}\right) \left (\frac{r^2 - 0^2}{r} \right) \right) \\ L &= 2r^2 (\arcsin(1) + 0 - (\arcsin(0) + 0)) \\ L &= 2r^2 \left (\frac{\pi}{2} \right) \\ L &= \pi r^2 \end{aligned} </math> </div></div> # Buktikan luas elips <math>L = \pi ab</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \frac{b \sqrt{a^2 - x^2}}{a} \text{ serta (-''a'', 0) dan (''a'', 0), } \\ L &= 4 \int_{0}^{r} \frac{b \sqrt{a^2 - x^2}}{a}\,dx \\ L &= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx \\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, } \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{\frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx}{4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx} \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{b}{a} \\ L_\text{elips} &= \frac{b}{a} L_\text{ling} \\ L_\text{elips} &= \frac{b}{a} \pi a^2 \\ L_\text{elips} &= \pi ab \end{aligned} </math> </div></div> # Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ x^-2x &= 4x+7 \\ x^2-6x-7 &= 0 \\ (x+1)(x-7) &= 0 \\ x=-1 &\text{ atau } x=7 \\ y &= 4(-1)+7 = 3 \\ y &= 4(7)+7 = 35 \\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut } \\ L &= \int_{-1}^{7} 4x+7-(x^2-2x)\,dx \\ &= \int_{-1}^{7} -x^2+6x+7\,dx \\ &= -\frac{x^3}{3}+3x^2+7x|_{-1}^{7} \\ &= -\frac{7^3}{3}+3(7)^2+7(7)-(-\frac{(-1)^3}{3}+3(-1)^2+7(-1)) \\ &= \frac{245}{3}-(-\frac{13}{3}) \\ &= \frac{258}{3} \\ &= 86 \\ \end{aligned} </math> </div></div> # Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ 6-1,5x &= x-4 \\ 2,5x &= 10 \\ x &= 4 \\ y &= 4-4 = 0 \\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut } \\ \text{untuk x=0 } \\ y &= 6-1,5(0) = 6 \\ y &= 0-4 = -4 \\ L &= \pi \int_{0}^{4} ((6-1,5x)^2-(x-4)^2)\,dx \\ &= \pi \int_{0}^{4} (36-18x+2,25x^2-(x^2-8x+16))\,dx \\ &= \pi \int_{0}^{4} (1,25x^2-10x+20)\,dx \\ &= \pi (\frac{1,25x^3}{3}-5x^2+20x)|_{0}^{4} \\ &= \pi (\frac{1,25(4)^3}{3}-5(4)^2+20(4)-(\frac{1,25(0)^3}{3}-5(0)^2+20(0))) \\ &= \pi (\frac{30}{3}-0) \\ &= 10\pi \\ \end{aligned} </math> </div></div> [[Kategori:Soal-Soal Matematika]] ort6wyfamhi9frdbuossztdekf7f1dj 115021 115020 2026-04-27T00:07:17Z ~2026-25546-23 43043 /* integral pecahan parsial */ 115021 wikitext text/x-wiki == Kaidah umum == :# <math>\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ konstan)}\,\!</math> :# <math>\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx</math> :# <math>\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx</math> :# <math>\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx</math> :# <math>\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\! </math> :# <math>\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C </math> :# <math>\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C </math> :# <math>\int f(x) \, dx = F(x) + C</math> :# <math>\int_a^b f(x) \, dx = F(b) - F(a)</math> :# <math>\int f(x) \circ g(x) dx</math> = f’(g(x)) g’(x) :# <math>\int f^{-1}(x) dx = \frac{1}{f'(f'(x))}</math> == rumus sederhana == :<math>\int \, dx = x + C</math> :<math>\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int {dx \over x} = \ln{\left|x\right|} + C</math> <math>\int {dx \over \sqrt{a^2-b^2 x^2}} = \frac{1}{b} \arcsin \, {bx \over a} + C</math> :<math>\int {dx \over {a^2+b^2x^2}} = {1 \over ab}\arctan \, {bx \over a} + C</math> <math>\int {dx \over x\sqrt{b^2x^2-a^2}} = {1 \over a}\arcsec \, {bx \over a} + C</math> ; Eksponen dan logaritma :<math>\int e^x\,dx = e^x + C</math> :<math>\int a^x\,dx = \frac{a^x}{\ln{a}} + C</math> :<math>\int \ln \, x\,dx = x \cdot \, \ln {x} - x + C</math> :<math>\int \, ^b\!\log \, x\,dx = x \cdot \, ^b\!\log \, x - x \cdot \,^b\!\log e + C</math> ; Trigonometri :<math>\int \sin \, x \, dx = -\cos \, x + C</math> :<math>\int \cos \, x \, dx = \sin \, x + C</math> :<math>\int \tan \, x \, dx = -\ln{\left| \cos \, x \right|} + C</math> :<math>\int \cot \, x \, dx = \ln{\left| \sin \, x \right|} + C</math> :<math>\int \sec \, x \, dx = \ln{\left| \sec \, x + \tan \, x \right|} + C</math> :<math>\int \csc \, x \, dx = -\ln{\left| \csc \, x + \cot \, x \right|} + C</math> ; Hiperbolik :<math>\int \sinh \, x \, dx = \cosh \, x + C</math> :<math>\int \cosh \, x \, dx = \sinh \, x + C</math> :<math>\int \tanh \, x \, dx = \ln{\left| \cosh \, x \right|} + C</math> :<math>\int \coth \, x \, dx = \ln{\left| \sinh \, x \right|} + C</math> :<math>\int \mbox{sech} \, x \, dx = \arctan \,(\sinh \, x) + C</math> :<math>\int \mbox{csch} \, x \, dx = \ln{\left| \tanh \, {x \over 2} \right|} + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : 1. <math>\int x^3 - cos 3x + e^{5x+2} - \frac{1}{2x+5}\,dx</math> : <math>= \frac{1}{3+1}x^{3+1} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : <math>= \frac{x^4}{4} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : 2. <math>\int x(x-5)^4\,dx</math> : <math>= \int x(x^4-4x^3(5)+6x^2(25)-4x(125)+625)\,dx</math> : <math>= \int x(x^4-20x^3+150x^2-500x+625)\,dx</math> : <math>= \int x^5-20x^4+150x^3-500x^2+625x\,dx</math> : <math>= \frac{x^6}{6}-4x^5+\frac{150x^4}{4}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> : <math>= \frac{x^6}{6}-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : 1. <math>\int \frac{\ln \, x}{x}\,dx</math> : <math>u = \ln \, x\, du = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{\ln(x)}{x} dx \\ \int u du \\ \frac{1}{2} u^2 + C \\ \frac{1}{2} ln^2 x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x-5,\, du = dx,\, x = u+5</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4\,dx \\ \int (u+5)u^4\,du \\ \int (u^5+5u^4)\,du \\ \frac{1}{6} u^6+u^5 + C \\ \frac{1}{6} (x-5)^6+(x-5)^5 + C \\ \frac{1}{6} (x^6-6x^5(5)+15x^4(25)-20x^3(125)+15x^2(625)-6x(3125)+15625)+(x^5-5x^4(5)+10x^3(25)-10x^2(125)+5x(625)+3125) + C \\ \frac{1}{6}x^6-5x^5+\frac{125x^4}{2}-\frac{1250x^3}{3}+\frac{3125x^2}{2}-3125x+\frac{15625}{6}+x^5-25x^4+250x^3-1250x^2+3125x+3125 + C \\ \frac{1}{6}x^6-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C \\ \end{aligned} </math> </div></div> === integral parsial === ;Cara 1: Rumus Integral parsial diselesaikan dengan rumus berikut. : <math>\int u\,dv = uv - \int v\,du </math> Berikut contoh penyelesaian cara parsial dengan rumus. : 1. <math>\int x \sin \, x \, dx</math> : <math>u = x, du = 1 \, dx, dv = \sin \, x \, dx, v = -\cos \, x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \,dx \\ (x)(-\cos \, x) - \int (-\cos \, x)(1 dx) \\ -x \cos \, x + \int \cos \, x dx \\ -x \cos \, x + \sin \, x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x,\, du = 1\,dx,\, dv = (x-5)^4\,dx,\, v = \frac{1}{5}(x-5)^5</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - \int (\frac{1}{5}(x-5)^5)(1\,dx) \\ \frac{x}{5}(x-5)^5 - \int \frac{1}{5}(x-5)^5\,dx \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6\,dx \\ \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x\,dx</math> : <math>u = \ln \,x,\, du = \frac{1}{x}\,dx,\, dv = x\,dx,\, v = \frac{x^2}{2}</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x\, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{x^2}{2})(\frac{1}{x}\,dx) \\ \frac{x^2}{2} \ln \,x - \int \frac{x^2}{2}(\frac{1}{x})\,dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \,dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x\,dx</math> ini terjadi dua kali integral parsial : <math>u = \sin \,x,\, du = \cos \,x\,dx,\, dv = e^x\,dx,\, v = e^x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x\, dx \\ (\sin \,x)(e^x) - \int (e^x)(\cos \,x\,dx) \\ e^x \sin \,x - \int e^x \cos \,x\,dx \\ u = \cos \,x,\, du = -\sin \,x\,dx,\, dv = e^x\,dx,\, v = e^x \\ (\cos \,x)(e^x) - \int (e^x)(-\sin \,x\,dx) \\ e^x \cos \,x + \int e^x \sin \,x\,dx \\ e^x \sin \,x - (e^x \cos \,x + \int e^x \sin \,x\,dx) \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x\,dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> ;Cara 2: Tabel Untuk <math display="inline">\int u\,dv</math>, berlaku ketentuan sebagai berikut. {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>u</math> || <math>dv</math> |- | - || <math>\frac{du}{dx}</math> || <math>v</math> |- | + || <math>\frac{d^2u}{dx^2}</math> || <math>\int v\,dx</math> |} Berikut contoh penyelesaian cara parsial dengan tabel. : 1. <math>\int x \sin \, x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>\sin \, x</math> |- | - || <math>1</math> || <math>-\cos \, x</math> |- | + || <math>0</math> || <math>-\sin \, x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \, dx \\ (x)(-\cos \, x) - (1)(-\sin \, x) + C \\ -x \cos \, x + \sin \, x + C \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4 \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>(x-5)^4</math> |- | - || <math>1</math> || <math>\frac{1}{5}(x-5)^5</math> |- | + || <math>0</math> || <math>\frac{1}{30}(x-5)^6</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - (1)(\frac{1}{30}(x-5)^6) + C \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6 + C \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\ln \,x</math> || <math>x</math> |- | - || <math>\frac{1}{x}</math> || <math>\frac{x^2}{2}</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x \, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{1}{x})(\frac{x^2}{2}) \, dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \, dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x \, dx</math> ini terjadi dua kali integral parsial {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\sin \,x</math> || <math>e^x</math> |- | - || <math>\cos \, x</math> || <math>e^x</math> |- | + || <math>-\sin \,x</math> || <math>e^x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x \, dx \\ (\sin \,x)(e^x) - (\cos \,x)(e^x) + \int (-\sin \,x)(e^x) \, dx \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x) \, dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{1}{x^2 - 4}\,dx</math> Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{1}{x^2 - 4} \\ =&\; \frac{1}{(x + 2)(x - 2)} \\ =&\; \frac{A}{x + 2} + \frac{B}{x - 2} \\ =&\; \frac{A(x - 2) + B(x + 2)}{x^2 - 4} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int -\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)}\,dx \\ =&\; \frac{1}{4} \int \frac{1}{x - 2} - \frac{1}{x + 2}\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> : <math>\int \frac{x^2+5x+4}{(x-3)^3}\,dx</math> : cara 1 Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^3} \\ : \text{untuk mencari C yaitu } (3)^2+5(3)+4=28 \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x - 3)^2 + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x^2 - 6x + 9) + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 - 6Ax + 9A + Bx - 3B}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 + (- 6A + B)x +(9 - 3B)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ \end{aligned} </math> Kita tahu bahwa <math display="inline">A = 1</math> dan <math display="inline">-6A + B = 5</math> dapat diselesaikan, yaitu <math display="inline">A = 1</math> dan {{nowrap|<math display="inline">B = 11</math>.}} : <math> \begin{aligned} &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int \frac{1}{x - 3} + \frac{11}{(x - 3)^2} + \frac{28}{(x - 3)^3}\,dx \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : cara 2 Pertama, ubah u=x-3 menjadi x=u+3 dan du=dx tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{(u+3)^2+5(u+3)+4}{u^3} \\ =&\; \frac{u^2+11u+28}{u^3} \\ =&\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int \frac{u^2+11u+28}{u^3}\,du \\ =&\; \int (\frac{u^2}{u^3} + \frac{11u}{u^3} + \frac{28}{u^3})\,du \\ =&\; \int (\frac{1}{u} + \frac{11}{u^2} + \frac{28}{u^3})\,du \\ =&\; \ln u + (-11 u^{-1}) + (-14 u^{-2}) + C \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> === integral substitusi trigonometri === {| class="wikitable" |- ! Bentuk !! Substitusi Trigonometri |- | <math>\sqrt{a^2 - b^2 x^2}</math> || <math>x = \frac{a}{b} sin \, \alpha </math> |- | <math>\sqrt{a^2 + b^2 x^2}</math> || <math>x = \frac{a}{b} tan \, \alpha</math> |- | <math>\sqrt{b^2 x^2 - a^2}</math> || <math>x = \frac{a}{b} sec \, \alpha</math> |} Berikut contoh penyelesaian cara substitusi trigonometri. : <math>\int \frac{dx}{x^2 \sqrt{x^2 + 4}}</math> : <math>x = 2 \tan \, A, dx = 2 \sec^2 \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \int \frac{dx}{x^2\sqrt{x^2+4}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(2 \tan \, A)^2 \sqrt{4 + (2 \tan \, A)^2}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 + 4 \tan^2 \, A}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4(1 + \tan^2 \, A)}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 \sec^2 \, A }} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(4 \tan^2 \, A)(2 \sec \, A)} \\ =&\; \int \frac{\sec \, A \,dA}{4 \tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\sec \, A \,dA}{\tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A} \,dA \end{aligned} </math> Substitusi berikut dapat dibuat. : <math>\int \frac{\cos \, A}{\sin^2 \, A} \,dA</math> : <math>t = \sin \, A, dt = \cos \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A}\,dA \\ =&\; \frac{1}{4} \int \frac{dt}{t^2} \\ =&\; \frac{1}{4} \left(-\frac{1}{t}\right) + C \\ =&\; -\frac{1}{4 t} + C \\ =&\; -\frac{1}{4 \sin \, A} + C \end{aligned} </math> Ingat bahwa <math display="inline">\sin \, A = \frac{x}{\sqrt{x^2 + 4}}</math> berlaku. : <math> \begin{aligned} =&\; -\frac{1}{4 \sin \, A} + C \\ =&\; -\frac{\sqrt{x^2 + 4}}{4x} + C \end{aligned} </math> === integral mutlak === : <math>\int |f(x)| \,dA</math> buatlah <math>f(x) \ge 0</math> jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x) : <math>\int_a^c |f(x)| \, dx = \int_a^b f(x) \, dx + \int_b^c -f(x) \, dx = F(b) - F(a) - F(c) + F(b)</math> jika <math>|f(x)| = \begin{cases} f(x), & \mbox{jika } f(x) \ge b, \\ -f(x), & \mbox{jika } f(x) < b. \end{cases}</math>. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif. === integral fungsi ganjil dan integral fungsi genap === : <math>\int_{-a}^a f(x) \,dx</math> dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral: : integral fungsi ganjil : <math>\int_{-a}^a f(x) \,dx</math> = 0 : integral fungsi genap : <math>\int_{-a}^a f(x) \,dx</math> = <math>2 \int_{0}^a f(x) \,dx</math> === integral notasi === : <math>\int_a^b f(x) \,dx = \int_{a-p}^{b-p} f(x) \,dx = \int_{a+p}^{b+p} f(x) \,dx</math> : <math>\int_a^c f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx</math> : <math>\int_a^b f(x \pm k) \,dx = \int_{a \pm k}^{b \pm k} f(x) \,dx</math> === integral terbalik === : <math>\int_a^b f(x) \,dx = -\int_b^a f(x) \,dx</math> == Jenis integral lainnya == === panjang busur === ; Sumbu ''x'' : <math>S = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' : <math>S = \int_{y_1}^{y_2} \sqrt{1 + (f'(y))^2}\,dy</math> === luas daerah === ; Satu kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} f(x)\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} f(y)\,dy</math> ; Dua kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} (f(x_2) - f(x_1))\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} (f(y_2) - f(y_1))\,dy</math> : atau juga <math>L = \frac {D \sqrt{D}}{6 a^2}</math> === luas permukaan benda putar === ; Sumbu ''x'' sebagai poros : <math>L_p = 2 \pi \int_{x_1}^{x_2} f(x)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' sebagai poros : <math>L_p = 2 \pi \int_{y_1}^{y_2} f(y)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(y))^2}\,dy</math> === volume benda putar === ; Satu kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} (f(x))^2\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} (f(y))^2\,dy</math> ; Dua kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} ((f(x_2))^2 - (f(x_1))^2)\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} ((f(y_2))^2 - (f(y_1))^2)\,dy</math> : atau juga <math>V = |\frac {D^2 \sqrt{D}}{30 a^3}|</math> == Integral lipat == Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga. ;contoh # Tentukan hasil dari: * <math>\int (5x-1)(5x^2-2x+7)^4 dx</math> * <math>\int x^2 cos 3x dx</math> * <math>\int sin^2 x cos x dx</math> * <math>\int cos^2 x sin x dx</math> * <math>\int sec x dx</math> * <math>\int csc x dx</math> * <math>\int \frac{1}{2x^2-5x+3} dx</math> * <math>\int \frac{1}{\sqrt{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> * <math>\int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx</math> * <math>\int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx</math> * <math>\int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx</math> * <math>\int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx</math> * <math>\int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx \\ a=1, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-sin^2 x} \\ &= \frac{1}{cos^2 x} \\ &= sec^2 x \\ \int sec^2 x dx &= tan x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx \\ a=1, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-cos^2 x} \\ &= \frac{1}{sin^2 x} \\ &= csc^2 x \\ \int csc^2 x dx &= -cot x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx \\ a=sin x, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{sin x}{1-sin^2 x} \\ &= \frac{sin x}{cos^2 x} \\ &= sec x \cdot tan x \\ \int sec x \cdot tan x dx &= sec x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx \\ a=cos x, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{cos x}{1-cos^2 x} \\ &= \frac{cos x}{sin^2 x} \\ &= csc x \cdot cot x \\ \int csc x \cdot cot x dx &= -csc x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx \\ \text{misalkan } y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y &= \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y^2 &= \frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}} \\ &= \frac{x}{y} \\ y^3 &= x \\ 3y^2 dy &= dx \\ \int y \cdot 3y^2 dy \\ \int 3y^3 dy \\ \frac{3y^4}{4}+C \\ \end{aligned} </math> </div></div> # Tentukan hasil dari: * <math>\int_{1}^{5} 2x-3 dx</math> * <math>\int_{0}^{4} |x-2| dx</math> * <math>\int_{2}^{4} |3-x| dx</math> * <math>\int_{1}^{5} x^2-6x+8 dx</math> * <math>\int_{-3}^{5} |x^2-2x-8| dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} 2x-3 dx &= \left. x^2 - 3x \right|_{1}^{5} \\ &= 10 - (-2) \\ &= 12 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{0}^{4} |x-2| dx \\ \text{ tentukan nilai harga nol } \\ x-2 &= 0 \\ x &= 2 \\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti } \\ |x-2| = \begin{cases} x-2, & \mbox{jika } x \ge 2, \\ -(x-2), & \mbox{jika } x < 2. \end{cases} \\ \int_{0}^{4} |x-2| dx &= \int_{2}^{4} x-2 dx + \int_{0}^{2} -(x-2) dx \\ &= \left. (\frac{x^2}{2}-2x) \right|_{2}^{4} + \left. (\frac{-x^2}{2}+2x) \right|_{0}^{2} \\ &= 0 - (-2) + 2 - 0 \\ &= 4 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{2}^{4} |3-x| dx \\ \text{ tentukan nilai syarat-syarat } \\ 3-x &\ge 0 \\ \text{ harga nol } \\ x &= 3 \\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti } \\ |3-x| = \begin{cases} 3-x, & \mbox{jika } x \le 3, \\ -(3-x), & \mbox{jika } x > 3. \end{cases} \\ \int_{2}^{4} |3-x| dx &= \int_{2}^{3} 3-x dx + \int_{3}^{4} -(3-x) dx \\ &= \left. (3x-\frac{x^2}{2}) \right|_{2}^{3} + \left. (-3x+\frac{x^2}{2}) \right|_{3}^{4} \\ &= 4.5 - 4 - 4 - (-4.5) \\ &= 1 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} x^2-6x+8 dx &= \left. \frac{x^3}{3}-3x^2+8x \right|_{1}^{5} \\ &= \frac{20}{3} - \frac{16}{3} \\ &= \frac{4}{3} \\ &= 1\frac{1}{3} \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{-3}^{5} |x^2-2x-8| dx \\ \text{ tentukan nilai syarat-syarat } \\ x^2-2x-8 &\ge 0 \\ \text{ harga nol } \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x = 4 \\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti } \\ |x^2-2x-8| = \begin{cases} x^2-2x-8, & \mbox{jika } x \le -2, \\ -(x^2-2x-8), & \mbox{jika } -2 < x < 4, \\ x^2-2x-8, & \mbox{jika } x \ge 4 \end{cases} \\ \int_{-3}^{5} |x^2-2x-8| dx &= \int_{-3}^{-2} x^2-2x-8 dx + \int_{-2}^{4} -(x^2-2x-8) dx + \int_{4}^{5} x^2-2x-8 dx \\ &= \left. (\frac{x^3}{3}-x^2-8x) \right|_{-3}^{-2} + \left. (\frac{-x^3}{3}+x^2+8x) \right|_{-2}^{4} + \left. (\frac{x^3}{3}-x^2-8x) \right|_{4}^{5} \\ &= \frac{28}{3} - 6 + \frac{80}{3} - (-\frac{28}{3}) - \frac{70}{3} - (-\frac{80}{3}) \\ &= \frac{128}{3} \\ &= 42\frac{2}{3} \\ \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x\,dx \\ L &= \frac{1}{2} x^2 \\ L &= \frac{1}{2} xy \quad (y = x) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x^2</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x^2\,dx \\ L &= \frac{1}{3} x^3 \\ L &= \frac{1}{3} xy \quad (y = x^2) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = \sqrt{x}</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int \sqrt{x}\,dx \\ L &= \frac{2}{3} x^{\frac{3}{2}} \\ L &= \frac{2}{3} xy \quad (y = \sqrt{x}) \end{aligned} </math> </div></div> # Buktikan luas persegi <math>L = s^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = s \text{ dan titik (''s'', ''s''), } \\ L &= \int_{0}^{s} s\,dx \\ L &= sx |_{0}^{s} \\ L &= ss - 0 \\ L &= s^2 \end{aligned} </math> </div></div> # Buktikan luas persegi panjang <math>L = pl</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = l \text{dan titik (''p'', ''l''), } \\ L &= \int_{0}^{p} l\,dx \\ L &= lx |_{0}^{p} \\ L &= pl - 0 \\ L &= pl \end{aligned} </math> </div></div> * Buktikan luas segitiga <math>L = \frac{at}{2}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{-tx}{a} + t \text { dan titik (''a'', ''t''), } \\ L &= \int_{0}^{a} \left (\frac{-tx}{a} + t \right) \,dx \\ L &= \left. \frac{-tx^2}{2a} + tx \right|_{0}^{a} \\ L &= \frac{-ta^2}{2a} + ta - 0 + 0 \\ L &= \frac{-ta}{2} + ta \\ L &= \frac{at}{2} \end{aligned} </math> </div></div> # Buktikan volume tabung <math>V = \pi r^2t</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ Dengan posisi } y = r \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} r^2\,dx \\ V &= \pi \left. r^2x \right|_{0}^{t} \\ V &= \pi r^2t - 0 \\ V &= \pi r^2t \end{aligned} </math> </div></div> # Buktikan volume kerucut <math>V = \frac{\pi r^2t}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{rx}{t} \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} \left (\frac{rx}{t} \right)^2 \,dx \\ V &= \pi \left. \frac{r^2 x^3}{3t^2} \right|_{0}^{t} \\ V &= \pi \frac{r^2 t^3}{3t^2} - 0 \\ V &= \frac{\pi r^2t}{3} \end{aligned} </math> </div></div> # Buktikan volume bola <math>V = \frac{4 \pi r^3}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), } \\ V &= \pi \int_{-r}^{r} \left(\sqrt {r^2 - x^2}\right)^2\,dx \\ V &= \pi \int_{-r}^{r} r^2 - x^2 dx \\ V &= \pi \left. r^2x - \frac{x^3}{3} \right|_{-r}^{r} \\ V &= \pi \left (r^3 - \frac{r^3}{3} - \left (-r^3 + \frac{r^3}{3} \right) \right) \\ V &= \frac{4 \pi r^3}{3} \end{aligned} </math> </div></div> # Buktikan luas permukaan bola <math>L = 4 \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{selanjutnya } \\ ds &= \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ ds &= \sqrt{1 + \frac{x^2}{r^2 - x^2}}\,dx \\ ds &= \sqrt{\frac{r^2}{r^2 - x^2}}\,dx \\ ds &= \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ \text{sehingga } \\ L &= 2 \pi \int_{-r}^{r} \sqrt{r^2 - x^2} \cdot \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ L &= 2 \pi \int_{-r}^{r} r\,dx \\ L &= 2 \pi rx|_{-r}^{r} \\ L &= 2 \pi (r r - r (-r)) \\ L &= 2 \pi (r^2 + r^2) \\ L &= 4 \pi r^2 \end{aligned} </math> </div></div> # Buktikan keliling lingkaran <math>K = 2 \pi r</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0) } \\ \text{kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{sehingga } \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{x^2}{r^2 - x^2})}\,dx \\ K &= 4 \int_{0}^{r} \sqrt\frac{r^2}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \sqrt\frac{1}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \frac{1}{\sqrt{r^2 - x^2}}\,dx \\ K &= 4r \left. \arcsin \left (\frac{x}{r} \right) \right|_{0}^{r} \\ K &= 4r \left (\arcsin \left (\frac{r}{r}\right) - \arcsin \left (\frac{0}{r} \right) \right) \\ K &= 4r \left (\arcsin\left (1 \right) - \arcsin \left(0 \right) \right)) \\ K &= 4r \left (\frac{\pi}{2} \right) \\ K &= 2 \pi r \end{aligned} </math> </div></div> # Buktikan luas lingkaran <math>L = \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. } \\ \sin(\theta) &= \frac{x}{r} \\ x &= r \sin(\theta) \\ dx &= r \cos(\theta)\,d\theta \\ \text{dengan turunan di atas } \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - x^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - (r \sin(\theta))^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - r^2 \sin^2(\theta)}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 (1 - \sin^2(\theta))}\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta)\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta) (r \cos(\theta)\,d\theta) \\ L &= 4 \int_{0}^{r} r^2 \cos^2(\theta)\,d\theta \\ L &= 4r^2 \int_{0}^{r} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta \\ L &= 2r^2 \int_{0}^{r} (1 + \cos(2\theta))\,d\theta \\ L &= 2r^2 \left (\theta + \frac{1}{2} \sin(2\theta) \right)|_{0}^{r} \\ L &= 2r^2 (\theta + \sin(\theta) \cos(\theta))|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left (\frac{x}{r} \right) + \left (\frac{x}{r} \right) \left (\frac{r^2 - x^2}{r} \right) \right)|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left(\frac{r}{r} \right) + \left (\frac{r}{r} \right) \left (\frac {r^2 - r^2}{r} \right) \right) - \left(\arcsin \left (\frac{0}{r} \right) + \left (\frac{0}{r}\right) \left (\frac{r^2 - 0^2}{r} \right) \right) \\ L &= 2r^2 (\arcsin(1) + 0 - (\arcsin(0) + 0)) \\ L &= 2r^2 \left (\frac{\pi}{2} \right) \\ L &= \pi r^2 \end{aligned} </math> </div></div> # Buktikan luas elips <math>L = \pi ab</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \frac{b \sqrt{a^2 - x^2}}{a} \text{ serta (-''a'', 0) dan (''a'', 0), } \\ L &= 4 \int_{0}^{r} \frac{b \sqrt{a^2 - x^2}}{a}\,dx \\ L &= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx \\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, } \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{\frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx}{4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx} \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{b}{a} \\ L_\text{elips} &= \frac{b}{a} L_\text{ling} \\ L_\text{elips} &= \frac{b}{a} \pi a^2 \\ L_\text{elips} &= \pi ab \end{aligned} </math> </div></div> # Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ x^-2x &= 4x+7 \\ x^2-6x-7 &= 0 \\ (x+1)(x-7) &= 0 \\ x=-1 &\text{ atau } x=7 \\ y &= 4(-1)+7 = 3 \\ y &= 4(7)+7 = 35 \\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut } \\ L &= \int_{-1}^{7} 4x+7-(x^2-2x)\,dx \\ &= \int_{-1}^{7} -x^2+6x+7\,dx \\ &= -\frac{x^3}{3}+3x^2+7x|_{-1}^{7} \\ &= -\frac{7^3}{3}+3(7)^2+7(7)-(-\frac{(-1)^3}{3}+3(-1)^2+7(-1)) \\ &= \frac{245}{3}-(-\frac{13}{3}) \\ &= \frac{258}{3} \\ &= 86 \\ \end{aligned} </math> </div></div> # Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ 6-1,5x &= x-4 \\ 2,5x &= 10 \\ x &= 4 \\ y &= 4-4 = 0 \\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut } \\ \text{untuk x=0 } \\ y &= 6-1,5(0) = 6 \\ y &= 0-4 = -4 \\ L &= \pi \int_{0}^{4} ((6-1,5x)^2-(x-4)^2)\,dx \\ &= \pi \int_{0}^{4} (36-18x+2,25x^2-(x^2-8x+16))\,dx \\ &= \pi \int_{0}^{4} (1,25x^2-10x+20)\,dx \\ &= \pi (\frac{1,25x^3}{3}-5x^2+20x)|_{0}^{4} \\ &= \pi (\frac{1,25(4)^3}{3}-5(4)^2+20(4)-(\frac{1,25(0)^3}{3}-5(0)^2+20(0))) \\ &= \pi (\frac{30}{3}-0) \\ &= 10\pi \\ \end{aligned} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 644arrmynfhmb89ttybt92q890ahs8b 115022 115021 2026-04-27T00:09:52Z ~2026-25546-23 43043 /* integral pecahan parsial */ 115022 wikitext text/x-wiki == Kaidah umum == :# <math>\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ konstan)}\,\!</math> :# <math>\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx</math> :# <math>\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx</math> :# <math>\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx</math> :# <math>\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\! </math> :# <math>\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C </math> :# <math>\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C </math> :# <math>\int f(x) \, dx = F(x) + C</math> :# <math>\int_a^b f(x) \, dx = F(b) - F(a)</math> :# <math>\int f(x) \circ g(x) dx</math> = f’(g(x)) g’(x) :# <math>\int f^{-1}(x) dx = \frac{1}{f'(f'(x))}</math> == rumus sederhana == :<math>\int \, dx = x + C</math> :<math>\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int {dx \over x} = \ln{\left|x\right|} + C</math> <math>\int {dx \over \sqrt{a^2-b^2 x^2}} = \frac{1}{b} \arcsin \, {bx \over a} + C</math> :<math>\int {dx \over {a^2+b^2x^2}} = {1 \over ab}\arctan \, {bx \over a} + C</math> <math>\int {dx \over x\sqrt{b^2x^2-a^2}} = {1 \over a}\arcsec \, {bx \over a} + C</math> ; Eksponen dan logaritma :<math>\int e^x\,dx = e^x + C</math> :<math>\int a^x\,dx = \frac{a^x}{\ln{a}} + C</math> :<math>\int \ln \, x\,dx = x \cdot \, \ln {x} - x + C</math> :<math>\int \, ^b\!\log \, x\,dx = x \cdot \, ^b\!\log \, x - x \cdot \,^b\!\log e + C</math> ; Trigonometri :<math>\int \sin \, x \, dx = -\cos \, x + C</math> :<math>\int \cos \, x \, dx = \sin \, x + C</math> :<math>\int \tan \, x \, dx = -\ln{\left| \cos \, x \right|} + C</math> :<math>\int \cot \, x \, dx = \ln{\left| \sin \, x \right|} + C</math> :<math>\int \sec \, x \, dx = \ln{\left| \sec \, x + \tan \, x \right|} + C</math> :<math>\int \csc \, x \, dx = -\ln{\left| \csc \, x + \cot \, x \right|} + C</math> ; Hiperbolik :<math>\int \sinh \, x \, dx = \cosh \, x + C</math> :<math>\int \cosh \, x \, dx = \sinh \, x + C</math> :<math>\int \tanh \, x \, dx = \ln{\left| \cosh \, x \right|} + C</math> :<math>\int \coth \, x \, dx = \ln{\left| \sinh \, x \right|} + C</math> :<math>\int \mbox{sech} \, x \, dx = \arctan \,(\sinh \, x) + C</math> :<math>\int \mbox{csch} \, x \, dx = \ln{\left| \tanh \, {x \over 2} \right|} + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : 1. <math>\int x^3 - cos 3x + e^{5x+2} - \frac{1}{2x+5}\,dx</math> : <math>= \frac{1}{3+1}x^{3+1} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : <math>= \frac{x^4}{4} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : 2. <math>\int x(x-5)^4\,dx</math> : <math>= \int x(x^4-4x^3(5)+6x^2(25)-4x(125)+625)\,dx</math> : <math>= \int x(x^4-20x^3+150x^2-500x+625)\,dx</math> : <math>= \int x^5-20x^4+150x^3-500x^2+625x\,dx</math> : <math>= \frac{x^6}{6}-4x^5+\frac{150x^4}{4}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> : <math>= \frac{x^6}{6}-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : 1. <math>\int \frac{\ln \, x}{x}\,dx</math> : <math>u = \ln \, x\, du = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{\ln(x)}{x} dx \\ \int u du \\ \frac{1}{2} u^2 + C \\ \frac{1}{2} ln^2 x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x-5,\, du = dx,\, x = u+5</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4\,dx \\ \int (u+5)u^4\,du \\ \int (u^5+5u^4)\,du \\ \frac{1}{6} u^6+u^5 + C \\ \frac{1}{6} (x-5)^6+(x-5)^5 + C \\ \frac{1}{6} (x^6-6x^5(5)+15x^4(25)-20x^3(125)+15x^2(625)-6x(3125)+15625)+(x^5-5x^4(5)+10x^3(25)-10x^2(125)+5x(625)+3125) + C \\ \frac{1}{6}x^6-5x^5+\frac{125x^4}{2}-\frac{1250x^3}{3}+\frac{3125x^2}{2}-3125x+\frac{15625}{6}+x^5-25x^4+250x^3-1250x^2+3125x+3125 + C \\ \frac{1}{6}x^6-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C \\ \end{aligned} </math> </div></div> === integral parsial === ;Cara 1: Rumus Integral parsial diselesaikan dengan rumus berikut. : <math>\int u\,dv = uv - \int v\,du </math> Berikut contoh penyelesaian cara parsial dengan rumus. : 1. <math>\int x \sin \, x \, dx</math> : <math>u = x, du = 1 \, dx, dv = \sin \, x \, dx, v = -\cos \, x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \,dx \\ (x)(-\cos \, x) - \int (-\cos \, x)(1 dx) \\ -x \cos \, x + \int \cos \, x dx \\ -x \cos \, x + \sin \, x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x,\, du = 1\,dx,\, dv = (x-5)^4\,dx,\, v = \frac{1}{5}(x-5)^5</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - \int (\frac{1}{5}(x-5)^5)(1\,dx) \\ \frac{x}{5}(x-5)^5 - \int \frac{1}{5}(x-5)^5\,dx \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6\,dx \\ \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x\,dx</math> : <math>u = \ln \,x,\, du = \frac{1}{x}\,dx,\, dv = x\,dx,\, v = \frac{x^2}{2}</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x\, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{x^2}{2})(\frac{1}{x}\,dx) \\ \frac{x^2}{2} \ln \,x - \int \frac{x^2}{2}(\frac{1}{x})\,dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \,dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x\,dx</math> ini terjadi dua kali integral parsial : <math>u = \sin \,x,\, du = \cos \,x\,dx,\, dv = e^x\,dx,\, v = e^x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x\, dx \\ (\sin \,x)(e^x) - \int (e^x)(\cos \,x\,dx) \\ e^x \sin \,x - \int e^x \cos \,x\,dx \\ u = \cos \,x,\, du = -\sin \,x\,dx,\, dv = e^x\,dx,\, v = e^x \\ (\cos \,x)(e^x) - \int (e^x)(-\sin \,x\,dx) \\ e^x \cos \,x + \int e^x \sin \,x\,dx \\ e^x \sin \,x - (e^x \cos \,x + \int e^x \sin \,x\,dx) \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x\,dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> ;Cara 2: Tabel Untuk <math display="inline">\int u\,dv</math>, berlaku ketentuan sebagai berikut. {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>u</math> || <math>dv</math> |- | - || <math>\frac{du}{dx}</math> || <math>v</math> |- | + || <math>\frac{d^2u}{dx^2}</math> || <math>\int v\,dx</math> |} Berikut contoh penyelesaian cara parsial dengan tabel. : 1. <math>\int x \sin \, x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>\sin \, x</math> |- | - || <math>1</math> || <math>-\cos \, x</math> |- | + || <math>0</math> || <math>-\sin \, x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \, dx \\ (x)(-\cos \, x) - (1)(-\sin \, x) + C \\ -x \cos \, x + \sin \, x + C \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4 \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>(x-5)^4</math> |- | - || <math>1</math> || <math>\frac{1}{5}(x-5)^5</math> |- | + || <math>0</math> || <math>\frac{1}{30}(x-5)^6</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - (1)(\frac{1}{30}(x-5)^6) + C \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6 + C \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\ln \,x</math> || <math>x</math> |- | - || <math>\frac{1}{x}</math> || <math>\frac{x^2}{2}</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x \, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{1}{x})(\frac{x^2}{2}) \, dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \, dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x \, dx</math> ini terjadi dua kali integral parsial {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\sin \,x</math> || <math>e^x</math> |- | - || <math>\cos \, x</math> || <math>e^x</math> |- | + || <math>-\sin \,x</math> || <math>e^x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x \, dx \\ (\sin \,x)(e^x) - (\cos \,x)(e^x) + \int (-\sin \,x)(e^x) \, dx \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x) \, dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{1}{x^2 - 4}\,dx</math> Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{1}{x^2 - 4} \\ =&\; \frac{1}{(x + 2)(x - 2)} \\ =&\; \frac{A}{x + 2} + \frac{B}{x - 2} \\ =&\; \frac{A(x - 2) + B(x + 2)}{x^2 - 4} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int \frac{1}{x - 2} - \frac{1}{x + 2}\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> : <math>\int \frac{x^2+5x+4}{(x-3)^3}\,dx</math> : cara 1 Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^3} \\ : \text{untuk mencari C yaitu } (3)^2+5(3)+4=28 \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x - 3)^2 + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x^2 - 6x + 9) + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 - 6Ax + 9A + Bx - 3B}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 + (- 6A + B)x +(9 - 3B)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ \end{aligned} </math> Kita tahu bahwa <math display="inline">A = 1</math> dan <math display="inline">-6A + B = 5</math> dapat diselesaikan, yaitu <math display="inline">A = 1</math> dan {{nowrap|<math display="inline">B = 11</math>.}} : <math> \begin{aligned} &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int (\frac{1}{x - 3} + \frac{11}{(x - 3)^2} + \frac{28}{(x - 3)^3})\,dx \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : cara 2 Pertama, ubah u=x-3 menjadi x=u+3 dan du=dx tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{(u+3)^2+5(u+3)+4}{u^3} \\ =&\; \frac{u^2+11u+28}{u^3} \\ =&\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int \frac{u^2+11u+28}{u^3}\,du \\ =&\; \int (\frac{u^2}{u^3} + \frac{11u}{u^3} + \frac{28}{u^3})\,du \\ =&\; \int (\frac{1}{u} + \frac{11}{u^2} + \frac{28}{u^3})\,du \\ =&\; \ln u + (-11 u^{-1}) + (-14 u^{-2}) + C \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> === integral substitusi trigonometri === {| class="wikitable" |- ! Bentuk !! Substitusi Trigonometri |- | <math>\sqrt{a^2 - b^2 x^2}</math> || <math>x = \frac{a}{b} sin \, \alpha </math> |- | <math>\sqrt{a^2 + b^2 x^2}</math> || <math>x = \frac{a}{b} tan \, \alpha</math> |- | <math>\sqrt{b^2 x^2 - a^2}</math> || <math>x = \frac{a}{b} sec \, \alpha</math> |} Berikut contoh penyelesaian cara substitusi trigonometri. : <math>\int \frac{dx}{x^2 \sqrt{x^2 + 4}}</math> : <math>x = 2 \tan \, A, dx = 2 \sec^2 \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \int \frac{dx}{x^2\sqrt{x^2+4}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(2 \tan \, A)^2 \sqrt{4 + (2 \tan \, A)^2}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 + 4 \tan^2 \, A}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4(1 + \tan^2 \, A)}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 \sec^2 \, A }} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(4 \tan^2 \, A)(2 \sec \, A)} \\ =&\; \int \frac{\sec \, A \,dA}{4 \tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\sec \, A \,dA}{\tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A} \,dA \end{aligned} </math> Substitusi berikut dapat dibuat. : <math>\int \frac{\cos \, A}{\sin^2 \, A} \,dA</math> : <math>t = \sin \, A, dt = \cos \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A}\,dA \\ =&\; \frac{1}{4} \int \frac{dt}{t^2} \\ =&\; \frac{1}{4} \left(-\frac{1}{t}\right) + C \\ =&\; -\frac{1}{4 t} + C \\ =&\; -\frac{1}{4 \sin \, A} + C \end{aligned} </math> Ingat bahwa <math display="inline">\sin \, A = \frac{x}{\sqrt{x^2 + 4}}</math> berlaku. : <math> \begin{aligned} =&\; -\frac{1}{4 \sin \, A} + C \\ =&\; -\frac{\sqrt{x^2 + 4}}{4x} + C \end{aligned} </math> === integral mutlak === : <math>\int |f(x)| \,dA</math> buatlah <math>f(x) \ge 0</math> jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x) : <math>\int_a^c |f(x)| \, dx = \int_a^b f(x) \, dx + \int_b^c -f(x) \, dx = F(b) - F(a) - F(c) + F(b)</math> jika <math>|f(x)| = \begin{cases} f(x), & \mbox{jika } f(x) \ge b, \\ -f(x), & \mbox{jika } f(x) < b. \end{cases}</math>. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif. === integral fungsi ganjil dan integral fungsi genap === : <math>\int_{-a}^a f(x) \,dx</math> dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral: : integral fungsi ganjil : <math>\int_{-a}^a f(x) \,dx</math> = 0 : integral fungsi genap : <math>\int_{-a}^a f(x) \,dx</math> = <math>2 \int_{0}^a f(x) \,dx</math> === integral notasi === : <math>\int_a^b f(x) \,dx = \int_{a-p}^{b-p} f(x) \,dx = \int_{a+p}^{b+p} f(x) \,dx</math> : <math>\int_a^c f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx</math> : <math>\int_a^b f(x \pm k) \,dx = \int_{a \pm k}^{b \pm k} f(x) \,dx</math> === integral terbalik === : <math>\int_a^b f(x) \,dx = -\int_b^a f(x) \,dx</math> == Jenis integral lainnya == === panjang busur === ; Sumbu ''x'' : <math>S = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' : <math>S = \int_{y_1}^{y_2} \sqrt{1 + (f'(y))^2}\,dy</math> === luas daerah === ; Satu kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} f(x)\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} f(y)\,dy</math> ; Dua kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} (f(x_2) - f(x_1))\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} (f(y_2) - f(y_1))\,dy</math> : atau juga <math>L = \frac {D \sqrt{D}}{6 a^2}</math> === luas permukaan benda putar === ; Sumbu ''x'' sebagai poros : <math>L_p = 2 \pi \int_{x_1}^{x_2} f(x)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' sebagai poros : <math>L_p = 2 \pi \int_{y_1}^{y_2} f(y)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(y))^2}\,dy</math> === volume benda putar === ; Satu kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} (f(x))^2\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} (f(y))^2\,dy</math> ; Dua kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} ((f(x_2))^2 - (f(x_1))^2)\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} ((f(y_2))^2 - (f(y_1))^2)\,dy</math> : atau juga <math>V = |\frac {D^2 \sqrt{D}}{30 a^3}|</math> == Integral lipat == Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga. ;contoh # Tentukan hasil dari: * <math>\int (5x-1)(5x^2-2x+7)^4 dx</math> * <math>\int x^2 cos 3x dx</math> * <math>\int sin^2 x cos x dx</math> * <math>\int cos^2 x sin x dx</math> * <math>\int sec x dx</math> * <math>\int csc x dx</math> * <math>\int \frac{1}{2x^2-5x+3} dx</math> * <math>\int \frac{1}{\sqrt{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> * <math>\int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx</math> * <math>\int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx</math> * <math>\int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx</math> * <math>\int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx</math> * <math>\int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx \\ a=1, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-sin^2 x} \\ &= \frac{1}{cos^2 x} \\ &= sec^2 x \\ \int sec^2 x dx &= tan x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx \\ a=1, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-cos^2 x} \\ &= \frac{1}{sin^2 x} \\ &= csc^2 x \\ \int csc^2 x dx &= -cot x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx \\ a=sin x, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{sin x}{1-sin^2 x} \\ &= \frac{sin x}{cos^2 x} \\ &= sec x \cdot tan x \\ \int sec x \cdot tan x dx &= sec x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx \\ a=cos x, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{cos x}{1-cos^2 x} \\ &= \frac{cos x}{sin^2 x} \\ &= csc x \cdot cot x \\ \int csc x \cdot cot x dx &= -csc x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx \\ \text{misalkan } y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y &= \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y^2 &= \frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}} \\ &= \frac{x}{y} \\ y^3 &= x \\ 3y^2 dy &= dx \\ \int y \cdot 3y^2 dy \\ \int 3y^3 dy \\ \frac{3y^4}{4}+C \\ \end{aligned} </math> </div></div> # Tentukan hasil dari: * <math>\int_{1}^{5} 2x-3 dx</math> * <math>\int_{0}^{4} |x-2| dx</math> * <math>\int_{2}^{4} |3-x| dx</math> * <math>\int_{1}^{5} x^2-6x+8 dx</math> * <math>\int_{-3}^{5} |x^2-2x-8| dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} 2x-3 dx &= \left. x^2 - 3x \right|_{1}^{5} \\ &= 10 - (-2) \\ &= 12 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{0}^{4} |x-2| dx \\ \text{ tentukan nilai harga nol } \\ x-2 &= 0 \\ x &= 2 \\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti } \\ |x-2| = \begin{cases} x-2, & \mbox{jika } x \ge 2, \\ -(x-2), & \mbox{jika } x < 2. \end{cases} \\ \int_{0}^{4} |x-2| dx &= \int_{2}^{4} x-2 dx + \int_{0}^{2} -(x-2) dx \\ &= \left. (\frac{x^2}{2}-2x) \right|_{2}^{4} + \left. (\frac{-x^2}{2}+2x) \right|_{0}^{2} \\ &= 0 - (-2) + 2 - 0 \\ &= 4 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{2}^{4} |3-x| dx \\ \text{ tentukan nilai syarat-syarat } \\ 3-x &\ge 0 \\ \text{ harga nol } \\ x &= 3 \\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti } \\ |3-x| = \begin{cases} 3-x, & \mbox{jika } x \le 3, \\ -(3-x), & \mbox{jika } x > 3. \end{cases} \\ \int_{2}^{4} |3-x| dx &= \int_{2}^{3} 3-x dx + \int_{3}^{4} -(3-x) dx \\ &= \left. (3x-\frac{x^2}{2}) \right|_{2}^{3} + \left. (-3x+\frac{x^2}{2}) \right|_{3}^{4} \\ &= 4.5 - 4 - 4 - (-4.5) \\ &= 1 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} x^2-6x+8 dx &= \left. \frac{x^3}{3}-3x^2+8x \right|_{1}^{5} \\ &= \frac{20}{3} - \frac{16}{3} \\ &= \frac{4}{3} \\ &= 1\frac{1}{3} \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{-3}^{5} |x^2-2x-8| dx \\ \text{ tentukan nilai syarat-syarat } \\ x^2-2x-8 &\ge 0 \\ \text{ harga nol } \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x = 4 \\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti } \\ |x^2-2x-8| = \begin{cases} x^2-2x-8, & \mbox{jika } x \le -2, \\ -(x^2-2x-8), & \mbox{jika } -2 < x < 4, \\ x^2-2x-8, & \mbox{jika } x \ge 4 \end{cases} \\ \int_{-3}^{5} |x^2-2x-8| dx &= \int_{-3}^{-2} x^2-2x-8 dx + \int_{-2}^{4} -(x^2-2x-8) dx + \int_{4}^{5} x^2-2x-8 dx \\ &= \left. (\frac{x^3}{3}-x^2-8x) \right|_{-3}^{-2} + \left. (\frac{-x^3}{3}+x^2+8x) \right|_{-2}^{4} + \left. (\frac{x^3}{3}-x^2-8x) \right|_{4}^{5} \\ &= \frac{28}{3} - 6 + \frac{80}{3} - (-\frac{28}{3}) - \frac{70}{3} - (-\frac{80}{3}) \\ &= \frac{128}{3} \\ &= 42\frac{2}{3} \\ \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x\,dx \\ L &= \frac{1}{2} x^2 \\ L &= \frac{1}{2} xy \quad (y = x) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x^2</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x^2\,dx \\ L &= \frac{1}{3} x^3 \\ L &= \frac{1}{3} xy \quad (y = x^2) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = \sqrt{x}</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int \sqrt{x}\,dx \\ L &= \frac{2}{3} x^{\frac{3}{2}} \\ L &= \frac{2}{3} xy \quad (y = \sqrt{x}) \end{aligned} </math> </div></div> # Buktikan luas persegi <math>L = s^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = s \text{ dan titik (''s'', ''s''), } \\ L &= \int_{0}^{s} s\,dx \\ L &= sx |_{0}^{s} \\ L &= ss - 0 \\ L &= s^2 \end{aligned} </math> </div></div> # Buktikan luas persegi panjang <math>L = pl</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = l \text{dan titik (''p'', ''l''), } \\ L &= \int_{0}^{p} l\,dx \\ L &= lx |_{0}^{p} \\ L &= pl - 0 \\ L &= pl \end{aligned} </math> </div></div> * Buktikan luas segitiga <math>L = \frac{at}{2}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{-tx}{a} + t \text { dan titik (''a'', ''t''), } \\ L &= \int_{0}^{a} \left (\frac{-tx}{a} + t \right) \,dx \\ L &= \left. \frac{-tx^2}{2a} + tx \right|_{0}^{a} \\ L &= \frac{-ta^2}{2a} + ta - 0 + 0 \\ L &= \frac{-ta}{2} + ta \\ L &= \frac{at}{2} \end{aligned} </math> </div></div> # Buktikan volume tabung <math>V = \pi r^2t</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ Dengan posisi } y = r \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} r^2\,dx \\ V &= \pi \left. r^2x \right|_{0}^{t} \\ V &= \pi r^2t - 0 \\ V &= \pi r^2t \end{aligned} </math> </div></div> # Buktikan volume kerucut <math>V = \frac{\pi r^2t}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{rx}{t} \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} \left (\frac{rx}{t} \right)^2 \,dx \\ V &= \pi \left. \frac{r^2 x^3}{3t^2} \right|_{0}^{t} \\ V &= \pi \frac{r^2 t^3}{3t^2} - 0 \\ V &= \frac{\pi r^2t}{3} \end{aligned} </math> </div></div> # Buktikan volume bola <math>V = \frac{4 \pi r^3}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), } \\ V &= \pi \int_{-r}^{r} \left(\sqrt {r^2 - x^2}\right)^2\,dx \\ V &= \pi \int_{-r}^{r} r^2 - x^2 dx \\ V &= \pi \left. r^2x - \frac{x^3}{3} \right|_{-r}^{r} \\ V &= \pi \left (r^3 - \frac{r^3}{3} - \left (-r^3 + \frac{r^3}{3} \right) \right) \\ V &= \frac{4 \pi r^3}{3} \end{aligned} </math> </div></div> # Buktikan luas permukaan bola <math>L = 4 \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{selanjutnya } \\ ds &= \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ ds &= \sqrt{1 + \frac{x^2}{r^2 - x^2}}\,dx \\ ds &= \sqrt{\frac{r^2}{r^2 - x^2}}\,dx \\ ds &= \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ \text{sehingga } \\ L &= 2 \pi \int_{-r}^{r} \sqrt{r^2 - x^2} \cdot \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ L &= 2 \pi \int_{-r}^{r} r\,dx \\ L &= 2 \pi rx|_{-r}^{r} \\ L &= 2 \pi (r r - r (-r)) \\ L &= 2 \pi (r^2 + r^2) \\ L &= 4 \pi r^2 \end{aligned} </math> </div></div> # Buktikan keliling lingkaran <math>K = 2 \pi r</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0) } \\ \text{kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{sehingga } \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{x^2}{r^2 - x^2})}\,dx \\ K &= 4 \int_{0}^{r} \sqrt\frac{r^2}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \sqrt\frac{1}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \frac{1}{\sqrt{r^2 - x^2}}\,dx \\ K &= 4r \left. \arcsin \left (\frac{x}{r} \right) \right|_{0}^{r} \\ K &= 4r \left (\arcsin \left (\frac{r}{r}\right) - \arcsin \left (\frac{0}{r} \right) \right) \\ K &= 4r \left (\arcsin\left (1 \right) - \arcsin \left(0 \right) \right)) \\ K &= 4r \left (\frac{\pi}{2} \right) \\ K &= 2 \pi r \end{aligned} </math> </div></div> # Buktikan luas lingkaran <math>L = \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. } \\ \sin(\theta) &= \frac{x}{r} \\ x &= r \sin(\theta) \\ dx &= r \cos(\theta)\,d\theta \\ \text{dengan turunan di atas } \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - x^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - (r \sin(\theta))^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - r^2 \sin^2(\theta)}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 (1 - \sin^2(\theta))}\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta)\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta) (r \cos(\theta)\,d\theta) \\ L &= 4 \int_{0}^{r} r^2 \cos^2(\theta)\,d\theta \\ L &= 4r^2 \int_{0}^{r} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta \\ L &= 2r^2 \int_{0}^{r} (1 + \cos(2\theta))\,d\theta \\ L &= 2r^2 \left (\theta + \frac{1}{2} \sin(2\theta) \right)|_{0}^{r} \\ L &= 2r^2 (\theta + \sin(\theta) \cos(\theta))|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left (\frac{x}{r} \right) + \left (\frac{x}{r} \right) \left (\frac{r^2 - x^2}{r} \right) \right)|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left(\frac{r}{r} \right) + \left (\frac{r}{r} \right) \left (\frac {r^2 - r^2}{r} \right) \right) - \left(\arcsin \left (\frac{0}{r} \right) + \left (\frac{0}{r}\right) \left (\frac{r^2 - 0^2}{r} \right) \right) \\ L &= 2r^2 (\arcsin(1) + 0 - (\arcsin(0) + 0)) \\ L &= 2r^2 \left (\frac{\pi}{2} \right) \\ L &= \pi r^2 \end{aligned} </math> </div></div> # Buktikan luas elips <math>L = \pi ab</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \frac{b \sqrt{a^2 - x^2}}{a} \text{ serta (-''a'', 0) dan (''a'', 0), } \\ L &= 4 \int_{0}^{r} \frac{b \sqrt{a^2 - x^2}}{a}\,dx \\ L &= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx \\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, } \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{\frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx}{4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx} \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{b}{a} \\ L_\text{elips} &= \frac{b}{a} L_\text{ling} \\ L_\text{elips} &= \frac{b}{a} \pi a^2 \\ L_\text{elips} &= \pi ab \end{aligned} </math> </div></div> # Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ x^-2x &= 4x+7 \\ x^2-6x-7 &= 0 \\ (x+1)(x-7) &= 0 \\ x=-1 &\text{ atau } x=7 \\ y &= 4(-1)+7 = 3 \\ y &= 4(7)+7 = 35 \\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut } \\ L &= \int_{-1}^{7} 4x+7-(x^2-2x)\,dx \\ &= \int_{-1}^{7} -x^2+6x+7\,dx \\ &= -\frac{x^3}{3}+3x^2+7x|_{-1}^{7} \\ &= -\frac{7^3}{3}+3(7)^2+7(7)-(-\frac{(-1)^3}{3}+3(-1)^2+7(-1)) \\ &= \frac{245}{3}-(-\frac{13}{3}) \\ &= \frac{258}{3} \\ &= 86 \\ \end{aligned} </math> </div></div> # Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ 6-1,5x &= x-4 \\ 2,5x &= 10 \\ x &= 4 \\ y &= 4-4 = 0 \\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut } \\ \text{untuk x=0 } \\ y &= 6-1,5(0) = 6 \\ y &= 0-4 = -4 \\ L &= \pi \int_{0}^{4} ((6-1,5x)^2-(x-4)^2)\,dx \\ &= \pi \int_{0}^{4} (36-18x+2,25x^2-(x^2-8x+16))\,dx \\ &= \pi \int_{0}^{4} (1,25x^2-10x+20)\,dx \\ &= \pi (\frac{1,25x^3}{3}-5x^2+20x)|_{0}^{4} \\ &= \pi (\frac{1,25(4)^3}{3}-5(4)^2+20(4)-(\frac{1,25(0)^3}{3}-5(0)^2+20(0))) \\ &= \pi (\frac{30}{3}-0) \\ &= 10\pi \\ \end{aligned} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 4crm016gr8dyak906xah7jkuky40tt2 115023 115022 2026-04-27T00:11:07Z ~2026-25546-23 43043 /* integral pecahan parsial */ 115023 wikitext text/x-wiki == Kaidah umum == :# <math>\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ konstan)}\,\!</math> :# <math>\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx</math> :# <math>\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx</math> :# <math>\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx</math> :# <math>\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\! </math> :# <math>\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C </math> :# <math>\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C </math> :# <math>\int f(x) \, dx = F(x) + C</math> :# <math>\int_a^b f(x) \, dx = F(b) - F(a)</math> :# <math>\int f(x) \circ g(x) dx</math> = f’(g(x)) g’(x) :# <math>\int f^{-1}(x) dx = \frac{1}{f'(f'(x))}</math> == rumus sederhana == :<math>\int \, dx = x + C</math> :<math>\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int {dx \over x} = \ln{\left|x\right|} + C</math> <math>\int {dx \over \sqrt{a^2-b^2 x^2}} = \frac{1}{b} \arcsin \, {bx \over a} + C</math> :<math>\int {dx \over {a^2+b^2x^2}} = {1 \over ab}\arctan \, {bx \over a} + C</math> <math>\int {dx \over x\sqrt{b^2x^2-a^2}} = {1 \over a}\arcsec \, {bx \over a} + C</math> ; Eksponen dan logaritma :<math>\int e^x\,dx = e^x + C</math> :<math>\int a^x\,dx = \frac{a^x}{\ln{a}} + C</math> :<math>\int \ln \, x\,dx = x \cdot \, \ln {x} - x + C</math> :<math>\int \, ^b\!\log \, x\,dx = x \cdot \, ^b\!\log \, x - x \cdot \,^b\!\log e + C</math> ; Trigonometri :<math>\int \sin \, x \, dx = -\cos \, x + C</math> :<math>\int \cos \, x \, dx = \sin \, x + C</math> :<math>\int \tan \, x \, dx = -\ln{\left| \cos \, x \right|} + C</math> :<math>\int \cot \, x \, dx = \ln{\left| \sin \, x \right|} + C</math> :<math>\int \sec \, x \, dx = \ln{\left| \sec \, x + \tan \, x \right|} + C</math> :<math>\int \csc \, x \, dx = -\ln{\left| \csc \, x + \cot \, x \right|} + C</math> ; Hiperbolik :<math>\int \sinh \, x \, dx = \cosh \, x + C</math> :<math>\int \cosh \, x \, dx = \sinh \, x + C</math> :<math>\int \tanh \, x \, dx = \ln{\left| \cosh \, x \right|} + C</math> :<math>\int \coth \, x \, dx = \ln{\left| \sinh \, x \right|} + C</math> :<math>\int \mbox{sech} \, x \, dx = \arctan \,(\sinh \, x) + C</math> :<math>\int \mbox{csch} \, x \, dx = \ln{\left| \tanh \, {x \over 2} \right|} + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : 1. <math>\int x^3 - cos 3x + e^{5x+2} - \frac{1}{2x+5}\,dx</math> : <math>= \frac{1}{3+1}x^{3+1} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : <math>= \frac{x^4}{4} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : 2. <math>\int x(x-5)^4\,dx</math> : <math>= \int x(x^4-4x^3(5)+6x^2(25)-4x(125)+625)\,dx</math> : <math>= \int x(x^4-20x^3+150x^2-500x+625)\,dx</math> : <math>= \int x^5-20x^4+150x^3-500x^2+625x\,dx</math> : <math>= \frac{x^6}{6}-4x^5+\frac{150x^4}{4}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> : <math>= \frac{x^6}{6}-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : 1. <math>\int \frac{\ln \, x}{x}\,dx</math> : <math>u = \ln \, x\, du = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{\ln(x)}{x} dx \\ \int u du \\ \frac{1}{2} u^2 + C \\ \frac{1}{2} ln^2 x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x-5,\, du = dx,\, x = u+5</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4\,dx \\ \int (u+5)u^4\,du \\ \int (u^5+5u^4)\,du \\ \frac{1}{6} u^6+u^5 + C \\ \frac{1}{6} (x-5)^6+(x-5)^5 + C \\ \frac{1}{6} (x^6-6x^5(5)+15x^4(25)-20x^3(125)+15x^2(625)-6x(3125)+15625)+(x^5-5x^4(5)+10x^3(25)-10x^2(125)+5x(625)+3125) + C \\ \frac{1}{6}x^6-5x^5+\frac{125x^4}{2}-\frac{1250x^3}{3}+\frac{3125x^2}{2}-3125x+\frac{15625}{6}+x^5-25x^4+250x^3-1250x^2+3125x+3125 + C \\ \frac{1}{6}x^6-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C \\ \end{aligned} </math> </div></div> === integral parsial === ;Cara 1: Rumus Integral parsial diselesaikan dengan rumus berikut. : <math>\int u\,dv = uv - \int v\,du </math> Berikut contoh penyelesaian cara parsial dengan rumus. : 1. <math>\int x \sin \, x \, dx</math> : <math>u = x, du = 1 \, dx, dv = \sin \, x \, dx, v = -\cos \, x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \,dx \\ (x)(-\cos \, x) - \int (-\cos \, x)(1 dx) \\ -x \cos \, x + \int \cos \, x dx \\ -x \cos \, x + \sin \, x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x,\, du = 1\,dx,\, dv = (x-5)^4\,dx,\, v = \frac{1}{5}(x-5)^5</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - \int (\frac{1}{5}(x-5)^5)(1\,dx) \\ \frac{x}{5}(x-5)^5 - \int \frac{1}{5}(x-5)^5\,dx \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6\,dx \\ \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x\,dx</math> : <math>u = \ln \,x,\, du = \frac{1}{x}\,dx,\, dv = x\,dx,\, v = \frac{x^2}{2}</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x\, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{x^2}{2})(\frac{1}{x}\,dx) \\ \frac{x^2}{2} \ln \,x - \int \frac{x^2}{2}(\frac{1}{x})\,dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \,dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x\,dx</math> ini terjadi dua kali integral parsial : <math>u = \sin \,x,\, du = \cos \,x\,dx,\, dv = e^x\,dx,\, v = e^x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x\, dx \\ (\sin \,x)(e^x) - \int (e^x)(\cos \,x\,dx) \\ e^x \sin \,x - \int e^x \cos \,x\,dx \\ u = \cos \,x,\, du = -\sin \,x\,dx,\, dv = e^x\,dx,\, v = e^x \\ (\cos \,x)(e^x) - \int (e^x)(-\sin \,x\,dx) \\ e^x \cos \,x + \int e^x \sin \,x\,dx \\ e^x \sin \,x - (e^x \cos \,x + \int e^x \sin \,x\,dx) \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x\,dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> ;Cara 2: Tabel Untuk <math display="inline">\int u\,dv</math>, berlaku ketentuan sebagai berikut. {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>u</math> || <math>dv</math> |- | - || <math>\frac{du}{dx}</math> || <math>v</math> |- | + || <math>\frac{d^2u}{dx^2}</math> || <math>\int v\,dx</math> |} Berikut contoh penyelesaian cara parsial dengan tabel. : 1. <math>\int x \sin \, x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>\sin \, x</math> |- | - || <math>1</math> || <math>-\cos \, x</math> |- | + || <math>0</math> || <math>-\sin \, x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \, dx \\ (x)(-\cos \, x) - (1)(-\sin \, x) + C \\ -x \cos \, x + \sin \, x + C \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4 \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>(x-5)^4</math> |- | - || <math>1</math> || <math>\frac{1}{5}(x-5)^5</math> |- | + || <math>0</math> || <math>\frac{1}{30}(x-5)^6</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - (1)(\frac{1}{30}(x-5)^6) + C \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6 + C \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\ln \,x</math> || <math>x</math> |- | - || <math>\frac{1}{x}</math> || <math>\frac{x^2}{2}</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x \, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{1}{x})(\frac{x^2}{2}) \, dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \, dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x \, dx</math> ini terjadi dua kali integral parsial {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\sin \,x</math> || <math>e^x</math> |- | - || <math>\cos \, x</math> || <math>e^x</math> |- | + || <math>-\sin \,x</math> || <math>e^x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x \, dx \\ (\sin \,x)(e^x) - (\cos \,x)(e^x) + \int (-\sin \,x)(e^x) \, dx \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x) \, dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{1}{x^2 - 4}\,dx</math> Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{1}{x^2 - 4} \\ =&\; \frac{1}{(x + 2)(x - 2)} \\ =&\; \frac{A}{x + 2} + \frac{B}{x - 2} \\ =&\; \frac{A(x - 2) + B(x + 2)}{x^2 - 4} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> : <math>\int \frac{x^2+5x+4}{(x-3)^3}\,dx</math> : cara 1 Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^3} \\ : \text{untuk mencari C yaitu } (3)^2+5(3)+4=28 \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x - 3)^2 + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x^2 - 6x + 9) + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 - 6Ax + 9A + Bx - 3B}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 + (- 6A + B)x +(9 - 3B)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ \end{aligned} </math> Kita tahu bahwa <math display="inline">A = 1</math> dan <math display="inline">-6A + B = 5</math> dapat diselesaikan, yaitu <math display="inline">A = 1</math> dan {{nowrap|<math display="inline">B = 11</math>.}} : <math> \begin{aligned} &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int (\frac{1}{x - 3} + \frac{11}{(x - 3)^2} + \frac{28}{(x - 3)^3})\,dx \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : cara 2 Pertama, ubah u=x-3 menjadi x=u+3 dan du=dx tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{(u+3)^2+5(u+3)+4}{u^3} \\ =&\; \frac{u^2+11u+28}{u^3} \\ =&\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int \frac{u^2+11u+28}{u^3}\,du \\ =&\; \int (\frac{u^2}{u^3} + \frac{11u}{u^3} + \frac{28}{u^3})\,du \\ =&\; \int (\frac{1}{u} + \frac{11}{u^2} + \frac{28}{u^3})\,du \\ =&\; \ln u + (-11 u^{-1}) + (-14 u^{-2}) + C \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> === integral substitusi trigonometri === {| class="wikitable" |- ! Bentuk !! Substitusi Trigonometri |- | <math>\sqrt{a^2 - b^2 x^2}</math> || <math>x = \frac{a}{b} sin \, \alpha </math> |- | <math>\sqrt{a^2 + b^2 x^2}</math> || <math>x = \frac{a}{b} tan \, \alpha</math> |- | <math>\sqrt{b^2 x^2 - a^2}</math> || <math>x = \frac{a}{b} sec \, \alpha</math> |} Berikut contoh penyelesaian cara substitusi trigonometri. : <math>\int \frac{dx}{x^2 \sqrt{x^2 + 4}}</math> : <math>x = 2 \tan \, A, dx = 2 \sec^2 \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \int \frac{dx}{x^2\sqrt{x^2+4}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(2 \tan \, A)^2 \sqrt{4 + (2 \tan \, A)^2}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 + 4 \tan^2 \, A}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4(1 + \tan^2 \, A)}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 \sec^2 \, A }} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(4 \tan^2 \, A)(2 \sec \, A)} \\ =&\; \int \frac{\sec \, A \,dA}{4 \tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\sec \, A \,dA}{\tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A} \,dA \end{aligned} </math> Substitusi berikut dapat dibuat. : <math>\int \frac{\cos \, A}{\sin^2 \, A} \,dA</math> : <math>t = \sin \, A, dt = \cos \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A}\,dA \\ =&\; \frac{1}{4} \int \frac{dt}{t^2} \\ =&\; \frac{1}{4} \left(-\frac{1}{t}\right) + C \\ =&\; -\frac{1}{4 t} + C \\ =&\; -\frac{1}{4 \sin \, A} + C \end{aligned} </math> Ingat bahwa <math display="inline">\sin \, A = \frac{x}{\sqrt{x^2 + 4}}</math> berlaku. : <math> \begin{aligned} =&\; -\frac{1}{4 \sin \, A} + C \\ =&\; -\frac{\sqrt{x^2 + 4}}{4x} + C \end{aligned} </math> === integral mutlak === : <math>\int |f(x)| \,dA</math> buatlah <math>f(x) \ge 0</math> jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x) : <math>\int_a^c |f(x)| \, dx = \int_a^b f(x) \, dx + \int_b^c -f(x) \, dx = F(b) - F(a) - F(c) + F(b)</math> jika <math>|f(x)| = \begin{cases} f(x), & \mbox{jika } f(x) \ge b, \\ -f(x), & \mbox{jika } f(x) < b. \end{cases}</math>. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif. === integral fungsi ganjil dan integral fungsi genap === : <math>\int_{-a}^a f(x) \,dx</math> dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral: : integral fungsi ganjil : <math>\int_{-a}^a f(x) \,dx</math> = 0 : integral fungsi genap : <math>\int_{-a}^a f(x) \,dx</math> = <math>2 \int_{0}^a f(x) \,dx</math> === integral notasi === : <math>\int_a^b f(x) \,dx = \int_{a-p}^{b-p} f(x) \,dx = \int_{a+p}^{b+p} f(x) \,dx</math> : <math>\int_a^c f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx</math> : <math>\int_a^b f(x \pm k) \,dx = \int_{a \pm k}^{b \pm k} f(x) \,dx</math> === integral terbalik === : <math>\int_a^b f(x) \,dx = -\int_b^a f(x) \,dx</math> == Jenis integral lainnya == === panjang busur === ; Sumbu ''x'' : <math>S = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' : <math>S = \int_{y_1}^{y_2} \sqrt{1 + (f'(y))^2}\,dy</math> === luas daerah === ; Satu kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} f(x)\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} f(y)\,dy</math> ; Dua kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} (f(x_2) - f(x_1))\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} (f(y_2) - f(y_1))\,dy</math> : atau juga <math>L = \frac {D \sqrt{D}}{6 a^2}</math> === luas permukaan benda putar === ; Sumbu ''x'' sebagai poros : <math>L_p = 2 \pi \int_{x_1}^{x_2} f(x)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' sebagai poros : <math>L_p = 2 \pi \int_{y_1}^{y_2} f(y)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(y))^2}\,dy</math> === volume benda putar === ; Satu kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} (f(x))^2\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} (f(y))^2\,dy</math> ; Dua kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} ((f(x_2))^2 - (f(x_1))^2)\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} ((f(y_2))^2 - (f(y_1))^2)\,dy</math> : atau juga <math>V = |\frac {D^2 \sqrt{D}}{30 a^3}|</math> == Integral lipat == Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga. ;contoh # Tentukan hasil dari: * <math>\int (5x-1)(5x^2-2x+7)^4 dx</math> * <math>\int x^2 cos 3x dx</math> * <math>\int sin^2 x cos x dx</math> * <math>\int cos^2 x sin x dx</math> * <math>\int sec x dx</math> * <math>\int csc x dx</math> * <math>\int \frac{1}{2x^2-5x+3} dx</math> * <math>\int \frac{1}{\sqrt{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> * <math>\int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx</math> * <math>\int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx</math> * <math>\int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx</math> * <math>\int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx</math> * <math>\int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx \\ a=1, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-sin^2 x} \\ &= \frac{1}{cos^2 x} \\ &= sec^2 x \\ \int sec^2 x dx &= tan x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx \\ a=1, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-cos^2 x} \\ &= \frac{1}{sin^2 x} \\ &= csc^2 x \\ \int csc^2 x dx &= -cot x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx \\ a=sin x, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{sin x}{1-sin^2 x} \\ &= \frac{sin x}{cos^2 x} \\ &= sec x \cdot tan x \\ \int sec x \cdot tan x dx &= sec x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx \\ a=cos x, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{cos x}{1-cos^2 x} \\ &= \frac{cos x}{sin^2 x} \\ &= csc x \cdot cot x \\ \int csc x \cdot cot x dx &= -csc x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx \\ \text{misalkan } y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y &= \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y^2 &= \frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}} \\ &= \frac{x}{y} \\ y^3 &= x \\ 3y^2 dy &= dx \\ \int y \cdot 3y^2 dy \\ \int 3y^3 dy \\ \frac{3y^4}{4}+C \\ \end{aligned} </math> </div></div> # Tentukan hasil dari: * <math>\int_{1}^{5} 2x-3 dx</math> * <math>\int_{0}^{4} |x-2| dx</math> * <math>\int_{2}^{4} |3-x| dx</math> * <math>\int_{1}^{5} x^2-6x+8 dx</math> * <math>\int_{-3}^{5} |x^2-2x-8| dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} 2x-3 dx &= \left. x^2 - 3x \right|_{1}^{5} \\ &= 10 - (-2) \\ &= 12 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{0}^{4} |x-2| dx \\ \text{ tentukan nilai harga nol } \\ x-2 &= 0 \\ x &= 2 \\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti } \\ |x-2| = \begin{cases} x-2, & \mbox{jika } x \ge 2, \\ -(x-2), & \mbox{jika } x < 2. \end{cases} \\ \int_{0}^{4} |x-2| dx &= \int_{2}^{4} x-2 dx + \int_{0}^{2} -(x-2) dx \\ &= \left. (\frac{x^2}{2}-2x) \right|_{2}^{4} + \left. (\frac{-x^2}{2}+2x) \right|_{0}^{2} \\ &= 0 - (-2) + 2 - 0 \\ &= 4 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{2}^{4} |3-x| dx \\ \text{ tentukan nilai syarat-syarat } \\ 3-x &\ge 0 \\ \text{ harga nol } \\ x &= 3 \\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti } \\ |3-x| = \begin{cases} 3-x, & \mbox{jika } x \le 3, \\ -(3-x), & \mbox{jika } x > 3. \end{cases} \\ \int_{2}^{4} |3-x| dx &= \int_{2}^{3} 3-x dx + \int_{3}^{4} -(3-x) dx \\ &= \left. (3x-\frac{x^2}{2}) \right|_{2}^{3} + \left. (-3x+\frac{x^2}{2}) \right|_{3}^{4} \\ &= 4.5 - 4 - 4 - (-4.5) \\ &= 1 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} x^2-6x+8 dx &= \left. \frac{x^3}{3}-3x^2+8x \right|_{1}^{5} \\ &= \frac{20}{3} - \frac{16}{3} \\ &= \frac{4}{3} \\ &= 1\frac{1}{3} \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{-3}^{5} |x^2-2x-8| dx \\ \text{ tentukan nilai syarat-syarat } \\ x^2-2x-8 &\ge 0 \\ \text{ harga nol } \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x = 4 \\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti } \\ |x^2-2x-8| = \begin{cases} x^2-2x-8, & \mbox{jika } x \le -2, \\ -(x^2-2x-8), & \mbox{jika } -2 < x < 4, \\ x^2-2x-8, & \mbox{jika } x \ge 4 \end{cases} \\ \int_{-3}^{5} |x^2-2x-8| dx &= \int_{-3}^{-2} x^2-2x-8 dx + \int_{-2}^{4} -(x^2-2x-8) dx + \int_{4}^{5} x^2-2x-8 dx \\ &= \left. (\frac{x^3}{3}-x^2-8x) \right|_{-3}^{-2} + \left. (\frac{-x^3}{3}+x^2+8x) \right|_{-2}^{4} + \left. (\frac{x^3}{3}-x^2-8x) \right|_{4}^{5} \\ &= \frac{28}{3} - 6 + \frac{80}{3} - (-\frac{28}{3}) - \frac{70}{3} - (-\frac{80}{3}) \\ &= \frac{128}{3} \\ &= 42\frac{2}{3} \\ \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x\,dx \\ L &= \frac{1}{2} x^2 \\ L &= \frac{1}{2} xy \quad (y = x) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x^2</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x^2\,dx \\ L &= \frac{1}{3} x^3 \\ L &= \frac{1}{3} xy \quad (y = x^2) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = \sqrt{x}</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int \sqrt{x}\,dx \\ L &= \frac{2}{3} x^{\frac{3}{2}} \\ L &= \frac{2}{3} xy \quad (y = \sqrt{x}) \end{aligned} </math> </div></div> # Buktikan luas persegi <math>L = s^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = s \text{ dan titik (''s'', ''s''), } \\ L &= \int_{0}^{s} s\,dx \\ L &= sx |_{0}^{s} \\ L &= ss - 0 \\ L &= s^2 \end{aligned} </math> </div></div> # Buktikan luas persegi panjang <math>L = pl</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = l \text{dan titik (''p'', ''l''), } \\ L &= \int_{0}^{p} l\,dx \\ L &= lx |_{0}^{p} \\ L &= pl - 0 \\ L &= pl \end{aligned} </math> </div></div> * Buktikan luas segitiga <math>L = \frac{at}{2}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{-tx}{a} + t \text { dan titik (''a'', ''t''), } \\ L &= \int_{0}^{a} \left (\frac{-tx}{a} + t \right) \,dx \\ L &= \left. \frac{-tx^2}{2a} + tx \right|_{0}^{a} \\ L &= \frac{-ta^2}{2a} + ta - 0 + 0 \\ L &= \frac{-ta}{2} + ta \\ L &= \frac{at}{2} \end{aligned} </math> </div></div> # Buktikan volume tabung <math>V = \pi r^2t</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ Dengan posisi } y = r \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} r^2\,dx \\ V &= \pi \left. r^2x \right|_{0}^{t} \\ V &= \pi r^2t - 0 \\ V &= \pi r^2t \end{aligned} </math> </div></div> # Buktikan volume kerucut <math>V = \frac{\pi r^2t}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{rx}{t} \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} \left (\frac{rx}{t} \right)^2 \,dx \\ V &= \pi \left. \frac{r^2 x^3}{3t^2} \right|_{0}^{t} \\ V &= \pi \frac{r^2 t^3}{3t^2} - 0 \\ V &= \frac{\pi r^2t}{3} \end{aligned} </math> </div></div> # Buktikan volume bola <math>V = \frac{4 \pi r^3}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), } \\ V &= \pi \int_{-r}^{r} \left(\sqrt {r^2 - x^2}\right)^2\,dx \\ V &= \pi \int_{-r}^{r} r^2 - x^2 dx \\ V &= \pi \left. r^2x - \frac{x^3}{3} \right|_{-r}^{r} \\ V &= \pi \left (r^3 - \frac{r^3}{3} - \left (-r^3 + \frac{r^3}{3} \right) \right) \\ V &= \frac{4 \pi r^3}{3} \end{aligned} </math> </div></div> # Buktikan luas permukaan bola <math>L = 4 \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{selanjutnya } \\ ds &= \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ ds &= \sqrt{1 + \frac{x^2}{r^2 - x^2}}\,dx \\ ds &= \sqrt{\frac{r^2}{r^2 - x^2}}\,dx \\ ds &= \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ \text{sehingga } \\ L &= 2 \pi \int_{-r}^{r} \sqrt{r^2 - x^2} \cdot \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ L &= 2 \pi \int_{-r}^{r} r\,dx \\ L &= 2 \pi rx|_{-r}^{r} \\ L &= 2 \pi (r r - r (-r)) \\ L &= 2 \pi (r^2 + r^2) \\ L &= 4 \pi r^2 \end{aligned} </math> </div></div> # Buktikan keliling lingkaran <math>K = 2 \pi r</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0) } \\ \text{kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{sehingga } \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{x^2}{r^2 - x^2})}\,dx \\ K &= 4 \int_{0}^{r} \sqrt\frac{r^2}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \sqrt\frac{1}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \frac{1}{\sqrt{r^2 - x^2}}\,dx \\ K &= 4r \left. \arcsin \left (\frac{x}{r} \right) \right|_{0}^{r} \\ K &= 4r \left (\arcsin \left (\frac{r}{r}\right) - \arcsin \left (\frac{0}{r} \right) \right) \\ K &= 4r \left (\arcsin\left (1 \right) - \arcsin \left(0 \right) \right)) \\ K &= 4r \left (\frac{\pi}{2} \right) \\ K &= 2 \pi r \end{aligned} </math> </div></div> # Buktikan luas lingkaran <math>L = \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. } \\ \sin(\theta) &= \frac{x}{r} \\ x &= r \sin(\theta) \\ dx &= r \cos(\theta)\,d\theta \\ \text{dengan turunan di atas } \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - x^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - (r \sin(\theta))^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - r^2 \sin^2(\theta)}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 (1 - \sin^2(\theta))}\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta)\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta) (r \cos(\theta)\,d\theta) \\ L &= 4 \int_{0}^{r} r^2 \cos^2(\theta)\,d\theta \\ L &= 4r^2 \int_{0}^{r} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta \\ L &= 2r^2 \int_{0}^{r} (1 + \cos(2\theta))\,d\theta \\ L &= 2r^2 \left (\theta + \frac{1}{2} \sin(2\theta) \right)|_{0}^{r} \\ L &= 2r^2 (\theta + \sin(\theta) \cos(\theta))|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left (\frac{x}{r} \right) + \left (\frac{x}{r} \right) \left (\frac{r^2 - x^2}{r} \right) \right)|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left(\frac{r}{r} \right) + \left (\frac{r}{r} \right) \left (\frac {r^2 - r^2}{r} \right) \right) - \left(\arcsin \left (\frac{0}{r} \right) + \left (\frac{0}{r}\right) \left (\frac{r^2 - 0^2}{r} \right) \right) \\ L &= 2r^2 (\arcsin(1) + 0 - (\arcsin(0) + 0)) \\ L &= 2r^2 \left (\frac{\pi}{2} \right) \\ L &= \pi r^2 \end{aligned} </math> </div></div> # Buktikan luas elips <math>L = \pi ab</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \frac{b \sqrt{a^2 - x^2}}{a} \text{ serta (-''a'', 0) dan (''a'', 0), } \\ L &= 4 \int_{0}^{r} \frac{b \sqrt{a^2 - x^2}}{a}\,dx \\ L &= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx \\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, } \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{\frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx}{4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx} \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{b}{a} \\ L_\text{elips} &= \frac{b}{a} L_\text{ling} \\ L_\text{elips} &= \frac{b}{a} \pi a^2 \\ L_\text{elips} &= \pi ab \end{aligned} </math> </div></div> # Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ x^-2x &= 4x+7 \\ x^2-6x-7 &= 0 \\ (x+1)(x-7) &= 0 \\ x=-1 &\text{ atau } x=7 \\ y &= 4(-1)+7 = 3 \\ y &= 4(7)+7 = 35 \\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut } \\ L &= \int_{-1}^{7} 4x+7-(x^2-2x)\,dx \\ &= \int_{-1}^{7} -x^2+6x+7\,dx \\ &= -\frac{x^3}{3}+3x^2+7x|_{-1}^{7} \\ &= -\frac{7^3}{3}+3(7)^2+7(7)-(-\frac{(-1)^3}{3}+3(-1)^2+7(-1)) \\ &= \frac{245}{3}-(-\frac{13}{3}) \\ &= \frac{258}{3} \\ &= 86 \\ \end{aligned} </math> </div></div> # Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ 6-1,5x &= x-4 \\ 2,5x &= 10 \\ x &= 4 \\ y &= 4-4 = 0 \\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut } \\ \text{untuk x=0 } \\ y &= 6-1,5(0) = 6 \\ y &= 0-4 = -4 \\ L &= \pi \int_{0}^{4} ((6-1,5x)^2-(x-4)^2)\,dx \\ &= \pi \int_{0}^{4} (36-18x+2,25x^2-(x^2-8x+16))\,dx \\ &= \pi \int_{0}^{4} (1,25x^2-10x+20)\,dx \\ &= \pi (\frac{1,25x^3}{3}-5x^2+20x)|_{0}^{4} \\ &= \pi (\frac{1,25(4)^3}{3}-5(4)^2+20(4)-(\frac{1,25(0)^3}{3}-5(0)^2+20(0))) \\ &= \pi (\frac{30}{3}-0) \\ &= 10\pi \\ \end{aligned} </math> </div></div> [[Kategori:Soal-Soal Matematika]] si54p2t22ek3kl0wcpy3pnc17uern7p 115024 115023 2026-04-27T00:32:23Z ~2026-25546-23 43043 /* integral pecahan parsial */ 115024 wikitext text/x-wiki == Kaidah umum == :# <math>\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ konstan)}\,\!</math> :# <math>\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx</math> :# <math>\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx</math> :# <math>\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx</math> :# <math>\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\! </math> :# <math>\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C </math> :# <math>\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C </math> :# <math>\int f(x) \, dx = F(x) + C</math> :# <math>\int_a^b f(x) \, dx = F(b) - F(a)</math> :# <math>\int f(x) \circ g(x) dx</math> = f’(g(x)) g’(x) :# <math>\int f^{-1}(x) dx = \frac{1}{f'(f'(x))}</math> == rumus sederhana == :<math>\int \, dx = x + C</math> :<math>\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int {dx \over x} = \ln{\left|x\right|} + C</math> <math>\int {dx \over \sqrt{a^2-b^2 x^2}} = \frac{1}{b} \arcsin \, {bx \over a} + C</math> :<math>\int {dx \over {a^2+b^2x^2}} = {1 \over ab}\arctan \, {bx \over a} + C</math> <math>\int {dx \over x\sqrt{b^2x^2-a^2}} = {1 \over a}\arcsec \, {bx \over a} + C</math> ; Eksponen dan logaritma :<math>\int e^x\,dx = e^x + C</math> :<math>\int a^x\,dx = \frac{a^x}{\ln{a}} + C</math> :<math>\int \ln \, x\,dx = x \cdot \, \ln {x} - x + C</math> :<math>\int \, ^b\!\log \, x\,dx = x \cdot \, ^b\!\log \, x - x \cdot \,^b\!\log e + C</math> ; Trigonometri :<math>\int \sin \, x \, dx = -\cos \, x + C</math> :<math>\int \cos \, x \, dx = \sin \, x + C</math> :<math>\int \tan \, x \, dx = -\ln{\left| \cos \, x \right|} + C</math> :<math>\int \cot \, x \, dx = \ln{\left| \sin \, x \right|} + C</math> :<math>\int \sec \, x \, dx = \ln{\left| \sec \, x + \tan \, x \right|} + C</math> :<math>\int \csc \, x \, dx = -\ln{\left| \csc \, x + \cot \, x \right|} + C</math> ; Hiperbolik :<math>\int \sinh \, x \, dx = \cosh \, x + C</math> :<math>\int \cosh \, x \, dx = \sinh \, x + C</math> :<math>\int \tanh \, x \, dx = \ln{\left| \cosh \, x \right|} + C</math> :<math>\int \coth \, x \, dx = \ln{\left| \sinh \, x \right|} + C</math> :<math>\int \mbox{sech} \, x \, dx = \arctan \,(\sinh \, x) + C</math> :<math>\int \mbox{csch} \, x \, dx = \ln{\left| \tanh \, {x \over 2} \right|} + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : 1. <math>\int x^3 - cos 3x + e^{5x+2} - \frac{1}{2x+5}\,dx</math> : <math>= \frac{1}{3+1}x^{3+1} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : <math>= \frac{x^4}{4} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : 2. <math>\int x(x-5)^4\,dx</math> : <math>= \int x(x^4-4x^3(5)+6x^2(25)-4x(125)+625)\,dx</math> : <math>= \int x(x^4-20x^3+150x^2-500x+625)\,dx</math> : <math>= \int x^5-20x^4+150x^3-500x^2+625x\,dx</math> : <math>= \frac{x^6}{6}-4x^5+\frac{150x^4}{4}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> : <math>= \frac{x^6}{6}-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : 1. <math>\int \frac{\ln \, x}{x}\,dx</math> : <math>u = \ln \, x\, du = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{\ln(x)}{x} dx \\ \int u du \\ \frac{1}{2} u^2 + C \\ \frac{1}{2} ln^2 x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x-5,\, du = dx,\, x = u+5</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4\,dx \\ \int (u+5)u^4\,du \\ \int (u^5+5u^4)\,du \\ \frac{1}{6} u^6+u^5 + C \\ \frac{1}{6} (x-5)^6+(x-5)^5 + C \\ \frac{1}{6} (x^6-6x^5(5)+15x^4(25)-20x^3(125)+15x^2(625)-6x(3125)+15625)+(x^5-5x^4(5)+10x^3(25)-10x^2(125)+5x(625)+3125) + C \\ \frac{1}{6}x^6-5x^5+\frac{125x^4}{2}-\frac{1250x^3}{3}+\frac{3125x^2}{2}-3125x+\frac{15625}{6}+x^5-25x^4+250x^3-1250x^2+3125x+3125 + C \\ \frac{1}{6}x^6-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C \\ \end{aligned} </math> </div></div> === integral parsial === ;Cara 1: Rumus Integral parsial diselesaikan dengan rumus berikut. : <math>\int u\,dv = uv - \int v\,du </math> Berikut contoh penyelesaian cara parsial dengan rumus. : 1. <math>\int x \sin \, x \, dx</math> : <math>u = x, du = 1 \, dx, dv = \sin \, x \, dx, v = -\cos \, x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \,dx \\ (x)(-\cos \, x) - \int (-\cos \, x)(1 dx) \\ -x \cos \, x + \int \cos \, x dx \\ -x \cos \, x + \sin \, x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x,\, du = 1\,dx,\, dv = (x-5)^4\,dx,\, v = \frac{1}{5}(x-5)^5</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - \int (\frac{1}{5}(x-5)^5)(1\,dx) \\ \frac{x}{5}(x-5)^5 - \int \frac{1}{5}(x-5)^5\,dx \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6\,dx \\ \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x\,dx</math> : <math>u = \ln \,x,\, du = \frac{1}{x}\,dx,\, dv = x\,dx,\, v = \frac{x^2}{2}</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x\, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{x^2}{2})(\frac{1}{x}\,dx) \\ \frac{x^2}{2} \ln \,x - \int \frac{x^2}{2}(\frac{1}{x})\,dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \,dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x\,dx</math> ini terjadi dua kali integral parsial : <math>u = \sin \,x,\, du = \cos \,x\,dx,\, dv = e^x\,dx,\, v = e^x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x\, dx \\ (\sin \,x)(e^x) - \int (e^x)(\cos \,x\,dx) \\ e^x \sin \,x - \int e^x \cos \,x\,dx \\ u = \cos \,x,\, du = -\sin \,x\,dx,\, dv = e^x\,dx,\, v = e^x \\ (\cos \,x)(e^x) - \int (e^x)(-\sin \,x\,dx) \\ e^x \cos \,x + \int e^x \sin \,x\,dx \\ e^x \sin \,x - (e^x \cos \,x + \int e^x \sin \,x\,dx) \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x\,dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> ;Cara 2: Tabel Untuk <math display="inline">\int u\,dv</math>, berlaku ketentuan sebagai berikut. {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>u</math> || <math>dv</math> |- | - || <math>\frac{du}{dx}</math> || <math>v</math> |- | + || <math>\frac{d^2u}{dx^2}</math> || <math>\int v\,dx</math> |} Berikut contoh penyelesaian cara parsial dengan tabel. : 1. <math>\int x \sin \, x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>\sin \, x</math> |- | - || <math>1</math> || <math>-\cos \, x</math> |- | + || <math>0</math> || <math>-\sin \, x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \, dx \\ (x)(-\cos \, x) - (1)(-\sin \, x) + C \\ -x \cos \, x + \sin \, x + C \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4 \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>(x-5)^4</math> |- | - || <math>1</math> || <math>\frac{1}{5}(x-5)^5</math> |- | + || <math>0</math> || <math>\frac{1}{30}(x-5)^6</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - (1)(\frac{1}{30}(x-5)^6) + C \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6 + C \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\ln \,x</math> || <math>x</math> |- | - || <math>\frac{1}{x}</math> || <math>\frac{x^2}{2}</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x \, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{1}{x})(\frac{x^2}{2}) \, dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \, dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x \, dx</math> ini terjadi dua kali integral parsial {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\sin \,x</math> || <math>e^x</math> |- | - || <math>\cos \, x</math> || <math>e^x</math> |- | + || <math>-\sin \,x</math> || <math>e^x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x \, dx \\ (\sin \,x)(e^x) - (\cos \,x)(e^x) + \int (-\sin \,x)(e^x) \, dx \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x) \, dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{1}{x^2 - 4}\,dx</math> Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{1}{x^2 - 4} \\ =&\; \frac{1}{(x + 2)(x - 2)} \\ =&\; \frac{A}{x + 2} + \frac{B}{x - 2} \\ =&\; \frac{A(x - 2) + B(x + 2)}{x^2 - 4} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> : <math>\int \frac{x^2+5x+4}{(x-3)^3}\,dx</math> : cara 1 Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^3} \\ : \text{untuk mencari C yaitu } (3)^2+5(3)+4=28 \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x - 3)^2 + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x^2 - 6x + 9) + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 - 6Ax + 9A + Bx - 3B}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 + (- 6A + B)x +(9 - 3B)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ \end{aligned} </math> Kita tahu bahwa <math display="inline">A = 1</math> dan <math display="inline">-6A + B = 5</math> dapat diselesaikan, yaitu <math display="inline">A = 1</math> dan {{nowrap|<math display="inline">B = 11</math>.}} : <math> \begin{aligned} &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int (\frac{1}{x - 3} + \frac{11}{(x - 3)^2} + \frac{28}{(x - 3)^3})\,dx \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : cara 2 Pertama, ubah u=x-3 menjadi x=u+3 dan du=dx tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{(u+3)^2+5(u+3)+4}{u^3} \\ =&\; \frac{u^2+11u+28}{u^3} \\ &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int \frac{u^2+11u+28}{u^3}\,du \\ =&\; \int (\frac{u^2}{u^3} + \frac{11u}{u^3} + \frac{28}{u^3})\,du \\ =&\; \int (\frac{1}{u} + \frac{11}{u^2} + \frac{28}{u^3})\,du \\ =&\; \ln u + (-11 u^{-1}) + (-14 u^{-2}) + C \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> === integral substitusi trigonometri === {| class="wikitable" |- ! Bentuk !! Substitusi Trigonometri |- | <math>\sqrt{a^2 - b^2 x^2}</math> || <math>x = \frac{a}{b} sin \, \alpha </math> |- | <math>\sqrt{a^2 + b^2 x^2}</math> || <math>x = \frac{a}{b} tan \, \alpha</math> |- | <math>\sqrt{b^2 x^2 - a^2}</math> || <math>x = \frac{a}{b} sec \, \alpha</math> |} Berikut contoh penyelesaian cara substitusi trigonometri. : <math>\int \frac{dx}{x^2 \sqrt{x^2 + 4}}</math> : <math>x = 2 \tan \, A, dx = 2 \sec^2 \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \int \frac{dx}{x^2\sqrt{x^2+4}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(2 \tan \, A)^2 \sqrt{4 + (2 \tan \, A)^2}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 + 4 \tan^2 \, A}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4(1 + \tan^2 \, A)}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 \sec^2 \, A }} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(4 \tan^2 \, A)(2 \sec \, A)} \\ =&\; \int \frac{\sec \, A \,dA}{4 \tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\sec \, A \,dA}{\tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A} \,dA \end{aligned} </math> Substitusi berikut dapat dibuat. : <math>\int \frac{\cos \, A}{\sin^2 \, A} \,dA</math> : <math>t = \sin \, A, dt = \cos \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A}\,dA \\ =&\; \frac{1}{4} \int \frac{dt}{t^2} \\ =&\; \frac{1}{4} \left(-\frac{1}{t}\right) + C \\ =&\; -\frac{1}{4 t} + C \\ =&\; -\frac{1}{4 \sin \, A} + C \end{aligned} </math> Ingat bahwa <math display="inline">\sin \, A = \frac{x}{\sqrt{x^2 + 4}}</math> berlaku. : <math> \begin{aligned} =&\; -\frac{1}{4 \sin \, A} + C \\ =&\; -\frac{\sqrt{x^2 + 4}}{4x} + C \end{aligned} </math> === integral mutlak === : <math>\int |f(x)| \,dA</math> buatlah <math>f(x) \ge 0</math> jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x) : <math>\int_a^c |f(x)| \, dx = \int_a^b f(x) \, dx + \int_b^c -f(x) \, dx = F(b) - F(a) - F(c) + F(b)</math> jika <math>|f(x)| = \begin{cases} f(x), & \mbox{jika } f(x) \ge b, \\ -f(x), & \mbox{jika } f(x) < b. \end{cases}</math>. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif. === integral fungsi ganjil dan integral fungsi genap === : <math>\int_{-a}^a f(x) \,dx</math> dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral: : integral fungsi ganjil : <math>\int_{-a}^a f(x) \,dx</math> = 0 : integral fungsi genap : <math>\int_{-a}^a f(x) \,dx</math> = <math>2 \int_{0}^a f(x) \,dx</math> === integral notasi === : <math>\int_a^b f(x) \,dx = \int_{a-p}^{b-p} f(x) \,dx = \int_{a+p}^{b+p} f(x) \,dx</math> : <math>\int_a^c f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx</math> : <math>\int_a^b f(x \pm k) \,dx = \int_{a \pm k}^{b \pm k} f(x) \,dx</math> === integral terbalik === : <math>\int_a^b f(x) \,dx = -\int_b^a f(x) \,dx</math> == Jenis integral lainnya == === panjang busur === ; Sumbu ''x'' : <math>S = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' : <math>S = \int_{y_1}^{y_2} \sqrt{1 + (f'(y))^2}\,dy</math> === luas daerah === ; Satu kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} f(x)\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} f(y)\,dy</math> ; Dua kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} (f(x_2) - f(x_1))\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} (f(y_2) - f(y_1))\,dy</math> : atau juga <math>L = \frac {D \sqrt{D}}{6 a^2}</math> === luas permukaan benda putar === ; Sumbu ''x'' sebagai poros : <math>L_p = 2 \pi \int_{x_1}^{x_2} f(x)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' sebagai poros : <math>L_p = 2 \pi \int_{y_1}^{y_2} f(y)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(y))^2}\,dy</math> === volume benda putar === ; Satu kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} (f(x))^2\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} (f(y))^2\,dy</math> ; Dua kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} ((f(x_2))^2 - (f(x_1))^2)\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} ((f(y_2))^2 - (f(y_1))^2)\,dy</math> : atau juga <math>V = |\frac {D^2 \sqrt{D}}{30 a^3}|</math> == Integral lipat == Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga. ;contoh # Tentukan hasil dari: * <math>\int (5x-1)(5x^2-2x+7)^4 dx</math> * <math>\int x^2 cos 3x dx</math> * <math>\int sin^2 x cos x dx</math> * <math>\int cos^2 x sin x dx</math> * <math>\int sec x dx</math> * <math>\int csc x dx</math> * <math>\int \frac{1}{2x^2-5x+3} dx</math> * <math>\int \frac{1}{\sqrt{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> * <math>\int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx</math> * <math>\int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx</math> * <math>\int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx</math> * <math>\int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx</math> * <math>\int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx \\ a=1, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-sin^2 x} \\ &= \frac{1}{cos^2 x} \\ &= sec^2 x \\ \int sec^2 x dx &= tan x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx \\ a=1, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-cos^2 x} \\ &= \frac{1}{sin^2 x} \\ &= csc^2 x \\ \int csc^2 x dx &= -cot x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx \\ a=sin x, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{sin x}{1-sin^2 x} \\ &= \frac{sin x}{cos^2 x} \\ &= sec x \cdot tan x \\ \int sec x \cdot tan x dx &= sec x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx \\ a=cos x, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{cos x}{1-cos^2 x} \\ &= \frac{cos x}{sin^2 x} \\ &= csc x \cdot cot x \\ \int csc x \cdot cot x dx &= -csc x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx \\ \text{misalkan } y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y &= \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y^2 &= \frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}} \\ &= \frac{x}{y} \\ y^3 &= x \\ 3y^2 dy &= dx \\ \int y \cdot 3y^2 dy \\ \int 3y^3 dy \\ \frac{3y^4}{4}+C \\ \end{aligned} </math> </div></div> # Tentukan hasil dari: * <math>\int_{1}^{5} 2x-3 dx</math> * <math>\int_{0}^{4} |x-2| dx</math> * <math>\int_{2}^{4} |3-x| dx</math> * <math>\int_{1}^{5} x^2-6x+8 dx</math> * <math>\int_{-3}^{5} |x^2-2x-8| dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} 2x-3 dx &= \left. x^2 - 3x \right|_{1}^{5} \\ &= 10 - (-2) \\ &= 12 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{0}^{4} |x-2| dx \\ \text{ tentukan nilai harga nol } \\ x-2 &= 0 \\ x &= 2 \\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti } \\ |x-2| = \begin{cases} x-2, & \mbox{jika } x \ge 2, \\ -(x-2), & \mbox{jika } x < 2. \end{cases} \\ \int_{0}^{4} |x-2| dx &= \int_{2}^{4} x-2 dx + \int_{0}^{2} -(x-2) dx \\ &= \left. (\frac{x^2}{2}-2x) \right|_{2}^{4} + \left. (\frac{-x^2}{2}+2x) \right|_{0}^{2} \\ &= 0 - (-2) + 2 - 0 \\ &= 4 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{2}^{4} |3-x| dx \\ \text{ tentukan nilai syarat-syarat } \\ 3-x &\ge 0 \\ \text{ harga nol } \\ x &= 3 \\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti } \\ |3-x| = \begin{cases} 3-x, & \mbox{jika } x \le 3, \\ -(3-x), & \mbox{jika } x > 3. \end{cases} \\ \int_{2}^{4} |3-x| dx &= \int_{2}^{3} 3-x dx + \int_{3}^{4} -(3-x) dx \\ &= \left. (3x-\frac{x^2}{2}) \right|_{2}^{3} + \left. (-3x+\frac{x^2}{2}) \right|_{3}^{4} \\ &= 4.5 - 4 - 4 - (-4.5) \\ &= 1 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} x^2-6x+8 dx &= \left. \frac{x^3}{3}-3x^2+8x \right|_{1}^{5} \\ &= \frac{20}{3} - \frac{16}{3} \\ &= \frac{4}{3} \\ &= 1\frac{1}{3} \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{-3}^{5} |x^2-2x-8| dx \\ \text{ tentukan nilai syarat-syarat } \\ x^2-2x-8 &\ge 0 \\ \text{ harga nol } \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x = 4 \\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti } \\ |x^2-2x-8| = \begin{cases} x^2-2x-8, & \mbox{jika } x \le -2, \\ -(x^2-2x-8), & \mbox{jika } -2 < x < 4, \\ x^2-2x-8, & \mbox{jika } x \ge 4 \end{cases} \\ \int_{-3}^{5} |x^2-2x-8| dx &= \int_{-3}^{-2} x^2-2x-8 dx + \int_{-2}^{4} -(x^2-2x-8) dx + \int_{4}^{5} x^2-2x-8 dx \\ &= \left. (\frac{x^3}{3}-x^2-8x) \right|_{-3}^{-2} + \left. (\frac{-x^3}{3}+x^2+8x) \right|_{-2}^{4} + \left. (\frac{x^3}{3}-x^2-8x) \right|_{4}^{5} \\ &= \frac{28}{3} - 6 + \frac{80}{3} - (-\frac{28}{3}) - \frac{70}{3} - (-\frac{80}{3}) \\ &= \frac{128}{3} \\ &= 42\frac{2}{3} \\ \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x\,dx \\ L &= \frac{1}{2} x^2 \\ L &= \frac{1}{2} xy \quad (y = x) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x^2</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x^2\,dx \\ L &= \frac{1}{3} x^3 \\ L &= \frac{1}{3} xy \quad (y = x^2) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = \sqrt{x}</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int \sqrt{x}\,dx \\ L &= \frac{2}{3} x^{\frac{3}{2}} \\ L &= \frac{2}{3} xy \quad (y = \sqrt{x}) \end{aligned} </math> </div></div> # Buktikan luas persegi <math>L = s^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = s \text{ dan titik (''s'', ''s''), } \\ L &= \int_{0}^{s} s\,dx \\ L &= sx |_{0}^{s} \\ L &= ss - 0 \\ L &= s^2 \end{aligned} </math> </div></div> # Buktikan luas persegi panjang <math>L = pl</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = l \text{dan titik (''p'', ''l''), } \\ L &= \int_{0}^{p} l\,dx \\ L &= lx |_{0}^{p} \\ L &= pl - 0 \\ L &= pl \end{aligned} </math> </div></div> * Buktikan luas segitiga <math>L = \frac{at}{2}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{-tx}{a} + t \text { dan titik (''a'', ''t''), } \\ L &= \int_{0}^{a} \left (\frac{-tx}{a} + t \right) \,dx \\ L &= \left. \frac{-tx^2}{2a} + tx \right|_{0}^{a} \\ L &= \frac{-ta^2}{2a} + ta - 0 + 0 \\ L &= \frac{-ta}{2} + ta \\ L &= \frac{at}{2} \end{aligned} </math> </div></div> # Buktikan volume tabung <math>V = \pi r^2t</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ Dengan posisi } y = r \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} r^2\,dx \\ V &= \pi \left. r^2x \right|_{0}^{t} \\ V &= \pi r^2t - 0 \\ V &= \pi r^2t \end{aligned} </math> </div></div> # Buktikan volume kerucut <math>V = \frac{\pi r^2t}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{rx}{t} \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} \left (\frac{rx}{t} \right)^2 \,dx \\ V &= \pi \left. \frac{r^2 x^3}{3t^2} \right|_{0}^{t} \\ V &= \pi \frac{r^2 t^3}{3t^2} - 0 \\ V &= \frac{\pi r^2t}{3} \end{aligned} </math> </div></div> # Buktikan volume bola <math>V = \frac{4 \pi r^3}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), } \\ V &= \pi \int_{-r}^{r} \left(\sqrt {r^2 - x^2}\right)^2\,dx \\ V &= \pi \int_{-r}^{r} r^2 - x^2 dx \\ V &= \pi \left. r^2x - \frac{x^3}{3} \right|_{-r}^{r} \\ V &= \pi \left (r^3 - \frac{r^3}{3} - \left (-r^3 + \frac{r^3}{3} \right) \right) \\ V &= \frac{4 \pi r^3}{3} \end{aligned} </math> </div></div> # Buktikan luas permukaan bola <math>L = 4 \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{selanjutnya } \\ ds &= \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ ds &= \sqrt{1 + \frac{x^2}{r^2 - x^2}}\,dx \\ ds &= \sqrt{\frac{r^2}{r^2 - x^2}}\,dx \\ ds &= \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ \text{sehingga } \\ L &= 2 \pi \int_{-r}^{r} \sqrt{r^2 - x^2} \cdot \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ L &= 2 \pi \int_{-r}^{r} r\,dx \\ L &= 2 \pi rx|_{-r}^{r} \\ L &= 2 \pi (r r - r (-r)) \\ L &= 2 \pi (r^2 + r^2) \\ L &= 4 \pi r^2 \end{aligned} </math> </div></div> # Buktikan keliling lingkaran <math>K = 2 \pi r</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0) } \\ \text{kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{sehingga } \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{x^2}{r^2 - x^2})}\,dx \\ K &= 4 \int_{0}^{r} \sqrt\frac{r^2}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \sqrt\frac{1}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \frac{1}{\sqrt{r^2 - x^2}}\,dx \\ K &= 4r \left. \arcsin \left (\frac{x}{r} \right) \right|_{0}^{r} \\ K &= 4r \left (\arcsin \left (\frac{r}{r}\right) - \arcsin \left (\frac{0}{r} \right) \right) \\ K &= 4r \left (\arcsin\left (1 \right) - \arcsin \left(0 \right) \right)) \\ K &= 4r \left (\frac{\pi}{2} \right) \\ K &= 2 \pi r \end{aligned} </math> </div></div> # Buktikan luas lingkaran <math>L = \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. } \\ \sin(\theta) &= \frac{x}{r} \\ x &= r \sin(\theta) \\ dx &= r \cos(\theta)\,d\theta \\ \text{dengan turunan di atas } \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - x^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - (r \sin(\theta))^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - r^2 \sin^2(\theta)}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 (1 - \sin^2(\theta))}\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta)\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta) (r \cos(\theta)\,d\theta) \\ L &= 4 \int_{0}^{r} r^2 \cos^2(\theta)\,d\theta \\ L &= 4r^2 \int_{0}^{r} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta \\ L &= 2r^2 \int_{0}^{r} (1 + \cos(2\theta))\,d\theta \\ L &= 2r^2 \left (\theta + \frac{1}{2} \sin(2\theta) \right)|_{0}^{r} \\ L &= 2r^2 (\theta + \sin(\theta) \cos(\theta))|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left (\frac{x}{r} \right) + \left (\frac{x}{r} \right) \left (\frac{r^2 - x^2}{r} \right) \right)|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left(\frac{r}{r} \right) + \left (\frac{r}{r} \right) \left (\frac {r^2 - r^2}{r} \right) \right) - \left(\arcsin \left (\frac{0}{r} \right) + \left (\frac{0}{r}\right) \left (\frac{r^2 - 0^2}{r} \right) \right) \\ L &= 2r^2 (\arcsin(1) + 0 - (\arcsin(0) + 0)) \\ L &= 2r^2 \left (\frac{\pi}{2} \right) \\ L &= \pi r^2 \end{aligned} </math> </div></div> # Buktikan luas elips <math>L = \pi ab</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \frac{b \sqrt{a^2 - x^2}}{a} \text{ serta (-''a'', 0) dan (''a'', 0), } \\ L &= 4 \int_{0}^{r} \frac{b \sqrt{a^2 - x^2}}{a}\,dx \\ L &= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx \\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, } \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{\frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx}{4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx} \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{b}{a} \\ L_\text{elips} &= \frac{b}{a} L_\text{ling} \\ L_\text{elips} &= \frac{b}{a} \pi a^2 \\ L_\text{elips} &= \pi ab \end{aligned} </math> </div></div> # Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ x^-2x &= 4x+7 \\ x^2-6x-7 &= 0 \\ (x+1)(x-7) &= 0 \\ x=-1 &\text{ atau } x=7 \\ y &= 4(-1)+7 = 3 \\ y &= 4(7)+7 = 35 \\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut } \\ L &= \int_{-1}^{7} 4x+7-(x^2-2x)\,dx \\ &= \int_{-1}^{7} -x^2+6x+7\,dx \\ &= -\frac{x^3}{3}+3x^2+7x|_{-1}^{7} \\ &= -\frac{7^3}{3}+3(7)^2+7(7)-(-\frac{(-1)^3}{3}+3(-1)^2+7(-1)) \\ &= \frac{245}{3}-(-\frac{13}{3}) \\ &= \frac{258}{3} \\ &= 86 \\ \end{aligned} </math> </div></div> # Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ 6-1,5x &= x-4 \\ 2,5x &= 10 \\ x &= 4 \\ y &= 4-4 = 0 \\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut } \\ \text{untuk x=0 } \\ y &= 6-1,5(0) = 6 \\ y &= 0-4 = -4 \\ L &= \pi \int_{0}^{4} ((6-1,5x)^2-(x-4)^2)\,dx \\ &= \pi \int_{0}^{4} (36-18x+2,25x^2-(x^2-8x+16))\,dx \\ &= \pi \int_{0}^{4} (1,25x^2-10x+20)\,dx \\ &= \pi (\frac{1,25x^3}{3}-5x^2+20x)|_{0}^{4} \\ &= \pi (\frac{1,25(4)^3}{3}-5(4)^2+20(4)-(\frac{1,25(0)^3}{3}-5(0)^2+20(0))) \\ &= \pi (\frac{30}{3}-0) \\ &= 10\pi \\ \end{aligned} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 43fyk8jhx9ds74cdcm4xduq6cg74ple 115025 115024 2026-04-27T01:49:24Z ~2026-25546-23 43043 /* integral pecahan parsial */ 115025 wikitext text/x-wiki == Kaidah umum == :# <math>\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ konstan)}\,\!</math> :# <math>\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx</math> :# <math>\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx</math> :# <math>\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx</math> :# <math>\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\! </math> :# <math>\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C </math> :# <math>\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C </math> :# <math>\int f(x) \, dx = F(x) + C</math> :# <math>\int_a^b f(x) \, dx = F(b) - F(a)</math> :# <math>\int f(x) \circ g(x) dx</math> = f’(g(x)) g’(x) :# <math>\int f^{-1}(x) dx = \frac{1}{f'(f'(x))}</math> == rumus sederhana == :<math>\int \, dx = x + C</math> :<math>\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int {dx \over x} = \ln{\left|x\right|} + C</math> <math>\int {dx \over \sqrt{a^2-b^2 x^2}} = \frac{1}{b} \arcsin \, {bx \over a} + C</math> :<math>\int {dx \over {a^2+b^2x^2}} = {1 \over ab}\arctan \, {bx \over a} + C</math> <math>\int {dx \over x\sqrt{b^2x^2-a^2}} = {1 \over a}\arcsec \, {bx \over a} + C</math> ; Eksponen dan logaritma :<math>\int e^x\,dx = e^x + C</math> :<math>\int a^x\,dx = \frac{a^x}{\ln{a}} + C</math> :<math>\int \ln \, x\,dx = x \cdot \, \ln {x} - x + C</math> :<math>\int \, ^b\!\log \, x\,dx = x \cdot \, ^b\!\log \, x - x \cdot \,^b\!\log e + C</math> ; Trigonometri :<math>\int \sin \, x \, dx = -\cos \, x + C</math> :<math>\int \cos \, x \, dx = \sin \, x + C</math> :<math>\int \tan \, x \, dx = -\ln{\left| \cos \, x \right|} + C</math> :<math>\int \cot \, x \, dx = \ln{\left| \sin \, x \right|} + C</math> :<math>\int \sec \, x \, dx = \ln{\left| \sec \, x + \tan \, x \right|} + C</math> :<math>\int \csc \, x \, dx = -\ln{\left| \csc \, x + \cot \, x \right|} + C</math> ; Hiperbolik :<math>\int \sinh \, x \, dx = \cosh \, x + C</math> :<math>\int \cosh \, x \, dx = \sinh \, x + C</math> :<math>\int \tanh \, x \, dx = \ln{\left| \cosh \, x \right|} + C</math> :<math>\int \coth \, x \, dx = \ln{\left| \sinh \, x \right|} + C</math> :<math>\int \mbox{sech} \, x \, dx = \arctan \,(\sinh \, x) + C</math> :<math>\int \mbox{csch} \, x \, dx = \ln{\left| \tanh \, {x \over 2} \right|} + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : 1. <math>\int x^3 - cos 3x + e^{5x+2} - \frac{1}{2x+5}\,dx</math> : <math>= \frac{1}{3+1}x^{3+1} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : <math>= \frac{x^4}{4} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : 2. <math>\int x(x-5)^4\,dx</math> : <math>= \int x(x^4-4x^3(5)+6x^2(25)-4x(125)+625)\,dx</math> : <math>= \int x(x^4-20x^3+150x^2-500x+625)\,dx</math> : <math>= \int x^5-20x^4+150x^3-500x^2+625x\,dx</math> : <math>= \frac{x^6}{6}-4x^5+\frac{150x^4}{4}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> : <math>= \frac{x^6}{6}-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : 1. <math>\int \frac{\ln \, x}{x}\,dx</math> : <math>u = \ln \, x\, du = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{\ln(x)}{x} dx \\ \int u du \\ \frac{1}{2} u^2 + C \\ \frac{1}{2} ln^2 x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x-5,\, du = dx,\, x = u+5</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4\,dx \\ \int (u+5)u^4\,du \\ \int (u^5+5u^4)\,du \\ \frac{1}{6} u^6+u^5 + C \\ \frac{1}{6} (x-5)^6+(x-5)^5 + C \\ \frac{1}{6} (x^6-6x^5(5)+15x^4(25)-20x^3(125)+15x^2(625)-6x(3125)+15625)+(x^5-5x^4(5)+10x^3(25)-10x^2(125)+5x(625)+3125) + C \\ \frac{1}{6}x^6-5x^5+\frac{125x^4}{2}-\frac{1250x^3}{3}+\frac{3125x^2}{2}-3125x+\frac{15625}{6}+x^5-25x^4+250x^3-1250x^2+3125x+3125 + C \\ \frac{1}{6}x^6-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C \\ \end{aligned} </math> </div></div> === integral parsial === ;Cara 1: Rumus Integral parsial diselesaikan dengan rumus berikut. : <math>\int u\,dv = uv - \int v\,du </math> Berikut contoh penyelesaian cara parsial dengan rumus. : 1. <math>\int x \sin \, x \, dx</math> : <math>u = x, du = 1 \, dx, dv = \sin \, x \, dx, v = -\cos \, x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \,dx \\ (x)(-\cos \, x) - \int (-\cos \, x)(1 dx) \\ -x \cos \, x + \int \cos \, x dx \\ -x \cos \, x + \sin \, x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x,\, du = 1\,dx,\, dv = (x-5)^4\,dx,\, v = \frac{1}{5}(x-5)^5</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - \int (\frac{1}{5}(x-5)^5)(1\,dx) \\ \frac{x}{5}(x-5)^5 - \int \frac{1}{5}(x-5)^5\,dx \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6\,dx \\ \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x\,dx</math> : <math>u = \ln \,x,\, du = \frac{1}{x}\,dx,\, dv = x\,dx,\, v = \frac{x^2}{2}</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x\, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{x^2}{2})(\frac{1}{x}\,dx) \\ \frac{x^2}{2} \ln \,x - \int \frac{x^2}{2}(\frac{1}{x})\,dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \,dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x\,dx</math> ini terjadi dua kali integral parsial : <math>u = \sin \,x,\, du = \cos \,x\,dx,\, dv = e^x\,dx,\, v = e^x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x\, dx \\ (\sin \,x)(e^x) - \int (e^x)(\cos \,x\,dx) \\ e^x \sin \,x - \int e^x \cos \,x\,dx \\ u = \cos \,x,\, du = -\sin \,x\,dx,\, dv = e^x\,dx,\, v = e^x \\ (\cos \,x)(e^x) - \int (e^x)(-\sin \,x\,dx) \\ e^x \cos \,x + \int e^x \sin \,x\,dx \\ e^x \sin \,x - (e^x \cos \,x + \int e^x \sin \,x\,dx) \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x\,dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> ;Cara 2: Tabel Untuk <math display="inline">\int u\,dv</math>, berlaku ketentuan sebagai berikut. {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>u</math> || <math>dv</math> |- | - || <math>\frac{du}{dx}</math> || <math>v</math> |- | + || <math>\frac{d^2u}{dx^2}</math> || <math>\int v\,dx</math> |} Berikut contoh penyelesaian cara parsial dengan tabel. : 1. <math>\int x \sin \, x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>\sin \, x</math> |- | - || <math>1</math> || <math>-\cos \, x</math> |- | + || <math>0</math> || <math>-\sin \, x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \, dx \\ (x)(-\cos \, x) - (1)(-\sin \, x) + C \\ -x \cos \, x + \sin \, x + C \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4 \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>(x-5)^4</math> |- | - || <math>1</math> || <math>\frac{1}{5}(x-5)^5</math> |- | + || <math>0</math> || <math>\frac{1}{30}(x-5)^6</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - (1)(\frac{1}{30}(x-5)^6) + C \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6 + C \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\ln \,x</math> || <math>x</math> |- | - || <math>\frac{1}{x}</math> || <math>\frac{x^2}{2}</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x \, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{1}{x})(\frac{x^2}{2}) \, dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \, dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x \, dx</math> ini terjadi dua kali integral parsial {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\sin \,x</math> || <math>e^x</math> |- | - || <math>\cos \, x</math> || <math>e^x</math> |- | + || <math>-\sin \,x</math> || <math>e^x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x \, dx \\ (\sin \,x)(e^x) - (\cos \,x)(e^x) + \int (-\sin \,x)(e^x) \, dx \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x) \, dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : 1. <math>\int \frac{1}{x^2 - 4}\,dx</math> Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{1}{x^2 - 4} \\ =&\; \frac{1}{(x + 2)(x - 2)} \\ =&\; \frac{A}{x + 2} + \frac{B}{x - 2} \\ =&\; \frac{A(x - 2) + B(x + 2)}{x^2 - 4} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> : 2. <math>\int \frac{x^2+5x+4}{(x-3)^3}\,dx</math> : cara 1 Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^3} \\ \text{untuk mencari C yaitu } (3)^2+5(3)+4=28 \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x - 3)^2 + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x^2 - 6x + 9) + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 - 6Ax + 9A + Bx - 3B}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 + (- 6A + B)x +(9 - 3B)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ \end{aligned} </math> Kita tahu bahwa <math display="inline">A = 1</math> dan <math display="inline">-6A + B = 5</math> dapat diselesaikan, yaitu <math display="inline">A = 1</math> dan {{nowrap|<math display="inline">B = 11</math>.}} : <math> \begin{aligned} &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int (\frac{1}{x - 3} + \frac{11}{(x - 3)^2} + \frac{28}{(x - 3)^3})\,dx \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : cara 2 Pertama, ubah u=x-3 menjadi x=u+3 dan du=dx tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{(u+3)^2+5(u+3)+4}{u^3} \\ =&\; \frac{u^2+11u+28}{u^3} \\ &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int \frac{u^2+11u+28}{u^3}\,du \\ =&\; \int (\frac{u^2}{u^3} + \frac{11u}{u^3} + \frac{28}{u^3})\,du \\ =&\; \int (\frac{1}{u} + \frac{11}{u^2} + \frac{28}{u^3})\,du \\ =&\; \ln u + (-11 u^{-1}) + (-14 u^{-2}) + C \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : 3. <math>\int \frac{𝑥^4+2𝑥^3−13𝑥^2−7𝑥+21}{x^2-4x+3}\,dx</math> Pertama, penyelesaian dengan persamaan polinomial tersebut. : <math> \begin{aligned} =&\; \frac{𝑥^4+2𝑥^3−13𝑥^2−7𝑥+21}{x^2-4x+3} \\ =&\; x^2+6x+8 + \frac{7x-3}{x^2-4x+3} \\ \int \frac{𝑥^4+2𝑥^3−13𝑥^2−7𝑥+21}{x^2-4x+3}\,dx =&\; \int (x^2+6x+8 + \frac{7x-3}{x^2-4x+3}\,dx \\ =&\; \frac{1}{3}x^3+3x^2+8x+ \int \frac{7x-3}{x^2-4x+3}\,dx \\ Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{7x - 3}{x^2 - 4x + 3} \\ =&\; \frac{7x - 3}{(x - 1)(x - 3)} \\ =&\; \frac{A}{x - 1} + \frac{B}{x - 3} \\ =&\; \frac{A(x - 3) + B(x - 1)}{x^2 - 4x + 3} \\ =&\; \frac{(A + B)x + (-3A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 7</math> dan <math display="inline">-3A - B = -3</math> dapat diselesaikan, yaitu <math display="inline">A = 9</math> dan {{nowrap|<math display="inline">B = -2</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> === integral substitusi trigonometri === {| class="wikitable" |- ! Bentuk !! Substitusi Trigonometri |- | <math>\sqrt{a^2 - b^2 x^2}</math> || <math>x = \frac{a}{b} sin \, \alpha </math> |- | <math>\sqrt{a^2 + b^2 x^2}</math> || <math>x = \frac{a}{b} tan \, \alpha</math> |- | <math>\sqrt{b^2 x^2 - a^2}</math> || <math>x = \frac{a}{b} sec \, \alpha</math> |} Berikut contoh penyelesaian cara substitusi trigonometri. : <math>\int \frac{dx}{x^2 \sqrt{x^2 + 4}}</math> : <math>x = 2 \tan \, A, dx = 2 \sec^2 \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \int \frac{dx}{x^2\sqrt{x^2+4}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(2 \tan \, A)^2 \sqrt{4 + (2 \tan \, A)^2}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 + 4 \tan^2 \, A}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4(1 + \tan^2 \, A)}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 \sec^2 \, A }} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(4 \tan^2 \, A)(2 \sec \, A)} \\ =&\; \int \frac{\sec \, A \,dA}{4 \tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\sec \, A \,dA}{\tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A} \,dA \end{aligned} </math> Substitusi berikut dapat dibuat. : <math>\int \frac{\cos \, A}{\sin^2 \, A} \,dA</math> : <math>t = \sin \, A, dt = \cos \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A}\,dA \\ =&\; \frac{1}{4} \int \frac{dt}{t^2} \\ =&\; \frac{1}{4} \left(-\frac{1}{t}\right) + C \\ =&\; -\frac{1}{4 t} + C \\ =&\; -\frac{1}{4 \sin \, A} + C \end{aligned} </math> Ingat bahwa <math display="inline">\sin \, A = \frac{x}{\sqrt{x^2 + 4}}</math> berlaku. : <math> \begin{aligned} =&\; -\frac{1}{4 \sin \, A} + C \\ =&\; -\frac{\sqrt{x^2 + 4}}{4x} + C \end{aligned} </math> === integral mutlak === : <math>\int |f(x)| \,dA</math> buatlah <math>f(x) \ge 0</math> jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x) : <math>\int_a^c |f(x)| \, dx = \int_a^b f(x) \, dx + \int_b^c -f(x) \, dx = F(b) - F(a) - F(c) + F(b)</math> jika <math>|f(x)| = \begin{cases} f(x), & \mbox{jika } f(x) \ge b, \\ -f(x), & \mbox{jika } f(x) < b. \end{cases}</math>. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif. === integral fungsi ganjil dan integral fungsi genap === : <math>\int_{-a}^a f(x) \,dx</math> dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral: : integral fungsi ganjil : <math>\int_{-a}^a f(x) \,dx</math> = 0 : integral fungsi genap : <math>\int_{-a}^a f(x) \,dx</math> = <math>2 \int_{0}^a f(x) \,dx</math> === integral notasi === : <math>\int_a^b f(x) \,dx = \int_{a-p}^{b-p} f(x) \,dx = \int_{a+p}^{b+p} f(x) \,dx</math> : <math>\int_a^c f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx</math> : <math>\int_a^b f(x \pm k) \,dx = \int_{a \pm k}^{b \pm k} f(x) \,dx</math> === integral terbalik === : <math>\int_a^b f(x) \,dx = -\int_b^a f(x) \,dx</math> == Jenis integral lainnya == === panjang busur === ; Sumbu ''x'' : <math>S = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' : <math>S = \int_{y_1}^{y_2} \sqrt{1 + (f'(y))^2}\,dy</math> === luas daerah === ; Satu kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} f(x)\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} f(y)\,dy</math> ; Dua kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} (f(x_2) - f(x_1))\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} (f(y_2) - f(y_1))\,dy</math> : atau juga <math>L = \frac {D \sqrt{D}}{6 a^2}</math> === luas permukaan benda putar === ; Sumbu ''x'' sebagai poros : <math>L_p = 2 \pi \int_{x_1}^{x_2} f(x)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' sebagai poros : <math>L_p = 2 \pi \int_{y_1}^{y_2} f(y)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(y))^2}\,dy</math> === volume benda putar === ; Satu kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} (f(x))^2\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} (f(y))^2\,dy</math> ; Dua kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} ((f(x_2))^2 - (f(x_1))^2)\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} ((f(y_2))^2 - (f(y_1))^2)\,dy</math> : atau juga <math>V = |\frac {D^2 \sqrt{D}}{30 a^3}|</math> == Integral lipat == Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga. ;contoh # Tentukan hasil dari: * <math>\int (5x-1)(5x^2-2x+7)^4 dx</math> * <math>\int x^2 cos 3x dx</math> * <math>\int sin^2 x cos x dx</math> * <math>\int cos^2 x sin x dx</math> * <math>\int sec x dx</math> * <math>\int csc x dx</math> * <math>\int \frac{1}{2x^2-5x+3} dx</math> * <math>\int \frac{1}{\sqrt{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> * <math>\int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx</math> * <math>\int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx</math> * <math>\int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx</math> * <math>\int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx</math> * <math>\int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx \\ a=1, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-sin^2 x} \\ &= \frac{1}{cos^2 x} \\ &= sec^2 x \\ \int sec^2 x dx &= tan x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx \\ a=1, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-cos^2 x} \\ &= \frac{1}{sin^2 x} \\ &= csc^2 x \\ \int csc^2 x dx &= -cot x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx \\ a=sin x, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{sin x}{1-sin^2 x} \\ &= \frac{sin x}{cos^2 x} \\ &= sec x \cdot tan x \\ \int sec x \cdot tan x dx &= sec x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx \\ a=cos x, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{cos x}{1-cos^2 x} \\ &= \frac{cos x}{sin^2 x} \\ &= csc x \cdot cot x \\ \int csc x \cdot cot x dx &= -csc x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx \\ \text{misalkan } y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y &= \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y^2 &= \frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}} \\ &= \frac{x}{y} \\ y^3 &= x \\ 3y^2 dy &= dx \\ \int y \cdot 3y^2 dy \\ \int 3y^3 dy \\ \frac{3y^4}{4}+C \\ \end{aligned} </math> </div></div> # Tentukan hasil dari: * <math>\int_{1}^{5} 2x-3 dx</math> * <math>\int_{0}^{4} |x-2| dx</math> * <math>\int_{2}^{4} |3-x| dx</math> * <math>\int_{1}^{5} x^2-6x+8 dx</math> * <math>\int_{-3}^{5} |x^2-2x-8| dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} 2x-3 dx &= \left. x^2 - 3x \right|_{1}^{5} \\ &= 10 - (-2) \\ &= 12 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{0}^{4} |x-2| dx \\ \text{ tentukan nilai harga nol } \\ x-2 &= 0 \\ x &= 2 \\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti } \\ |x-2| = \begin{cases} x-2, & \mbox{jika } x \ge 2, \\ -(x-2), & \mbox{jika } x < 2. \end{cases} \\ \int_{0}^{4} |x-2| dx &= \int_{2}^{4} x-2 dx + \int_{0}^{2} -(x-2) dx \\ &= \left. (\frac{x^2}{2}-2x) \right|_{2}^{4} + \left. (\frac{-x^2}{2}+2x) \right|_{0}^{2} \\ &= 0 - (-2) + 2 - 0 \\ &= 4 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{2}^{4} |3-x| dx \\ \text{ tentukan nilai syarat-syarat } \\ 3-x &\ge 0 \\ \text{ harga nol } \\ x &= 3 \\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti } \\ |3-x| = \begin{cases} 3-x, & \mbox{jika } x \le 3, \\ -(3-x), & \mbox{jika } x > 3. \end{cases} \\ \int_{2}^{4} |3-x| dx &= \int_{2}^{3} 3-x dx + \int_{3}^{4} -(3-x) dx \\ &= \left. (3x-\frac{x^2}{2}) \right|_{2}^{3} + \left. (-3x+\frac{x^2}{2}) \right|_{3}^{4} \\ &= 4.5 - 4 - 4 - (-4.5) \\ &= 1 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} x^2-6x+8 dx &= \left. \frac{x^3}{3}-3x^2+8x \right|_{1}^{5} \\ &= \frac{20}{3} - \frac{16}{3} \\ &= \frac{4}{3} \\ &= 1\frac{1}{3} \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{-3}^{5} |x^2-2x-8| dx \\ \text{ tentukan nilai syarat-syarat } \\ x^2-2x-8 &\ge 0 \\ \text{ harga nol } \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x = 4 \\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti } \\ |x^2-2x-8| = \begin{cases} x^2-2x-8, & \mbox{jika } x \le -2, \\ -(x^2-2x-8), & \mbox{jika } -2 < x < 4, \\ x^2-2x-8, & \mbox{jika } x \ge 4 \end{cases} \\ \int_{-3}^{5} |x^2-2x-8| dx &= \int_{-3}^{-2} x^2-2x-8 dx + \int_{-2}^{4} -(x^2-2x-8) dx + \int_{4}^{5} x^2-2x-8 dx \\ &= \left. (\frac{x^3}{3}-x^2-8x) \right|_{-3}^{-2} + \left. (\frac{-x^3}{3}+x^2+8x) \right|_{-2}^{4} + \left. (\frac{x^3}{3}-x^2-8x) \right|_{4}^{5} \\ &= \frac{28}{3} - 6 + \frac{80}{3} - (-\frac{28}{3}) - \frac{70}{3} - (-\frac{80}{3}) \\ &= \frac{128}{3} \\ &= 42\frac{2}{3} \\ \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x\,dx \\ L &= \frac{1}{2} x^2 \\ L &= \frac{1}{2} xy \quad (y = x) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x^2</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x^2\,dx \\ L &= \frac{1}{3} x^3 \\ L &= \frac{1}{3} xy \quad (y = x^2) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = \sqrt{x}</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int \sqrt{x}\,dx \\ L &= \frac{2}{3} x^{\frac{3}{2}} \\ L &= \frac{2}{3} xy \quad (y = \sqrt{x}) \end{aligned} </math> </div></div> # Buktikan luas persegi <math>L = s^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = s \text{ dan titik (''s'', ''s''), } \\ L &= \int_{0}^{s} s\,dx \\ L &= sx |_{0}^{s} \\ L &= ss - 0 \\ L &= s^2 \end{aligned} </math> </div></div> # Buktikan luas persegi panjang <math>L = pl</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = l \text{dan titik (''p'', ''l''), } \\ L &= \int_{0}^{p} l\,dx \\ L &= lx |_{0}^{p} \\ L &= pl - 0 \\ L &= pl \end{aligned} </math> </div></div> * Buktikan luas segitiga <math>L = \frac{at}{2}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{-tx}{a} + t \text { dan titik (''a'', ''t''), } \\ L &= \int_{0}^{a} \left (\frac{-tx}{a} + t \right) \,dx \\ L &= \left. \frac{-tx^2}{2a} + tx \right|_{0}^{a} \\ L &= \frac{-ta^2}{2a} + ta - 0 + 0 \\ L &= \frac{-ta}{2} + ta \\ L &= \frac{at}{2} \end{aligned} </math> </div></div> # Buktikan volume tabung <math>V = \pi r^2t</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ Dengan posisi } y = r \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} r^2\,dx \\ V &= \pi \left. r^2x \right|_{0}^{t} \\ V &= \pi r^2t - 0 \\ V &= \pi r^2t \end{aligned} </math> </div></div> # Buktikan volume kerucut <math>V = \frac{\pi r^2t}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{rx}{t} \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} \left (\frac{rx}{t} \right)^2 \,dx \\ V &= \pi \left. \frac{r^2 x^3}{3t^2} \right|_{0}^{t} \\ V &= \pi \frac{r^2 t^3}{3t^2} - 0 \\ V &= \frac{\pi r^2t}{3} \end{aligned} </math> </div></div> # Buktikan volume bola <math>V = \frac{4 \pi r^3}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), } \\ V &= \pi \int_{-r}^{r} \left(\sqrt {r^2 - x^2}\right)^2\,dx \\ V &= \pi \int_{-r}^{r} r^2 - x^2 dx \\ V &= \pi \left. r^2x - \frac{x^3}{3} \right|_{-r}^{r} \\ V &= \pi \left (r^3 - \frac{r^3}{3} - \left (-r^3 + \frac{r^3}{3} \right) \right) \\ V &= \frac{4 \pi r^3}{3} \end{aligned} </math> </div></div> # Buktikan luas permukaan bola <math>L = 4 \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{selanjutnya } \\ ds &= \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ ds &= \sqrt{1 + \frac{x^2}{r^2 - x^2}}\,dx \\ ds &= \sqrt{\frac{r^2}{r^2 - x^2}}\,dx \\ ds &= \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ \text{sehingga } \\ L &= 2 \pi \int_{-r}^{r} \sqrt{r^2 - x^2} \cdot \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ L &= 2 \pi \int_{-r}^{r} r\,dx \\ L &= 2 \pi rx|_{-r}^{r} \\ L &= 2 \pi (r r - r (-r)) \\ L &= 2 \pi (r^2 + r^2) \\ L &= 4 \pi r^2 \end{aligned} </math> </div></div> # Buktikan keliling lingkaran <math>K = 2 \pi r</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0) } \\ \text{kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{sehingga } \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{x^2}{r^2 - x^2})}\,dx \\ K &= 4 \int_{0}^{r} \sqrt\frac{r^2}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \sqrt\frac{1}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \frac{1}{\sqrt{r^2 - x^2}}\,dx \\ K &= 4r \left. \arcsin \left (\frac{x}{r} \right) \right|_{0}^{r} \\ K &= 4r \left (\arcsin \left (\frac{r}{r}\right) - \arcsin \left (\frac{0}{r} \right) \right) \\ K &= 4r \left (\arcsin\left (1 \right) - \arcsin \left(0 \right) \right)) \\ K &= 4r \left (\frac{\pi}{2} \right) \\ K &= 2 \pi r \end{aligned} </math> </div></div> # Buktikan luas lingkaran <math>L = \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. } \\ \sin(\theta) &= \frac{x}{r} \\ x &= r \sin(\theta) \\ dx &= r \cos(\theta)\,d\theta \\ \text{dengan turunan di atas } \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - x^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - (r \sin(\theta))^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - r^2 \sin^2(\theta)}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 (1 - \sin^2(\theta))}\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta)\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta) (r \cos(\theta)\,d\theta) \\ L &= 4 \int_{0}^{r} r^2 \cos^2(\theta)\,d\theta \\ L &= 4r^2 \int_{0}^{r} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta \\ L &= 2r^2 \int_{0}^{r} (1 + \cos(2\theta))\,d\theta \\ L &= 2r^2 \left (\theta + \frac{1}{2} \sin(2\theta) \right)|_{0}^{r} \\ L &= 2r^2 (\theta + \sin(\theta) \cos(\theta))|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left (\frac{x}{r} \right) + \left (\frac{x}{r} \right) \left (\frac{r^2 - x^2}{r} \right) \right)|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left(\frac{r}{r} \right) + \left (\frac{r}{r} \right) \left (\frac {r^2 - r^2}{r} \right) \right) - \left(\arcsin \left (\frac{0}{r} \right) + \left (\frac{0}{r}\right) \left (\frac{r^2 - 0^2}{r} \right) \right) \\ L &= 2r^2 (\arcsin(1) + 0 - (\arcsin(0) + 0)) \\ L &= 2r^2 \left (\frac{\pi}{2} \right) \\ L &= \pi r^2 \end{aligned} </math> </div></div> # Buktikan luas elips <math>L = \pi ab</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \frac{b \sqrt{a^2 - x^2}}{a} \text{ serta (-''a'', 0) dan (''a'', 0), } \\ L &= 4 \int_{0}^{r} \frac{b \sqrt{a^2 - x^2}}{a}\,dx \\ L &= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx \\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, } \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{\frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx}{4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx} \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{b}{a} \\ L_\text{elips} &= \frac{b}{a} L_\text{ling} \\ L_\text{elips} &= \frac{b}{a} \pi a^2 \\ L_\text{elips} &= \pi ab \end{aligned} </math> </div></div> # Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ x^-2x &= 4x+7 \\ x^2-6x-7 &= 0 \\ (x+1)(x-7) &= 0 \\ x=-1 &\text{ atau } x=7 \\ y &= 4(-1)+7 = 3 \\ y &= 4(7)+7 = 35 \\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut } \\ L &= \int_{-1}^{7} 4x+7-(x^2-2x)\,dx \\ &= \int_{-1}^{7} -x^2+6x+7\,dx \\ &= -\frac{x^3}{3}+3x^2+7x|_{-1}^{7} \\ &= -\frac{7^3}{3}+3(7)^2+7(7)-(-\frac{(-1)^3}{3}+3(-1)^2+7(-1)) \\ &= \frac{245}{3}-(-\frac{13}{3}) \\ &= \frac{258}{3} \\ &= 86 \\ \end{aligned} </math> </div></div> # Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ 6-1,5x &= x-4 \\ 2,5x &= 10 \\ x &= 4 \\ y &= 4-4 = 0 \\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut } \\ \text{untuk x=0 } \\ y &= 6-1,5(0) = 6 \\ y &= 0-4 = -4 \\ L &= \pi \int_{0}^{4} ((6-1,5x)^2-(x-4)^2)\,dx \\ &= \pi \int_{0}^{4} (36-18x+2,25x^2-(x^2-8x+16))\,dx \\ &= \pi \int_{0}^{4} (1,25x^2-10x+20)\,dx \\ &= \pi (\frac{1,25x^3}{3}-5x^2+20x)|_{0}^{4} \\ &= \pi (\frac{1,25(4)^3}{3}-5(4)^2+20(4)-(\frac{1,25(0)^3}{3}-5(0)^2+20(0))) \\ &= \pi (\frac{30}{3}-0) \\ &= 10\pi \\ \end{aligned} </math> </div></div> [[Kategori:Soal-Soal Matematika]] e37qixuqqp89jv8mqt5vpn45qo02v5x 115026 115025 2026-04-27T01:52:04Z ~2026-25546-23 43043 /* integral pecahan parsial */ 115026 wikitext text/x-wiki == Kaidah umum == :# <math>\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ konstan)}\,\!</math> :# <math>\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx</math> :# <math>\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx</math> :# <math>\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx</math> :# <math>\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\! </math> :# <math>\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C </math> :# <math>\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C </math> :# <math>\int f(x) \, dx = F(x) + C</math> :# <math>\int_a^b f(x) \, dx = F(b) - F(a)</math> :# <math>\int f(x) \circ g(x) dx</math> = f’(g(x)) g’(x) :# <math>\int f^{-1}(x) dx = \frac{1}{f'(f'(x))}</math> == rumus sederhana == :<math>\int \, dx = x + C</math> :<math>\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int {dx \over x} = \ln{\left|x\right|} + C</math> <math>\int {dx \over \sqrt{a^2-b^2 x^2}} = \frac{1}{b} \arcsin \, {bx \over a} + C</math> :<math>\int {dx \over {a^2+b^2x^2}} = {1 \over ab}\arctan \, {bx \over a} + C</math> <math>\int {dx \over x\sqrt{b^2x^2-a^2}} = {1 \over a}\arcsec \, {bx \over a} + C</math> ; Eksponen dan logaritma :<math>\int e^x\,dx = e^x + C</math> :<math>\int a^x\,dx = \frac{a^x}{\ln{a}} + C</math> :<math>\int \ln \, x\,dx = x \cdot \, \ln {x} - x + C</math> :<math>\int \, ^b\!\log \, x\,dx = x \cdot \, ^b\!\log \, x - x \cdot \,^b\!\log e + C</math> ; Trigonometri :<math>\int \sin \, x \, dx = -\cos \, x + C</math> :<math>\int \cos \, x \, dx = \sin \, x + C</math> :<math>\int \tan \, x \, dx = -\ln{\left| \cos \, x \right|} + C</math> :<math>\int \cot \, x \, dx = \ln{\left| \sin \, x \right|} + C</math> :<math>\int \sec \, x \, dx = \ln{\left| \sec \, x + \tan \, x \right|} + C</math> :<math>\int \csc \, x \, dx = -\ln{\left| \csc \, x + \cot \, x \right|} + C</math> ; Hiperbolik :<math>\int \sinh \, x \, dx = \cosh \, x + C</math> :<math>\int \cosh \, x \, dx = \sinh \, x + C</math> :<math>\int \tanh \, x \, dx = \ln{\left| \cosh \, x \right|} + C</math> :<math>\int \coth \, x \, dx = \ln{\left| \sinh \, x \right|} + C</math> :<math>\int \mbox{sech} \, x \, dx = \arctan \,(\sinh \, x) + C</math> :<math>\int \mbox{csch} \, x \, dx = \ln{\left| \tanh \, {x \over 2} \right|} + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : 1. <math>\int x^3 - cos 3x + e^{5x+2} - \frac{1}{2x+5}\,dx</math> : <math>= \frac{1}{3+1}x^{3+1} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : <math>= \frac{x^4}{4} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : 2. <math>\int x(x-5)^4\,dx</math> : <math>= \int x(x^4-4x^3(5)+6x^2(25)-4x(125)+625)\,dx</math> : <math>= \int x(x^4-20x^3+150x^2-500x+625)\,dx</math> : <math>= \int x^5-20x^4+150x^3-500x^2+625x\,dx</math> : <math>= \frac{x^6}{6}-4x^5+\frac{150x^4}{4}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> : <math>= \frac{x^6}{6}-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : 1. <math>\int \frac{\ln \, x}{x}\,dx</math> : <math>u = \ln \, x\, du = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{\ln(x)}{x} dx \\ \int u du \\ \frac{1}{2} u^2 + C \\ \frac{1}{2} ln^2 x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x-5,\, du = dx,\, x = u+5</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4\,dx \\ \int (u+5)u^4\,du \\ \int (u^5+5u^4)\,du \\ \frac{1}{6} u^6+u^5 + C \\ \frac{1}{6} (x-5)^6+(x-5)^5 + C \\ \frac{1}{6} (x^6-6x^5(5)+15x^4(25)-20x^3(125)+15x^2(625)-6x(3125)+15625)+(x^5-5x^4(5)+10x^3(25)-10x^2(125)+5x(625)+3125) + C \\ \frac{1}{6}x^6-5x^5+\frac{125x^4}{2}-\frac{1250x^3}{3}+\frac{3125x^2}{2}-3125x+\frac{15625}{6}+x^5-25x^4+250x^3-1250x^2+3125x+3125 + C \\ \frac{1}{6}x^6-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C \\ \end{aligned} </math> </div></div> === integral parsial === ;Cara 1: Rumus Integral parsial diselesaikan dengan rumus berikut. : <math>\int u\,dv = uv - \int v\,du </math> Berikut contoh penyelesaian cara parsial dengan rumus. : 1. <math>\int x \sin \, x \, dx</math> : <math>u = x, du = 1 \, dx, dv = \sin \, x \, dx, v = -\cos \, x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \,dx \\ (x)(-\cos \, x) - \int (-\cos \, x)(1 dx) \\ -x \cos \, x + \int \cos \, x dx \\ -x \cos \, x + \sin \, x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x,\, du = 1\,dx,\, dv = (x-5)^4\,dx,\, v = \frac{1}{5}(x-5)^5</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - \int (\frac{1}{5}(x-5)^5)(1\,dx) \\ \frac{x}{5}(x-5)^5 - \int \frac{1}{5}(x-5)^5\,dx \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6\,dx \\ \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x\,dx</math> : <math>u = \ln \,x,\, du = \frac{1}{x}\,dx,\, dv = x\,dx,\, v = \frac{x^2}{2}</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x\, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{x^2}{2})(\frac{1}{x}\,dx) \\ \frac{x^2}{2} \ln \,x - \int \frac{x^2}{2}(\frac{1}{x})\,dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \,dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x\,dx</math> ini terjadi dua kali integral parsial : <math>u = \sin \,x,\, du = \cos \,x\,dx,\, dv = e^x\,dx,\, v = e^x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x\, dx \\ (\sin \,x)(e^x) - \int (e^x)(\cos \,x\,dx) \\ e^x \sin \,x - \int e^x \cos \,x\,dx \\ u = \cos \,x,\, du = -\sin \,x\,dx,\, dv = e^x\,dx,\, v = e^x \\ (\cos \,x)(e^x) - \int (e^x)(-\sin \,x\,dx) \\ e^x \cos \,x + \int e^x \sin \,x\,dx \\ e^x \sin \,x - (e^x \cos \,x + \int e^x \sin \,x\,dx) \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x\,dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> ;Cara 2: Tabel Untuk <math display="inline">\int u\,dv</math>, berlaku ketentuan sebagai berikut. {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>u</math> || <math>dv</math> |- | - || <math>\frac{du}{dx}</math> || <math>v</math> |- | + || <math>\frac{d^2u}{dx^2}</math> || <math>\int v\,dx</math> |} Berikut contoh penyelesaian cara parsial dengan tabel. : 1. <math>\int x \sin \, x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>\sin \, x</math> |- | - || <math>1</math> || <math>-\cos \, x</math> |- | + || <math>0</math> || <math>-\sin \, x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \, dx \\ (x)(-\cos \, x) - (1)(-\sin \, x) + C \\ -x \cos \, x + \sin \, x + C \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4 \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>(x-5)^4</math> |- | - || <math>1</math> || <math>\frac{1}{5}(x-5)^5</math> |- | + || <math>0</math> || <math>\frac{1}{30}(x-5)^6</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - (1)(\frac{1}{30}(x-5)^6) + C \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6 + C \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\ln \,x</math> || <math>x</math> |- | - || <math>\frac{1}{x}</math> || <math>\frac{x^2}{2}</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x \, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{1}{x})(\frac{x^2}{2}) \, dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \, dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x \, dx</math> ini terjadi dua kali integral parsial {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\sin \,x</math> || <math>e^x</math> |- | - || <math>\cos \, x</math> || <math>e^x</math> |- | + || <math>-\sin \,x</math> || <math>e^x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x \, dx \\ (\sin \,x)(e^x) - (\cos \,x)(e^x) + \int (-\sin \,x)(e^x) \, dx \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x) \, dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : 1. <math>\int \frac{1}{x^2 - 4}\,dx</math> Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{1}{x^2 - 4} \\ =&\; \frac{1}{(x + 2)(x - 2)} \\ =&\; \frac{A}{x + 2} + \frac{B}{x - 2} \\ =&\; \frac{A(x - 2) + B(x + 2)}{x^2 - 4} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> : 2. <math>\int \frac{x^2+5x+4}{(x-3)^3}\,dx</math> : cara 1 Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^3} \\ \text{untuk mencari C yaitu } (3)^2+5(3)+4=28 \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x - 3)^2 + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x^2 - 6x + 9) + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 - 6Ax + 9A + Bx - 3B}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 + (- 6A + B)x +(9 - 3B)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ \end{aligned} </math> Kita tahu bahwa <math display="inline">A = 1</math> dan <math display="inline">-6A + B = 5</math> dapat diselesaikan, yaitu <math display="inline">A = 1</math> dan {{nowrap|<math display="inline">B = 11</math>.}} : <math> \begin{aligned} &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int (\frac{1}{x - 3} + \frac{11}{(x - 3)^2} + \frac{28}{(x - 3)^3})\,dx \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : cara 2 Pertama, ubah u=x-3 menjadi x=u+3 dan du=dx tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{(u+3)^2+5(u+3)+4}{u^3} \\ =&\; \frac{u^2+11u+28}{u^3} \\ &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int \frac{u^2+11u+28}{u^3}\,du \\ =&\; \int (\frac{u^2}{u^3} + \frac{11u}{u^3} + \frac{28}{u^3})\,du \\ =&\; \int (\frac{1}{u} + \frac{11}{u^2} + \frac{28}{u^3})\,du \\ =&\; \ln u + (-11 u^{-1}) + (-14 u^{-2}) + C \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : 3. <math>\int \frac{x^4+2x^3−13x^2−7x+21}{x^2-4x+3}\,dx</math> Pertama, penyelesaian dengan persamaan polinomial tersebut. : <math> \begin{aligned} =&\; \frac{x^4+2x^3−13x^2−7x+21}{x^2-4x+3} \\ =&\; x^2+6x+8 + \frac{7x-3}{x^2-4x+3} \\ \int \frac{x^4+2x^3−13x^2−7x+21}{x^2-4x+3}\,dx =&\; \int (x^2+6x+8 + \frac{7x-3}{x^2-4x+3}\,dx \\ =&\; \frac{1}{3}x^3+3x^2+8x+ \int \frac{7x-3}{x^2-4x+3}\,dx \\ Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{7x - 3}{x^2 - 4x + 3} \\ =&\; \frac{7x - 3}{(x - 1)(x - 3)} \\ =&\; \frac{A}{x - 1} + \frac{B}{x - 3} \\ =&\; \frac{A(x - 3) + B(x - 1)}{x^2 - 4x + 3} \\ =&\; \frac{(A + B)x + (-3A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 7</math> dan <math display="inline">-3A - B = -3</math> dapat diselesaikan, yaitu <math display="inline">A = 9</math> dan {{nowrap|<math display="inline">B = -2</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> === integral substitusi trigonometri === {| class="wikitable" |- ! Bentuk !! Substitusi Trigonometri |- | <math>\sqrt{a^2 - b^2 x^2}</math> || <math>x = \frac{a}{b} sin \, \alpha </math> |- | <math>\sqrt{a^2 + b^2 x^2}</math> || <math>x = \frac{a}{b} tan \, \alpha</math> |- | <math>\sqrt{b^2 x^2 - a^2}</math> || <math>x = \frac{a}{b} sec \, \alpha</math> |} Berikut contoh penyelesaian cara substitusi trigonometri. : <math>\int \frac{dx}{x^2 \sqrt{x^2 + 4}}</math> : <math>x = 2 \tan \, A, dx = 2 \sec^2 \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \int \frac{dx}{x^2\sqrt{x^2+4}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(2 \tan \, A)^2 \sqrt{4 + (2 \tan \, A)^2}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 + 4 \tan^2 \, A}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4(1 + \tan^2 \, A)}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 \sec^2 \, A }} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(4 \tan^2 \, A)(2 \sec \, A)} \\ =&\; \int \frac{\sec \, A \,dA}{4 \tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\sec \, A \,dA}{\tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A} \,dA \end{aligned} </math> Substitusi berikut dapat dibuat. : <math>\int \frac{\cos \, A}{\sin^2 \, A} \,dA</math> : <math>t = \sin \, A, dt = \cos \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A}\,dA \\ =&\; \frac{1}{4} \int \frac{dt}{t^2} \\ =&\; \frac{1}{4} \left(-\frac{1}{t}\right) + C \\ =&\; -\frac{1}{4 t} + C \\ =&\; -\frac{1}{4 \sin \, A} + C \end{aligned} </math> Ingat bahwa <math display="inline">\sin \, A = \frac{x}{\sqrt{x^2 + 4}}</math> berlaku. : <math> \begin{aligned} =&\; -\frac{1}{4 \sin \, A} + C \\ =&\; -\frac{\sqrt{x^2 + 4}}{4x} + C \end{aligned} </math> === integral mutlak === : <math>\int |f(x)| \,dA</math> buatlah <math>f(x) \ge 0</math> jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x) : <math>\int_a^c |f(x)| \, dx = \int_a^b f(x) \, dx + \int_b^c -f(x) \, dx = F(b) - F(a) - F(c) + F(b)</math> jika <math>|f(x)| = \begin{cases} f(x), & \mbox{jika } f(x) \ge b, \\ -f(x), & \mbox{jika } f(x) < b. \end{cases}</math>. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif. === integral fungsi ganjil dan integral fungsi genap === : <math>\int_{-a}^a f(x) \,dx</math> dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral: : integral fungsi ganjil : <math>\int_{-a}^a f(x) \,dx</math> = 0 : integral fungsi genap : <math>\int_{-a}^a f(x) \,dx</math> = <math>2 \int_{0}^a f(x) \,dx</math> === integral notasi === : <math>\int_a^b f(x) \,dx = \int_{a-p}^{b-p} f(x) \,dx = \int_{a+p}^{b+p} f(x) \,dx</math> : <math>\int_a^c f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx</math> : <math>\int_a^b f(x \pm k) \,dx = \int_{a \pm k}^{b \pm k} f(x) \,dx</math> === integral terbalik === : <math>\int_a^b f(x) \,dx = -\int_b^a f(x) \,dx</math> == Jenis integral lainnya == === panjang busur === ; Sumbu ''x'' : <math>S = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' : <math>S = \int_{y_1}^{y_2} \sqrt{1 + (f'(y))^2}\,dy</math> === luas daerah === ; Satu kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} f(x)\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} f(y)\,dy</math> ; Dua kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} (f(x_2) - f(x_1))\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} (f(y_2) - f(y_1))\,dy</math> : atau juga <math>L = \frac {D \sqrt{D}}{6 a^2}</math> === luas permukaan benda putar === ; Sumbu ''x'' sebagai poros : <math>L_p = 2 \pi \int_{x_1}^{x_2} f(x)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' sebagai poros : <math>L_p = 2 \pi \int_{y_1}^{y_2} f(y)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(y))^2}\,dy</math> === volume benda putar === ; Satu kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} (f(x))^2\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} (f(y))^2\,dy</math> ; Dua kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} ((f(x_2))^2 - (f(x_1))^2)\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} ((f(y_2))^2 - (f(y_1))^2)\,dy</math> : atau juga <math>V = |\frac {D^2 \sqrt{D}}{30 a^3}|</math> == Integral lipat == Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga. ;contoh # Tentukan hasil dari: * <math>\int (5x-1)(5x^2-2x+7)^4 dx</math> * <math>\int x^2 cos 3x dx</math> * <math>\int sin^2 x cos x dx</math> * <math>\int cos^2 x sin x dx</math> * <math>\int sec x dx</math> * <math>\int csc x dx</math> * <math>\int \frac{1}{2x^2-5x+3} dx</math> * <math>\int \frac{1}{\sqrt{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> * <math>\int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx</math> * <math>\int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx</math> * <math>\int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx</math> * <math>\int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx</math> * <math>\int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx \\ a=1, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-sin^2 x} \\ &= \frac{1}{cos^2 x} \\ &= sec^2 x \\ \int sec^2 x dx &= tan x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx \\ a=1, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-cos^2 x} \\ &= \frac{1}{sin^2 x} \\ &= csc^2 x \\ \int csc^2 x dx &= -cot x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx \\ a=sin x, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{sin x}{1-sin^2 x} \\ &= \frac{sin x}{cos^2 x} \\ &= sec x \cdot tan x \\ \int sec x \cdot tan x dx &= sec x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx \\ a=cos x, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{cos x}{1-cos^2 x} \\ &= \frac{cos x}{sin^2 x} \\ &= csc x \cdot cot x \\ \int csc x \cdot cot x dx &= -csc x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx \\ \text{misalkan } y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y &= \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y^2 &= \frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}} \\ &= \frac{x}{y} \\ y^3 &= x \\ 3y^2 dy &= dx \\ \int y \cdot 3y^2 dy \\ \int 3y^3 dy \\ \frac{3y^4}{4}+C \\ \end{aligned} </math> </div></div> # Tentukan hasil dari: * <math>\int_{1}^{5} 2x-3 dx</math> * <math>\int_{0}^{4} |x-2| dx</math> * <math>\int_{2}^{4} |3-x| dx</math> * <math>\int_{1}^{5} x^2-6x+8 dx</math> * <math>\int_{-3}^{5} |x^2-2x-8| dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} 2x-3 dx &= \left. x^2 - 3x \right|_{1}^{5} \\ &= 10 - (-2) \\ &= 12 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{0}^{4} |x-2| dx \\ \text{ tentukan nilai harga nol } \\ x-2 &= 0 \\ x &= 2 \\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti } \\ |x-2| = \begin{cases} x-2, & \mbox{jika } x \ge 2, \\ -(x-2), & \mbox{jika } x < 2. \end{cases} \\ \int_{0}^{4} |x-2| dx &= \int_{2}^{4} x-2 dx + \int_{0}^{2} -(x-2) dx \\ &= \left. (\frac{x^2}{2}-2x) \right|_{2}^{4} + \left. (\frac{-x^2}{2}+2x) \right|_{0}^{2} \\ &= 0 - (-2) + 2 - 0 \\ &= 4 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{2}^{4} |3-x| dx \\ \text{ tentukan nilai syarat-syarat } \\ 3-x &\ge 0 \\ \text{ harga nol } \\ x &= 3 \\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti } \\ |3-x| = \begin{cases} 3-x, & \mbox{jika } x \le 3, \\ -(3-x), & \mbox{jika } x > 3. \end{cases} \\ \int_{2}^{4} |3-x| dx &= \int_{2}^{3} 3-x dx + \int_{3}^{4} -(3-x) dx \\ &= \left. (3x-\frac{x^2}{2}) \right|_{2}^{3} + \left. (-3x+\frac{x^2}{2}) \right|_{3}^{4} \\ &= 4.5 - 4 - 4 - (-4.5) \\ &= 1 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} x^2-6x+8 dx &= \left. \frac{x^3}{3}-3x^2+8x \right|_{1}^{5} \\ &= \frac{20}{3} - \frac{16}{3} \\ &= \frac{4}{3} \\ &= 1\frac{1}{3} \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{-3}^{5} |x^2-2x-8| dx \\ \text{ tentukan nilai syarat-syarat } \\ x^2-2x-8 &\ge 0 \\ \text{ harga nol } \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x = 4 \\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti } \\ |x^2-2x-8| = \begin{cases} x^2-2x-8, & \mbox{jika } x \le -2, \\ -(x^2-2x-8), & \mbox{jika } -2 < x < 4, \\ x^2-2x-8, & \mbox{jika } x \ge 4 \end{cases} \\ \int_{-3}^{5} |x^2-2x-8| dx &= \int_{-3}^{-2} x^2-2x-8 dx + \int_{-2}^{4} -(x^2-2x-8) dx + \int_{4}^{5} x^2-2x-8 dx \\ &= \left. (\frac{x^3}{3}-x^2-8x) \right|_{-3}^{-2} + \left. (\frac{-x^3}{3}+x^2+8x) \right|_{-2}^{4} + \left. (\frac{x^3}{3}-x^2-8x) \right|_{4}^{5} \\ &= \frac{28}{3} - 6 + \frac{80}{3} - (-\frac{28}{3}) - \frac{70}{3} - (-\frac{80}{3}) \\ &= \frac{128}{3} \\ &= 42\frac{2}{3} \\ \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x\,dx \\ L &= \frac{1}{2} x^2 \\ L &= \frac{1}{2} xy \quad (y = x) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x^2</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x^2\,dx \\ L &= \frac{1}{3} x^3 \\ L &= \frac{1}{3} xy \quad (y = x^2) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = \sqrt{x}</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int \sqrt{x}\,dx \\ L &= \frac{2}{3} x^{\frac{3}{2}} \\ L &= \frac{2}{3} xy \quad (y = \sqrt{x}) \end{aligned} </math> </div></div> # Buktikan luas persegi <math>L = s^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = s \text{ dan titik (''s'', ''s''), } \\ L &= \int_{0}^{s} s\,dx \\ L &= sx |_{0}^{s} \\ L &= ss - 0 \\ L &= s^2 \end{aligned} </math> </div></div> # Buktikan luas persegi panjang <math>L = pl</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = l \text{dan titik (''p'', ''l''), } \\ L &= \int_{0}^{p} l\,dx \\ L &= lx |_{0}^{p} \\ L &= pl - 0 \\ L &= pl \end{aligned} </math> </div></div> * Buktikan luas segitiga <math>L = \frac{at}{2}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{-tx}{a} + t \text { dan titik (''a'', ''t''), } \\ L &= \int_{0}^{a} \left (\frac{-tx}{a} + t \right) \,dx \\ L &= \left. \frac{-tx^2}{2a} + tx \right|_{0}^{a} \\ L &= \frac{-ta^2}{2a} + ta - 0 + 0 \\ L &= \frac{-ta}{2} + ta \\ L &= \frac{at}{2} \end{aligned} </math> </div></div> # Buktikan volume tabung <math>V = \pi r^2t</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ Dengan posisi } y = r \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} r^2\,dx \\ V &= \pi \left. r^2x \right|_{0}^{t} \\ V &= \pi r^2t - 0 \\ V &= \pi r^2t \end{aligned} </math> </div></div> # Buktikan volume kerucut <math>V = \frac{\pi r^2t}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{rx}{t} \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} \left (\frac{rx}{t} \right)^2 \,dx \\ V &= \pi \left. \frac{r^2 x^3}{3t^2} \right|_{0}^{t} \\ V &= \pi \frac{r^2 t^3}{3t^2} - 0 \\ V &= \frac{\pi r^2t}{3} \end{aligned} </math> </div></div> # Buktikan volume bola <math>V = \frac{4 \pi r^3}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), } \\ V &= \pi \int_{-r}^{r} \left(\sqrt {r^2 - x^2}\right)^2\,dx \\ V &= \pi \int_{-r}^{r} r^2 - x^2 dx \\ V &= \pi \left. r^2x - \frac{x^3}{3} \right|_{-r}^{r} \\ V &= \pi \left (r^3 - \frac{r^3}{3} - \left (-r^3 + \frac{r^3}{3} \right) \right) \\ V &= \frac{4 \pi r^3}{3} \end{aligned} </math> </div></div> # Buktikan luas permukaan bola <math>L = 4 \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{selanjutnya } \\ ds &= \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ ds &= \sqrt{1 + \frac{x^2}{r^2 - x^2}}\,dx \\ ds &= \sqrt{\frac{r^2}{r^2 - x^2}}\,dx \\ ds &= \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ \text{sehingga } \\ L &= 2 \pi \int_{-r}^{r} \sqrt{r^2 - x^2} \cdot \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ L &= 2 \pi \int_{-r}^{r} r\,dx \\ L &= 2 \pi rx|_{-r}^{r} \\ L &= 2 \pi (r r - r (-r)) \\ L &= 2 \pi (r^2 + r^2) \\ L &= 4 \pi r^2 \end{aligned} </math> </div></div> # Buktikan keliling lingkaran <math>K = 2 \pi r</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0) } \\ \text{kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{sehingga } \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{x^2}{r^2 - x^2})}\,dx \\ K &= 4 \int_{0}^{r} \sqrt\frac{r^2}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \sqrt\frac{1}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \frac{1}{\sqrt{r^2 - x^2}}\,dx \\ K &= 4r \left. \arcsin \left (\frac{x}{r} \right) \right|_{0}^{r} \\ K &= 4r \left (\arcsin \left (\frac{r}{r}\right) - \arcsin \left (\frac{0}{r} \right) \right) \\ K &= 4r \left (\arcsin\left (1 \right) - \arcsin \left(0 \right) \right)) \\ K &= 4r \left (\frac{\pi}{2} \right) \\ K &= 2 \pi r \end{aligned} </math> </div></div> # Buktikan luas lingkaran <math>L = \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. } \\ \sin(\theta) &= \frac{x}{r} \\ x &= r \sin(\theta) \\ dx &= r \cos(\theta)\,d\theta \\ \text{dengan turunan di atas } \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - x^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - (r \sin(\theta))^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - r^2 \sin^2(\theta)}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 (1 - \sin^2(\theta))}\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta)\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta) (r \cos(\theta)\,d\theta) \\ L &= 4 \int_{0}^{r} r^2 \cos^2(\theta)\,d\theta \\ L &= 4r^2 \int_{0}^{r} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta \\ L &= 2r^2 \int_{0}^{r} (1 + \cos(2\theta))\,d\theta \\ L &= 2r^2 \left (\theta + \frac{1}{2} \sin(2\theta) \right)|_{0}^{r} \\ L &= 2r^2 (\theta + \sin(\theta) \cos(\theta))|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left (\frac{x}{r} \right) + \left (\frac{x}{r} \right) \left (\frac{r^2 - x^2}{r} \right) \right)|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left(\frac{r}{r} \right) + \left (\frac{r}{r} \right) \left (\frac {r^2 - r^2}{r} \right) \right) - \left(\arcsin \left (\frac{0}{r} \right) + \left (\frac{0}{r}\right) \left (\frac{r^2 - 0^2}{r} \right) \right) \\ L &= 2r^2 (\arcsin(1) + 0 - (\arcsin(0) + 0)) \\ L &= 2r^2 \left (\frac{\pi}{2} \right) \\ L &= \pi r^2 \end{aligned} </math> </div></div> # Buktikan luas elips <math>L = \pi ab</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \frac{b \sqrt{a^2 - x^2}}{a} \text{ serta (-''a'', 0) dan (''a'', 0), } \\ L &= 4 \int_{0}^{r} \frac{b \sqrt{a^2 - x^2}}{a}\,dx \\ L &= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx \\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, } \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{\frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx}{4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx} \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{b}{a} \\ L_\text{elips} &= \frac{b}{a} L_\text{ling} \\ L_\text{elips} &= \frac{b}{a} \pi a^2 \\ L_\text{elips} &= \pi ab \end{aligned} </math> </div></div> # Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ x^-2x &= 4x+7 \\ x^2-6x-7 &= 0 \\ (x+1)(x-7) &= 0 \\ x=-1 &\text{ atau } x=7 \\ y &= 4(-1)+7 = 3 \\ y &= 4(7)+7 = 35 \\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut } \\ L &= \int_{-1}^{7} 4x+7-(x^2-2x)\,dx \\ &= \int_{-1}^{7} -x^2+6x+7\,dx \\ &= -\frac{x^3}{3}+3x^2+7x|_{-1}^{7} \\ &= -\frac{7^3}{3}+3(7)^2+7(7)-(-\frac{(-1)^3}{3}+3(-1)^2+7(-1)) \\ &= \frac{245}{3}-(-\frac{13}{3}) \\ &= \frac{258}{3} \\ &= 86 \\ \end{aligned} </math> </div></div> # Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ 6-1,5x &= x-4 \\ 2,5x &= 10 \\ x &= 4 \\ y &= 4-4 = 0 \\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut } \\ \text{untuk x=0 } \\ y &= 6-1,5(0) = 6 \\ y &= 0-4 = -4 \\ L &= \pi \int_{0}^{4} ((6-1,5x)^2-(x-4)^2)\,dx \\ &= \pi \int_{0}^{4} (36-18x+2,25x^2-(x^2-8x+16))\,dx \\ &= \pi \int_{0}^{4} (1,25x^2-10x+20)\,dx \\ &= \pi (\frac{1,25x^3}{3}-5x^2+20x)|_{0}^{4} \\ &= \pi (\frac{1,25(4)^3}{3}-5(4)^2+20(4)-(\frac{1,25(0)^3}{3}-5(0)^2+20(0))) \\ &= \pi (\frac{30}{3}-0) \\ &= 10\pi \\ \end{aligned} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 4fds0eyzxzm9rz9bw57j5b9hzho0xd3 115027 115026 2026-04-27T01:54:45Z ~2026-25546-23 43043 /* integral pecahan parsial */ 115027 wikitext text/x-wiki == Kaidah umum == :# <math>\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ konstan)}\,\!</math> :# <math>\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx</math> :# <math>\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx</math> :# <math>\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx</math> :# <math>\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\! </math> :# <math>\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C </math> :# <math>\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C </math> :# <math>\int f(x) \, dx = F(x) + C</math> :# <math>\int_a^b f(x) \, dx = F(b) - F(a)</math> :# <math>\int f(x) \circ g(x) dx</math> = f’(g(x)) g’(x) :# <math>\int f^{-1}(x) dx = \frac{1}{f'(f'(x))}</math> == rumus sederhana == :<math>\int \, dx = x + C</math> :<math>\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int {dx \over x} = \ln{\left|x\right|} + C</math> <math>\int {dx \over \sqrt{a^2-b^2 x^2}} = \frac{1}{b} \arcsin \, {bx \over a} + C</math> :<math>\int {dx \over {a^2+b^2x^2}} = {1 \over ab}\arctan \, {bx \over a} + C</math> <math>\int {dx \over x\sqrt{b^2x^2-a^2}} = {1 \over a}\arcsec \, {bx \over a} + C</math> ; Eksponen dan logaritma :<math>\int e^x\,dx = e^x + C</math> :<math>\int a^x\,dx = \frac{a^x}{\ln{a}} + C</math> :<math>\int \ln \, x\,dx = x \cdot \, \ln {x} - x + C</math> :<math>\int \, ^b\!\log \, x\,dx = x \cdot \, ^b\!\log \, x - x \cdot \,^b\!\log e + C</math> ; Trigonometri :<math>\int \sin \, x \, dx = -\cos \, x + C</math> :<math>\int \cos \, x \, dx = \sin \, x + C</math> :<math>\int \tan \, x \, dx = -\ln{\left| \cos \, x \right|} + C</math> :<math>\int \cot \, x \, dx = \ln{\left| \sin \, x \right|} + C</math> :<math>\int \sec \, x \, dx = \ln{\left| \sec \, x + \tan \, x \right|} + C</math> :<math>\int \csc \, x \, dx = -\ln{\left| \csc \, x + \cot \, x \right|} + C</math> ; Hiperbolik :<math>\int \sinh \, x \, dx = \cosh \, x + C</math> :<math>\int \cosh \, x \, dx = \sinh \, x + C</math> :<math>\int \tanh \, x \, dx = \ln{\left| \cosh \, x \right|} + C</math> :<math>\int \coth \, x \, dx = \ln{\left| \sinh \, x \right|} + C</math> :<math>\int \mbox{sech} \, x \, dx = \arctan \,(\sinh \, x) + C</math> :<math>\int \mbox{csch} \, x \, dx = \ln{\left| \tanh \, {x \over 2} \right|} + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : 1. <math>\int x^3 - cos 3x + e^{5x+2} - \frac{1}{2x+5}\,dx</math> : <math>= \frac{1}{3+1}x^{3+1} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : <math>= \frac{x^4}{4} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : 2. <math>\int x(x-5)^4\,dx</math> : <math>= \int x(x^4-4x^3(5)+6x^2(25)-4x(125)+625)\,dx</math> : <math>= \int x(x^4-20x^3+150x^2-500x+625)\,dx</math> : <math>= \int x^5-20x^4+150x^3-500x^2+625x\,dx</math> : <math>= \frac{x^6}{6}-4x^5+\frac{150x^4}{4}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> : <math>= \frac{x^6}{6}-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : 1. <math>\int \frac{\ln \, x}{x}\,dx</math> : <math>u = \ln \, x\, du = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{\ln(x)}{x} dx \\ \int u du \\ \frac{1}{2} u^2 + C \\ \frac{1}{2} ln^2 x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x-5,\, du = dx,\, x = u+5</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4\,dx \\ \int (u+5)u^4\,du \\ \int (u^5+5u^4)\,du \\ \frac{1}{6} u^6+u^5 + C \\ \frac{1}{6} (x-5)^6+(x-5)^5 + C \\ \frac{1}{6} (x^6-6x^5(5)+15x^4(25)-20x^3(125)+15x^2(625)-6x(3125)+15625)+(x^5-5x^4(5)+10x^3(25)-10x^2(125)+5x(625)+3125) + C \\ \frac{1}{6}x^6-5x^5+\frac{125x^4}{2}-\frac{1250x^3}{3}+\frac{3125x^2}{2}-3125x+\frac{15625}{6}+x^5-25x^4+250x^3-1250x^2+3125x+3125 + C \\ \frac{1}{6}x^6-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C \\ \end{aligned} </math> </div></div> === integral parsial === ;Cara 1: Rumus Integral parsial diselesaikan dengan rumus berikut. : <math>\int u\,dv = uv - \int v\,du </math> Berikut contoh penyelesaian cara parsial dengan rumus. : 1. <math>\int x \sin \, x \, dx</math> : <math>u = x, du = 1 \, dx, dv = \sin \, x \, dx, v = -\cos \, x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \,dx \\ (x)(-\cos \, x) - \int (-\cos \, x)(1 dx) \\ -x \cos \, x + \int \cos \, x dx \\ -x \cos \, x + \sin \, x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x,\, du = 1\,dx,\, dv = (x-5)^4\,dx,\, v = \frac{1}{5}(x-5)^5</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - \int (\frac{1}{5}(x-5)^5)(1\,dx) \\ \frac{x}{5}(x-5)^5 - \int \frac{1}{5}(x-5)^5\,dx \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6\,dx \\ \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x\,dx</math> : <math>u = \ln \,x,\, du = \frac{1}{x}\,dx,\, dv = x\,dx,\, v = \frac{x^2}{2}</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x\, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{x^2}{2})(\frac{1}{x}\,dx) \\ \frac{x^2}{2} \ln \,x - \int \frac{x^2}{2}(\frac{1}{x})\,dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \,dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x\,dx</math> ini terjadi dua kali integral parsial : <math>u = \sin \,x,\, du = \cos \,x\,dx,\, dv = e^x\,dx,\, v = e^x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x\, dx \\ (\sin \,x)(e^x) - \int (e^x)(\cos \,x\,dx) \\ e^x \sin \,x - \int e^x \cos \,x\,dx \\ u = \cos \,x,\, du = -\sin \,x\,dx,\, dv = e^x\,dx,\, v = e^x \\ (\cos \,x)(e^x) - \int (e^x)(-\sin \,x\,dx) \\ e^x \cos \,x + \int e^x \sin \,x\,dx \\ e^x \sin \,x - (e^x \cos \,x + \int e^x \sin \,x\,dx) \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x\,dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> ;Cara 2: Tabel Untuk <math display="inline">\int u\,dv</math>, berlaku ketentuan sebagai berikut. {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>u</math> || <math>dv</math> |- | - || <math>\frac{du}{dx}</math> || <math>v</math> |- | + || <math>\frac{d^2u}{dx^2}</math> || <math>\int v\,dx</math> |} Berikut contoh penyelesaian cara parsial dengan tabel. : 1. <math>\int x \sin \, x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>\sin \, x</math> |- | - || <math>1</math> || <math>-\cos \, x</math> |- | + || <math>0</math> || <math>-\sin \, x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \, dx \\ (x)(-\cos \, x) - (1)(-\sin \, x) + C \\ -x \cos \, x + \sin \, x + C \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4 \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>(x-5)^4</math> |- | - || <math>1</math> || <math>\frac{1}{5}(x-5)^5</math> |- | + || <math>0</math> || <math>\frac{1}{30}(x-5)^6</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - (1)(\frac{1}{30}(x-5)^6) + C \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6 + C \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\ln \,x</math> || <math>x</math> |- | - || <math>\frac{1}{x}</math> || <math>\frac{x^2}{2}</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x \, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{1}{x})(\frac{x^2}{2}) \, dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \, dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x \, dx</math> ini terjadi dua kali integral parsial {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\sin \,x</math> || <math>e^x</math> |- | - || <math>\cos \, x</math> || <math>e^x</math> |- | + || <math>-\sin \,x</math> || <math>e^x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x \, dx \\ (\sin \,x)(e^x) - (\cos \,x)(e^x) + \int (-\sin \,x)(e^x) \, dx \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x) \, dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : 1. <math>\int \frac{1}{x^2 - 4}\,dx</math> Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{1}{x^2 - 4} \\ =&\; \frac{1}{(x + 2)(x - 2)} \\ =&\; \frac{A}{x + 2} + \frac{B}{x - 2} \\ =&\; \frac{A(x - 2) + B(x + 2)}{x^2 - 4} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> : 2. <math>\int \frac{x^2+5x+4}{(x-3)^3}\,dx</math> : cara 1 Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^3} \\ \text{untuk mencari C yaitu } (3)^2+5(3)+4=28 \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x - 3)^2 + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x^2 - 6x + 9) + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 - 6Ax + 9A + Bx - 3B}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 + (- 6A + B)x +(9 - 3B)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ \end{aligned} </math> Kita tahu bahwa <math display="inline">A = 1</math> dan <math display="inline">-6A + B = 5</math> dapat diselesaikan, yaitu <math display="inline">A = 1</math> dan {{nowrap|<math display="inline">B = 11</math>.}} : <math> \begin{aligned} &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int (\frac{1}{x - 3} + \frac{11}{(x - 3)^2} + \frac{28}{(x - 3)^3})\,dx \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : cara 2 Pertama, ubah u=x-3 menjadi x=u+3 dan du=dx tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{(u+3)^2+5(u+3)+4}{u^3} \\ =&\; \frac{u^2+11u+28}{u^3} \\ &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int \frac{u^2+11u+28}{u^3}\,du \\ =&\; \int (\frac{u^2}{u^3} + \frac{11u}{u^3} + \frac{28}{u^3})\,du \\ =&\; \int (\frac{1}{u} + \frac{11}{u^2} + \frac{28}{u^3})\,du \\ =&\; \ln u + (-11 u^{-1}) + (-14 u^{-2}) + C \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : 3. <math>\int \frac{x^4+2x^3-13x^2-7x+21}{x^2-4x+3}\,dx</math> Pertama, penyelesaian dengan persamaan polinomial tersebut. : <math> \begin{aligned} =&\; \frac{x^4+2x^3−13x^2−7x+21}{x^2-4x+3} \\ =&\; x^2+6x+8 + \frac{7x-3}{x^2-4x+3} \\ \int \frac{x^4+2x^3−13x^2−7x+21}{x^2-4x+3}\,dx =&\; \int (x^2+6x+8 + \frac{7x-3}{x^2-4x+3}\,dx \\ =&\; \frac{1}{3}x^3+3x^2+8x+ \int \frac{7x-3}{x^2-4x+3}\,dx \\ Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{7x - 3}{x^2 - 4x + 3} \\ =&\; \frac{7x - 3}{(x - 1)(x - 3)} \\ =&\; \frac{A}{x - 1} + \frac{B}{x - 3} \\ =&\; \frac{A(x - 3) + B(x - 1)}{x^2 - 4x + 3} \\ =&\; \frac{(A + B)x + (-3A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 7</math> dan <math display="inline">-3A - B = -3</math> dapat diselesaikan, yaitu <math display="inline">A = 9</math> dan {{nowrap|<math display="inline">B = -2</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> === integral substitusi trigonometri === {| class="wikitable" |- ! Bentuk !! Substitusi Trigonometri |- | <math>\sqrt{a^2 - b^2 x^2}</math> || <math>x = \frac{a}{b} sin \, \alpha </math> |- | <math>\sqrt{a^2 + b^2 x^2}</math> || <math>x = \frac{a}{b} tan \, \alpha</math> |- | <math>\sqrt{b^2 x^2 - a^2}</math> || <math>x = \frac{a}{b} sec \, \alpha</math> |} Berikut contoh penyelesaian cara substitusi trigonometri. : <math>\int \frac{dx}{x^2 \sqrt{x^2 + 4}}</math> : <math>x = 2 \tan \, A, dx = 2 \sec^2 \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \int \frac{dx}{x^2\sqrt{x^2+4}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(2 \tan \, A)^2 \sqrt{4 + (2 \tan \, A)^2}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 + 4 \tan^2 \, A}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4(1 + \tan^2 \, A)}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 \sec^2 \, A }} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(4 \tan^2 \, A)(2 \sec \, A)} \\ =&\; \int \frac{\sec \, A \,dA}{4 \tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\sec \, A \,dA}{\tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A} \,dA \end{aligned} </math> Substitusi berikut dapat dibuat. : <math>\int \frac{\cos \, A}{\sin^2 \, A} \,dA</math> : <math>t = \sin \, A, dt = \cos \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A}\,dA \\ =&\; \frac{1}{4} \int \frac{dt}{t^2} \\ =&\; \frac{1}{4} \left(-\frac{1}{t}\right) + C \\ =&\; -\frac{1}{4 t} + C \\ =&\; -\frac{1}{4 \sin \, A} + C \end{aligned} </math> Ingat bahwa <math display="inline">\sin \, A = \frac{x}{\sqrt{x^2 + 4}}</math> berlaku. : <math> \begin{aligned} =&\; -\frac{1}{4 \sin \, A} + C \\ =&\; -\frac{\sqrt{x^2 + 4}}{4x} + C \end{aligned} </math> === integral mutlak === : <math>\int |f(x)| \,dA</math> buatlah <math>f(x) \ge 0</math> jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x) : <math>\int_a^c |f(x)| \, dx = \int_a^b f(x) \, dx + \int_b^c -f(x) \, dx = F(b) - F(a) - F(c) + F(b)</math> jika <math>|f(x)| = \begin{cases} f(x), & \mbox{jika } f(x) \ge b, \\ -f(x), & \mbox{jika } f(x) < b. \end{cases}</math>. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif. === integral fungsi ganjil dan integral fungsi genap === : <math>\int_{-a}^a f(x) \,dx</math> dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral: : integral fungsi ganjil : <math>\int_{-a}^a f(x) \,dx</math> = 0 : integral fungsi genap : <math>\int_{-a}^a f(x) \,dx</math> = <math>2 \int_{0}^a f(x) \,dx</math> === integral notasi === : <math>\int_a^b f(x) \,dx = \int_{a-p}^{b-p} f(x) \,dx = \int_{a+p}^{b+p} f(x) \,dx</math> : <math>\int_a^c f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx</math> : <math>\int_a^b f(x \pm k) \,dx = \int_{a \pm k}^{b \pm k} f(x) \,dx</math> === integral terbalik === : <math>\int_a^b f(x) \,dx = -\int_b^a f(x) \,dx</math> == Jenis integral lainnya == === panjang busur === ; Sumbu ''x'' : <math>S = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' : <math>S = \int_{y_1}^{y_2} \sqrt{1 + (f'(y))^2}\,dy</math> === luas daerah === ; Satu kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} f(x)\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} f(y)\,dy</math> ; Dua kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} (f(x_2) - f(x_1))\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} (f(y_2) - f(y_1))\,dy</math> : atau juga <math>L = \frac {D \sqrt{D}}{6 a^2}</math> === luas permukaan benda putar === ; Sumbu ''x'' sebagai poros : <math>L_p = 2 \pi \int_{x_1}^{x_2} f(x)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' sebagai poros : <math>L_p = 2 \pi \int_{y_1}^{y_2} f(y)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(y))^2}\,dy</math> === volume benda putar === ; Satu kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} (f(x))^2\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} (f(y))^2\,dy</math> ; Dua kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} ((f(x_2))^2 - (f(x_1))^2)\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} ((f(y_2))^2 - (f(y_1))^2)\,dy</math> : atau juga <math>V = |\frac {D^2 \sqrt{D}}{30 a^3}|</math> == Integral lipat == Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga. ;contoh # Tentukan hasil dari: * <math>\int (5x-1)(5x^2-2x+7)^4 dx</math> * <math>\int x^2 cos 3x dx</math> * <math>\int sin^2 x cos x dx</math> * <math>\int cos^2 x sin x dx</math> * <math>\int sec x dx</math> * <math>\int csc x dx</math> * <math>\int \frac{1}{2x^2-5x+3} dx</math> * <math>\int \frac{1}{\sqrt{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> * <math>\int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx</math> * <math>\int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx</math> * <math>\int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx</math> * <math>\int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx</math> * <math>\int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx \\ a=1, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-sin^2 x} \\ &= \frac{1}{cos^2 x} \\ &= sec^2 x \\ \int sec^2 x dx &= tan x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx \\ a=1, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-cos^2 x} \\ &= \frac{1}{sin^2 x} \\ &= csc^2 x \\ \int csc^2 x dx &= -cot x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx \\ a=sin x, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{sin x}{1-sin^2 x} \\ &= \frac{sin x}{cos^2 x} \\ &= sec x \cdot tan x \\ \int sec x \cdot tan x dx &= sec x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx \\ a=cos x, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{cos x}{1-cos^2 x} \\ &= \frac{cos x}{sin^2 x} \\ &= csc x \cdot cot x \\ \int csc x \cdot cot x dx &= -csc x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx \\ \text{misalkan } y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y &= \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y^2 &= \frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}} \\ &= \frac{x}{y} \\ y^3 &= x \\ 3y^2 dy &= dx \\ \int y \cdot 3y^2 dy \\ \int 3y^3 dy \\ \frac{3y^4}{4}+C \\ \end{aligned} </math> </div></div> # Tentukan hasil dari: * <math>\int_{1}^{5} 2x-3 dx</math> * <math>\int_{0}^{4} |x-2| dx</math> * <math>\int_{2}^{4} |3-x| dx</math> * <math>\int_{1}^{5} x^2-6x+8 dx</math> * <math>\int_{-3}^{5} |x^2-2x-8| dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} 2x-3 dx &= \left. x^2 - 3x \right|_{1}^{5} \\ &= 10 - (-2) \\ &= 12 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{0}^{4} |x-2| dx \\ \text{ tentukan nilai harga nol } \\ x-2 &= 0 \\ x &= 2 \\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti } \\ |x-2| = \begin{cases} x-2, & \mbox{jika } x \ge 2, \\ -(x-2), & \mbox{jika } x < 2. \end{cases} \\ \int_{0}^{4} |x-2| dx &= \int_{2}^{4} x-2 dx + \int_{0}^{2} -(x-2) dx \\ &= \left. (\frac{x^2}{2}-2x) \right|_{2}^{4} + \left. (\frac{-x^2}{2}+2x) \right|_{0}^{2} \\ &= 0 - (-2) + 2 - 0 \\ &= 4 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{2}^{4} |3-x| dx \\ \text{ tentukan nilai syarat-syarat } \\ 3-x &\ge 0 \\ \text{ harga nol } \\ x &= 3 \\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti } \\ |3-x| = \begin{cases} 3-x, & \mbox{jika } x \le 3, \\ -(3-x), & \mbox{jika } x > 3. \end{cases} \\ \int_{2}^{4} |3-x| dx &= \int_{2}^{3} 3-x dx + \int_{3}^{4} -(3-x) dx \\ &= \left. (3x-\frac{x^2}{2}) \right|_{2}^{3} + \left. (-3x+\frac{x^2}{2}) \right|_{3}^{4} \\ &= 4.5 - 4 - 4 - (-4.5) \\ &= 1 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} x^2-6x+8 dx &= \left. \frac{x^3}{3}-3x^2+8x \right|_{1}^{5} \\ &= \frac{20}{3} - \frac{16}{3} \\ &= \frac{4}{3} \\ &= 1\frac{1}{3} \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{-3}^{5} |x^2-2x-8| dx \\ \text{ tentukan nilai syarat-syarat } \\ x^2-2x-8 &\ge 0 \\ \text{ harga nol } \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x = 4 \\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti } \\ |x^2-2x-8| = \begin{cases} x^2-2x-8, & \mbox{jika } x \le -2, \\ -(x^2-2x-8), & \mbox{jika } -2 < x < 4, \\ x^2-2x-8, & \mbox{jika } x \ge 4 \end{cases} \\ \int_{-3}^{5} |x^2-2x-8| dx &= \int_{-3}^{-2} x^2-2x-8 dx + \int_{-2}^{4} -(x^2-2x-8) dx + \int_{4}^{5} x^2-2x-8 dx \\ &= \left. (\frac{x^3}{3}-x^2-8x) \right|_{-3}^{-2} + \left. (\frac{-x^3}{3}+x^2+8x) \right|_{-2}^{4} + \left. (\frac{x^3}{3}-x^2-8x) \right|_{4}^{5} \\ &= \frac{28}{3} - 6 + \frac{80}{3} - (-\frac{28}{3}) - \frac{70}{3} - (-\frac{80}{3}) \\ &= \frac{128}{3} \\ &= 42\frac{2}{3} \\ \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x\,dx \\ L &= \frac{1}{2} x^2 \\ L &= \frac{1}{2} xy \quad (y = x) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x^2</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x^2\,dx \\ L &= \frac{1}{3} x^3 \\ L &= \frac{1}{3} xy \quad (y = x^2) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = \sqrt{x}</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int \sqrt{x}\,dx \\ L &= \frac{2}{3} x^{\frac{3}{2}} \\ L &= \frac{2}{3} xy \quad (y = \sqrt{x}) \end{aligned} </math> </div></div> # Buktikan luas persegi <math>L = s^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = s \text{ dan titik (''s'', ''s''), } \\ L &= \int_{0}^{s} s\,dx \\ L &= sx |_{0}^{s} \\ L &= ss - 0 \\ L &= s^2 \end{aligned} </math> </div></div> # Buktikan luas persegi panjang <math>L = pl</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = l \text{dan titik (''p'', ''l''), } \\ L &= \int_{0}^{p} l\,dx \\ L &= lx |_{0}^{p} \\ L &= pl - 0 \\ L &= pl \end{aligned} </math> </div></div> * Buktikan luas segitiga <math>L = \frac{at}{2}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{-tx}{a} + t \text { dan titik (''a'', ''t''), } \\ L &= \int_{0}^{a} \left (\frac{-tx}{a} + t \right) \,dx \\ L &= \left. \frac{-tx^2}{2a} + tx \right|_{0}^{a} \\ L &= \frac{-ta^2}{2a} + ta - 0 + 0 \\ L &= \frac{-ta}{2} + ta \\ L &= \frac{at}{2} \end{aligned} </math> </div></div> # Buktikan volume tabung <math>V = \pi r^2t</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ Dengan posisi } y = r \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} r^2\,dx \\ V &= \pi \left. r^2x \right|_{0}^{t} \\ V &= \pi r^2t - 0 \\ V &= \pi r^2t \end{aligned} </math> </div></div> # Buktikan volume kerucut <math>V = \frac{\pi r^2t}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{rx}{t} \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} \left (\frac{rx}{t} \right)^2 \,dx \\ V &= \pi \left. \frac{r^2 x^3}{3t^2} \right|_{0}^{t} \\ V &= \pi \frac{r^2 t^3}{3t^2} - 0 \\ V &= \frac{\pi r^2t}{3} \end{aligned} </math> </div></div> # Buktikan volume bola <math>V = \frac{4 \pi r^3}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), } \\ V &= \pi \int_{-r}^{r} \left(\sqrt {r^2 - x^2}\right)^2\,dx \\ V &= \pi \int_{-r}^{r} r^2 - x^2 dx \\ V &= \pi \left. r^2x - \frac{x^3}{3} \right|_{-r}^{r} \\ V &= \pi \left (r^3 - \frac{r^3}{3} - \left (-r^3 + \frac{r^3}{3} \right) \right) \\ V &= \frac{4 \pi r^3}{3} \end{aligned} </math> </div></div> # Buktikan luas permukaan bola <math>L = 4 \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{selanjutnya } \\ ds &= \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ ds &= \sqrt{1 + \frac{x^2}{r^2 - x^2}}\,dx \\ ds &= \sqrt{\frac{r^2}{r^2 - x^2}}\,dx \\ ds &= \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ \text{sehingga } \\ L &= 2 \pi \int_{-r}^{r} \sqrt{r^2 - x^2} \cdot \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ L &= 2 \pi \int_{-r}^{r} r\,dx \\ L &= 2 \pi rx|_{-r}^{r} \\ L &= 2 \pi (r r - r (-r)) \\ L &= 2 \pi (r^2 + r^2) \\ L &= 4 \pi r^2 \end{aligned} </math> </div></div> # Buktikan keliling lingkaran <math>K = 2 \pi r</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0) } \\ \text{kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{sehingga } \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{x^2}{r^2 - x^2})}\,dx \\ K &= 4 \int_{0}^{r} \sqrt\frac{r^2}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \sqrt\frac{1}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \frac{1}{\sqrt{r^2 - x^2}}\,dx \\ K &= 4r \left. \arcsin \left (\frac{x}{r} \right) \right|_{0}^{r} \\ K &= 4r \left (\arcsin \left (\frac{r}{r}\right) - \arcsin \left (\frac{0}{r} \right) \right) \\ K &= 4r \left (\arcsin\left (1 \right) - \arcsin \left(0 \right) \right)) \\ K &= 4r \left (\frac{\pi}{2} \right) \\ K &= 2 \pi r \end{aligned} </math> </div></div> # Buktikan luas lingkaran <math>L = \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. } \\ \sin(\theta) &= \frac{x}{r} \\ x &= r \sin(\theta) \\ dx &= r \cos(\theta)\,d\theta \\ \text{dengan turunan di atas } \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - x^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - (r \sin(\theta))^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - r^2 \sin^2(\theta)}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 (1 - \sin^2(\theta))}\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta)\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta) (r \cos(\theta)\,d\theta) \\ L &= 4 \int_{0}^{r} r^2 \cos^2(\theta)\,d\theta \\ L &= 4r^2 \int_{0}^{r} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta \\ L &= 2r^2 \int_{0}^{r} (1 + \cos(2\theta))\,d\theta \\ L &= 2r^2 \left (\theta + \frac{1}{2} \sin(2\theta) \right)|_{0}^{r} \\ L &= 2r^2 (\theta + \sin(\theta) \cos(\theta))|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left (\frac{x}{r} \right) + \left (\frac{x}{r} \right) \left (\frac{r^2 - x^2}{r} \right) \right)|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left(\frac{r}{r} \right) + \left (\frac{r}{r} \right) \left (\frac {r^2 - r^2}{r} \right) \right) - \left(\arcsin \left (\frac{0}{r} \right) + \left (\frac{0}{r}\right) \left (\frac{r^2 - 0^2}{r} \right) \right) \\ L &= 2r^2 (\arcsin(1) + 0 - (\arcsin(0) + 0)) \\ L &= 2r^2 \left (\frac{\pi}{2} \right) \\ L &= \pi r^2 \end{aligned} </math> </div></div> # Buktikan luas elips <math>L = \pi ab</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \frac{b \sqrt{a^2 - x^2}}{a} \text{ serta (-''a'', 0) dan (''a'', 0), } \\ L &= 4 \int_{0}^{r} \frac{b \sqrt{a^2 - x^2}}{a}\,dx \\ L &= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx \\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, } \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{\frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx}{4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx} \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{b}{a} \\ L_\text{elips} &= \frac{b}{a} L_\text{ling} \\ L_\text{elips} &= \frac{b}{a} \pi a^2 \\ L_\text{elips} &= \pi ab \end{aligned} </math> </div></div> # Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ x^-2x &= 4x+7 \\ x^2-6x-7 &= 0 \\ (x+1)(x-7) &= 0 \\ x=-1 &\text{ atau } x=7 \\ y &= 4(-1)+7 = 3 \\ y &= 4(7)+7 = 35 \\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut } \\ L &= \int_{-1}^{7} 4x+7-(x^2-2x)\,dx \\ &= \int_{-1}^{7} -x^2+6x+7\,dx \\ &= -\frac{x^3}{3}+3x^2+7x|_{-1}^{7} \\ &= -\frac{7^3}{3}+3(7)^2+7(7)-(-\frac{(-1)^3}{3}+3(-1)^2+7(-1)) \\ &= \frac{245}{3}-(-\frac{13}{3}) \\ &= \frac{258}{3} \\ &= 86 \\ \end{aligned} </math> </div></div> # Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ 6-1,5x &= x-4 \\ 2,5x &= 10 \\ x &= 4 \\ y &= 4-4 = 0 \\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut } \\ \text{untuk x=0 } \\ y &= 6-1,5(0) = 6 \\ y &= 0-4 = -4 \\ L &= \pi \int_{0}^{4} ((6-1,5x)^2-(x-4)^2)\,dx \\ &= \pi \int_{0}^{4} (36-18x+2,25x^2-(x^2-8x+16))\,dx \\ &= \pi \int_{0}^{4} (1,25x^2-10x+20)\,dx \\ &= \pi (\frac{1,25x^3}{3}-5x^2+20x)|_{0}^{4} \\ &= \pi (\frac{1,25(4)^3}{3}-5(4)^2+20(4)-(\frac{1,25(0)^3}{3}-5(0)^2+20(0))) \\ &= \pi (\frac{30}{3}-0) \\ &= 10\pi \\ \end{aligned} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 0dglacl4nbjmr3ixduffa8c39658olt 115028 115027 2026-04-27T01:56:02Z ~2026-25546-23 43043 /* integral pecahan parsial */ 115028 wikitext text/x-wiki == Kaidah umum == :# <math>\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ konstan)}\,\!</math> :# <math>\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx</math> :# <math>\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx</math> :# <math>\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx</math> :# <math>\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\! </math> :# <math>\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C </math> :# <math>\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C </math> :# <math>\int f(x) \, dx = F(x) + C</math> :# <math>\int_a^b f(x) \, dx = F(b) - F(a)</math> :# <math>\int f(x) \circ g(x) dx</math> = f’(g(x)) g’(x) :# <math>\int f^{-1}(x) dx = \frac{1}{f'(f'(x))}</math> == rumus sederhana == :<math>\int \, dx = x + C</math> :<math>\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int {dx \over x} = \ln{\left|x\right|} + C</math> <math>\int {dx \over \sqrt{a^2-b^2 x^2}} = \frac{1}{b} \arcsin \, {bx \over a} + C</math> :<math>\int {dx \over {a^2+b^2x^2}} = {1 \over ab}\arctan \, {bx \over a} + C</math> <math>\int {dx \over x\sqrt{b^2x^2-a^2}} = {1 \over a}\arcsec \, {bx \over a} + C</math> ; Eksponen dan logaritma :<math>\int e^x\,dx = e^x + C</math> :<math>\int a^x\,dx = \frac{a^x}{\ln{a}} + C</math> :<math>\int \ln \, x\,dx = x \cdot \, \ln {x} - x + C</math> :<math>\int \, ^b\!\log \, x\,dx = x \cdot \, ^b\!\log \, x - x \cdot \,^b\!\log e + C</math> ; Trigonometri :<math>\int \sin \, x \, dx = -\cos \, x + C</math> :<math>\int \cos \, x \, dx = \sin \, x + C</math> :<math>\int \tan \, x \, dx = -\ln{\left| \cos \, x \right|} + C</math> :<math>\int \cot \, x \, dx = \ln{\left| \sin \, x \right|} + C</math> :<math>\int \sec \, x \, dx = \ln{\left| \sec \, x + \tan \, x \right|} + C</math> :<math>\int \csc \, x \, dx = -\ln{\left| \csc \, x + \cot \, x \right|} + C</math> ; Hiperbolik :<math>\int \sinh \, x \, dx = \cosh \, x + C</math> :<math>\int \cosh \, x \, dx = \sinh \, x + C</math> :<math>\int \tanh \, x \, dx = \ln{\left| \cosh \, x \right|} + C</math> :<math>\int \coth \, x \, dx = \ln{\left| \sinh \, x \right|} + C</math> :<math>\int \mbox{sech} \, x \, dx = \arctan \,(\sinh \, x) + C</math> :<math>\int \mbox{csch} \, x \, dx = \ln{\left| \tanh \, {x \over 2} \right|} + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : 1. <math>\int x^3 - cos 3x + e^{5x+2} - \frac{1}{2x+5}\,dx</math> : <math>= \frac{1}{3+1}x^{3+1} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : <math>= \frac{x^4}{4} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : 2. <math>\int x(x-5)^4\,dx</math> : <math>= \int x(x^4-4x^3(5)+6x^2(25)-4x(125)+625)\,dx</math> : <math>= \int x(x^4-20x^3+150x^2-500x+625)\,dx</math> : <math>= \int x^5-20x^4+150x^3-500x^2+625x\,dx</math> : <math>= \frac{x^6}{6}-4x^5+\frac{150x^4}{4}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> : <math>= \frac{x^6}{6}-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : 1. <math>\int \frac{\ln \, x}{x}\,dx</math> : <math>u = \ln \, x\, du = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{\ln(x)}{x} dx \\ \int u du \\ \frac{1}{2} u^2 + C \\ \frac{1}{2} ln^2 x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x-5,\, du = dx,\, x = u+5</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4\,dx \\ \int (u+5)u^4\,du \\ \int (u^5+5u^4)\,du \\ \frac{1}{6} u^6+u^5 + C \\ \frac{1}{6} (x-5)^6+(x-5)^5 + C \\ \frac{1}{6} (x^6-6x^5(5)+15x^4(25)-20x^3(125)+15x^2(625)-6x(3125)+15625)+(x^5-5x^4(5)+10x^3(25)-10x^2(125)+5x(625)+3125) + C \\ \frac{1}{6}x^6-5x^5+\frac{125x^4}{2}-\frac{1250x^3}{3}+\frac{3125x^2}{2}-3125x+\frac{15625}{6}+x^5-25x^4+250x^3-1250x^2+3125x+3125 + C \\ \frac{1}{6}x^6-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C \\ \end{aligned} </math> </div></div> === integral parsial === ;Cara 1: Rumus Integral parsial diselesaikan dengan rumus berikut. : <math>\int u\,dv = uv - \int v\,du </math> Berikut contoh penyelesaian cara parsial dengan rumus. : 1. <math>\int x \sin \, x \, dx</math> : <math>u = x, du = 1 \, dx, dv = \sin \, x \, dx, v = -\cos \, x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \,dx \\ (x)(-\cos \, x) - \int (-\cos \, x)(1 dx) \\ -x \cos \, x + \int \cos \, x dx \\ -x \cos \, x + \sin \, x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x,\, du = 1\,dx,\, dv = (x-5)^4\,dx,\, v = \frac{1}{5}(x-5)^5</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - \int (\frac{1}{5}(x-5)^5)(1\,dx) \\ \frac{x}{5}(x-5)^5 - \int \frac{1}{5}(x-5)^5\,dx \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6\,dx \\ \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x\,dx</math> : <math>u = \ln \,x,\, du = \frac{1}{x}\,dx,\, dv = x\,dx,\, v = \frac{x^2}{2}</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x\, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{x^2}{2})(\frac{1}{x}\,dx) \\ \frac{x^2}{2} \ln \,x - \int \frac{x^2}{2}(\frac{1}{x})\,dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \,dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x\,dx</math> ini terjadi dua kali integral parsial : <math>u = \sin \,x,\, du = \cos \,x\,dx,\, dv = e^x\,dx,\, v = e^x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x\, dx \\ (\sin \,x)(e^x) - \int (e^x)(\cos \,x\,dx) \\ e^x \sin \,x - \int e^x \cos \,x\,dx \\ u = \cos \,x,\, du = -\sin \,x\,dx,\, dv = e^x\,dx,\, v = e^x \\ (\cos \,x)(e^x) - \int (e^x)(-\sin \,x\,dx) \\ e^x \cos \,x + \int e^x \sin \,x\,dx \\ e^x \sin \,x - (e^x \cos \,x + \int e^x \sin \,x\,dx) \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x\,dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> ;Cara 2: Tabel Untuk <math display="inline">\int u\,dv</math>, berlaku ketentuan sebagai berikut. {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>u</math> || <math>dv</math> |- | - || <math>\frac{du}{dx}</math> || <math>v</math> |- | + || <math>\frac{d^2u}{dx^2}</math> || <math>\int v\,dx</math> |} Berikut contoh penyelesaian cara parsial dengan tabel. : 1. <math>\int x \sin \, x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>\sin \, x</math> |- | - || <math>1</math> || <math>-\cos \, x</math> |- | + || <math>0</math> || <math>-\sin \, x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \, dx \\ (x)(-\cos \, x) - (1)(-\sin \, x) + C \\ -x \cos \, x + \sin \, x + C \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4 \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>(x-5)^4</math> |- | - || <math>1</math> || <math>\frac{1}{5}(x-5)^5</math> |- | + || <math>0</math> || <math>\frac{1}{30}(x-5)^6</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - (1)(\frac{1}{30}(x-5)^6) + C \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6 + C \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\ln \,x</math> || <math>x</math> |- | - || <math>\frac{1}{x}</math> || <math>\frac{x^2}{2}</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x \, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{1}{x})(\frac{x^2}{2}) \, dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \, dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x \, dx</math> ini terjadi dua kali integral parsial {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\sin \,x</math> || <math>e^x</math> |- | - || <math>\cos \, x</math> || <math>e^x</math> |- | + || <math>-\sin \,x</math> || <math>e^x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x \, dx \\ (\sin \,x)(e^x) - (\cos \,x)(e^x) + \int (-\sin \,x)(e^x) \, dx \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x) \, dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : 1. <math>\int \frac{1}{x^2 - 4}\,dx</math> Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{1}{x^2 - 4} \\ =&\; \frac{1}{(x + 2)(x - 2)} \\ =&\; \frac{A}{x + 2} + \frac{B}{x - 2} \\ =&\; \frac{A(x - 2) + B(x + 2)}{x^2 - 4} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> : 2. <math>\int \frac{x^2+5x+4}{(x-3)^3}\,dx</math> : cara 1 Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^3} \\ \text{untuk mencari C yaitu } (3)^2+5(3)+4=28 \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x - 3)^2 + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x^2 - 6x + 9) + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 - 6Ax + 9A + Bx - 3B}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 + (- 6A + B)x +(9 - 3B)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ \end{aligned} </math> Kita tahu bahwa <math display="inline">A = 1</math> dan <math display="inline">-6A + B = 5</math> dapat diselesaikan, yaitu <math display="inline">A = 1</math> dan {{nowrap|<math display="inline">B = 11</math>.}} : <math> \begin{aligned} &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int (\frac{1}{x - 3} + \frac{11}{(x - 3)^2} + \frac{28}{(x - 3)^3})\,dx \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : cara 2 Pertama, ubah u=x-3 menjadi x=u+3 dan du=dx tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{(u+3)^2+5(u+3)+4}{u^3} \\ =&\; \frac{u^2+11u+28}{u^3} \\ &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int \frac{u^2+11u+28}{u^3}\,du \\ =&\; \int (\frac{u^2}{u^3} + \frac{11u}{u^3} + \frac{28}{u^3})\,du \\ =&\; \int (\frac{1}{u} + \frac{11}{u^2} + \frac{28}{u^3})\,du \\ =&\; \ln u + (-11 u^{-1}) + (-14 u^{-2}) + C \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : 3. <math>\int \frac{x^4+2x^3-13x^2-7x+21}{x^2-4x+3}\,dx</math> Pertama, penyelesaian dengan persamaan polinomial tersebut. : <math> \begin{aligned} =&\; \frac{x^4+2x^3-13x^2-7x+21}{x^2-4x+3} \\ =&\; x^2+6x+8 + \frac{7x-3}{x^2-4x+3} \\ \int \frac{x^4+2x^3-13x^2-7x+21}{x^2-4x+3}\,dx =&\; \int (x^2+6x+8 + \frac{7x-3}{x^2-4x+3}\,dx \\ =&\; \frac{1}{3}x^3+3x^2+8x+ \int \frac{7x-3}{x^2-4x+3}\,dx \\ Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{7x - 3}{x^2 - 4x + 3} \\ =&\; \frac{7x - 3}{(x - 1)(x - 3)} \\ =&\; \frac{A}{x - 1} + \frac{B}{x - 3} \\ =&\; \frac{A(x - 3) + B(x - 1)}{x^2 - 4x + 3} \\ =&\; \frac{(A + B)x + (-3A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 7</math> dan <math display="inline">-3A - B = -3</math> dapat diselesaikan, yaitu <math display="inline">A = 9</math> dan {{nowrap|<math display="inline">B = -2</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> === integral substitusi trigonometri === {| class="wikitable" |- ! Bentuk !! Substitusi Trigonometri |- | <math>\sqrt{a^2 - b^2 x^2}</math> || <math>x = \frac{a}{b} sin \, \alpha </math> |- | <math>\sqrt{a^2 + b^2 x^2}</math> || <math>x = \frac{a}{b} tan \, \alpha</math> |- | <math>\sqrt{b^2 x^2 - a^2}</math> || <math>x = \frac{a}{b} sec \, \alpha</math> |} Berikut contoh penyelesaian cara substitusi trigonometri. : <math>\int \frac{dx}{x^2 \sqrt{x^2 + 4}}</math> : <math>x = 2 \tan \, A, dx = 2 \sec^2 \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \int \frac{dx}{x^2\sqrt{x^2+4}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(2 \tan \, A)^2 \sqrt{4 + (2 \tan \, A)^2}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 + 4 \tan^2 \, A}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4(1 + \tan^2 \, A)}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 \sec^2 \, A }} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(4 \tan^2 \, A)(2 \sec \, A)} \\ =&\; \int \frac{\sec \, A \,dA}{4 \tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\sec \, A \,dA}{\tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A} \,dA \end{aligned} </math> Substitusi berikut dapat dibuat. : <math>\int \frac{\cos \, A}{\sin^2 \, A} \,dA</math> : <math>t = \sin \, A, dt = \cos \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A}\,dA \\ =&\; \frac{1}{4} \int \frac{dt}{t^2} \\ =&\; \frac{1}{4} \left(-\frac{1}{t}\right) + C \\ =&\; -\frac{1}{4 t} + C \\ =&\; -\frac{1}{4 \sin \, A} + C \end{aligned} </math> Ingat bahwa <math display="inline">\sin \, A = \frac{x}{\sqrt{x^2 + 4}}</math> berlaku. : <math> \begin{aligned} =&\; -\frac{1}{4 \sin \, A} + C \\ =&\; -\frac{\sqrt{x^2 + 4}}{4x} + C \end{aligned} </math> === integral mutlak === : <math>\int |f(x)| \,dA</math> buatlah <math>f(x) \ge 0</math> jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x) : <math>\int_a^c |f(x)| \, dx = \int_a^b f(x) \, dx + \int_b^c -f(x) \, dx = F(b) - F(a) - F(c) + F(b)</math> jika <math>|f(x)| = \begin{cases} f(x), & \mbox{jika } f(x) \ge b, \\ -f(x), & \mbox{jika } f(x) < b. \end{cases}</math>. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif. === integral fungsi ganjil dan integral fungsi genap === : <math>\int_{-a}^a f(x) \,dx</math> dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral: : integral fungsi ganjil : <math>\int_{-a}^a f(x) \,dx</math> = 0 : integral fungsi genap : <math>\int_{-a}^a f(x) \,dx</math> = <math>2 \int_{0}^a f(x) \,dx</math> === integral notasi === : <math>\int_a^b f(x) \,dx = \int_{a-p}^{b-p} f(x) \,dx = \int_{a+p}^{b+p} f(x) \,dx</math> : <math>\int_a^c f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx</math> : <math>\int_a^b f(x \pm k) \,dx = \int_{a \pm k}^{b \pm k} f(x) \,dx</math> === integral terbalik === : <math>\int_a^b f(x) \,dx = -\int_b^a f(x) \,dx</math> == Jenis integral lainnya == === panjang busur === ; Sumbu ''x'' : <math>S = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' : <math>S = \int_{y_1}^{y_2} \sqrt{1 + (f'(y))^2}\,dy</math> === luas daerah === ; Satu kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} f(x)\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} f(y)\,dy</math> ; Dua kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} (f(x_2) - f(x_1))\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} (f(y_2) - f(y_1))\,dy</math> : atau juga <math>L = \frac {D \sqrt{D}}{6 a^2}</math> === luas permukaan benda putar === ; Sumbu ''x'' sebagai poros : <math>L_p = 2 \pi \int_{x_1}^{x_2} f(x)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' sebagai poros : <math>L_p = 2 \pi \int_{y_1}^{y_2} f(y)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(y))^2}\,dy</math> === volume benda putar === ; Satu kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} (f(x))^2\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} (f(y))^2\,dy</math> ; Dua kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} ((f(x_2))^2 - (f(x_1))^2)\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} ((f(y_2))^2 - (f(y_1))^2)\,dy</math> : atau juga <math>V = |\frac {D^2 \sqrt{D}}{30 a^3}|</math> == Integral lipat == Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga. ;contoh # Tentukan hasil dari: * <math>\int (5x-1)(5x^2-2x+7)^4 dx</math> * <math>\int x^2 cos 3x dx</math> * <math>\int sin^2 x cos x dx</math> * <math>\int cos^2 x sin x dx</math> * <math>\int sec x dx</math> * <math>\int csc x dx</math> * <math>\int \frac{1}{2x^2-5x+3} dx</math> * <math>\int \frac{1}{\sqrt{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> * <math>\int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx</math> * <math>\int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx</math> * <math>\int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx</math> * <math>\int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx</math> * <math>\int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx \\ a=1, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-sin^2 x} \\ &= \frac{1}{cos^2 x} \\ &= sec^2 x \\ \int sec^2 x dx &= tan x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx \\ a=1, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-cos^2 x} \\ &= \frac{1}{sin^2 x} \\ &= csc^2 x \\ \int csc^2 x dx &= -cot x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx \\ a=sin x, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{sin x}{1-sin^2 x} \\ &= \frac{sin x}{cos^2 x} \\ &= sec x \cdot tan x \\ \int sec x \cdot tan x dx &= sec x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx \\ a=cos x, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{cos x}{1-cos^2 x} \\ &= \frac{cos x}{sin^2 x} \\ &= csc x \cdot cot x \\ \int csc x \cdot cot x dx &= -csc x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx \\ \text{misalkan } y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y &= \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y^2 &= \frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}} \\ &= \frac{x}{y} \\ y^3 &= x \\ 3y^2 dy &= dx \\ \int y \cdot 3y^2 dy \\ \int 3y^3 dy \\ \frac{3y^4}{4}+C \\ \end{aligned} </math> </div></div> # Tentukan hasil dari: * <math>\int_{1}^{5} 2x-3 dx</math> * <math>\int_{0}^{4} |x-2| dx</math> * <math>\int_{2}^{4} |3-x| dx</math> * <math>\int_{1}^{5} x^2-6x+8 dx</math> * <math>\int_{-3}^{5} |x^2-2x-8| dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} 2x-3 dx &= \left. x^2 - 3x \right|_{1}^{5} \\ &= 10 - (-2) \\ &= 12 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{0}^{4} |x-2| dx \\ \text{ tentukan nilai harga nol } \\ x-2 &= 0 \\ x &= 2 \\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti } \\ |x-2| = \begin{cases} x-2, & \mbox{jika } x \ge 2, \\ -(x-2), & \mbox{jika } x < 2. \end{cases} \\ \int_{0}^{4} |x-2| dx &= \int_{2}^{4} x-2 dx + \int_{0}^{2} -(x-2) dx \\ &= \left. (\frac{x^2}{2}-2x) \right|_{2}^{4} + \left. (\frac{-x^2}{2}+2x) \right|_{0}^{2} \\ &= 0 - (-2) + 2 - 0 \\ &= 4 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{2}^{4} |3-x| dx \\ \text{ tentukan nilai syarat-syarat } \\ 3-x &\ge 0 \\ \text{ harga nol } \\ x &= 3 \\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti } \\ |3-x| = \begin{cases} 3-x, & \mbox{jika } x \le 3, \\ -(3-x), & \mbox{jika } x > 3. \end{cases} \\ \int_{2}^{4} |3-x| dx &= \int_{2}^{3} 3-x dx + \int_{3}^{4} -(3-x) dx \\ &= \left. (3x-\frac{x^2}{2}) \right|_{2}^{3} + \left. (-3x+\frac{x^2}{2}) \right|_{3}^{4} \\ &= 4.5 - 4 - 4 - (-4.5) \\ &= 1 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} x^2-6x+8 dx &= \left. \frac{x^3}{3}-3x^2+8x \right|_{1}^{5} \\ &= \frac{20}{3} - \frac{16}{3} \\ &= \frac{4}{3} \\ &= 1\frac{1}{3} \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{-3}^{5} |x^2-2x-8| dx \\ \text{ tentukan nilai syarat-syarat } \\ x^2-2x-8 &\ge 0 \\ \text{ harga nol } \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x = 4 \\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti } \\ |x^2-2x-8| = \begin{cases} x^2-2x-8, & \mbox{jika } x \le -2, \\ -(x^2-2x-8), & \mbox{jika } -2 < x < 4, \\ x^2-2x-8, & \mbox{jika } x \ge 4 \end{cases} \\ \int_{-3}^{5} |x^2-2x-8| dx &= \int_{-3}^{-2} x^2-2x-8 dx + \int_{-2}^{4} -(x^2-2x-8) dx + \int_{4}^{5} x^2-2x-8 dx \\ &= \left. (\frac{x^3}{3}-x^2-8x) \right|_{-3}^{-2} + \left. (\frac{-x^3}{3}+x^2+8x) \right|_{-2}^{4} + \left. (\frac{x^3}{3}-x^2-8x) \right|_{4}^{5} \\ &= \frac{28}{3} - 6 + \frac{80}{3} - (-\frac{28}{3}) - \frac{70}{3} - (-\frac{80}{3}) \\ &= \frac{128}{3} \\ &= 42\frac{2}{3} \\ \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x\,dx \\ L &= \frac{1}{2} x^2 \\ L &= \frac{1}{2} xy \quad (y = x) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x^2</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x^2\,dx \\ L &= \frac{1}{3} x^3 \\ L &= \frac{1}{3} xy \quad (y = x^2) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = \sqrt{x}</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int \sqrt{x}\,dx \\ L &= \frac{2}{3} x^{\frac{3}{2}} \\ L &= \frac{2}{3} xy \quad (y = \sqrt{x}) \end{aligned} </math> </div></div> # Buktikan luas persegi <math>L = s^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = s \text{ dan titik (''s'', ''s''), } \\ L &= \int_{0}^{s} s\,dx \\ L &= sx |_{0}^{s} \\ L &= ss - 0 \\ L &= s^2 \end{aligned} </math> </div></div> # Buktikan luas persegi panjang <math>L = pl</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = l \text{dan titik (''p'', ''l''), } \\ L &= \int_{0}^{p} l\,dx \\ L &= lx |_{0}^{p} \\ L &= pl - 0 \\ L &= pl \end{aligned} </math> </div></div> * Buktikan luas segitiga <math>L = \frac{at}{2}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{-tx}{a} + t \text { dan titik (''a'', ''t''), } \\ L &= \int_{0}^{a} \left (\frac{-tx}{a} + t \right) \,dx \\ L &= \left. \frac{-tx^2}{2a} + tx \right|_{0}^{a} \\ L &= \frac{-ta^2}{2a} + ta - 0 + 0 \\ L &= \frac{-ta}{2} + ta \\ L &= \frac{at}{2} \end{aligned} </math> </div></div> # Buktikan volume tabung <math>V = \pi r^2t</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ Dengan posisi } y = r \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} r^2\,dx \\ V &= \pi \left. r^2x \right|_{0}^{t} \\ V &= \pi r^2t - 0 \\ V &= \pi r^2t \end{aligned} </math> </div></div> # Buktikan volume kerucut <math>V = \frac{\pi r^2t}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{rx}{t} \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} \left (\frac{rx}{t} \right)^2 \,dx \\ V &= \pi \left. \frac{r^2 x^3}{3t^2} \right|_{0}^{t} \\ V &= \pi \frac{r^2 t^3}{3t^2} - 0 \\ V &= \frac{\pi r^2t}{3} \end{aligned} </math> </div></div> # Buktikan volume bola <math>V = \frac{4 \pi r^3}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), } \\ V &= \pi \int_{-r}^{r} \left(\sqrt {r^2 - x^2}\right)^2\,dx \\ V &= \pi \int_{-r}^{r} r^2 - x^2 dx \\ V &= \pi \left. r^2x - \frac{x^3}{3} \right|_{-r}^{r} \\ V &= \pi \left (r^3 - \frac{r^3}{3} - \left (-r^3 + \frac{r^3}{3} \right) \right) \\ V &= \frac{4 \pi r^3}{3} \end{aligned} </math> </div></div> # Buktikan luas permukaan bola <math>L = 4 \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{selanjutnya } \\ ds &= \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ ds &= \sqrt{1 + \frac{x^2}{r^2 - x^2}}\,dx \\ ds &= \sqrt{\frac{r^2}{r^2 - x^2}}\,dx \\ ds &= \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ \text{sehingga } \\ L &= 2 \pi \int_{-r}^{r} \sqrt{r^2 - x^2} \cdot \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ L &= 2 \pi \int_{-r}^{r} r\,dx \\ L &= 2 \pi rx|_{-r}^{r} \\ L &= 2 \pi (r r - r (-r)) \\ L &= 2 \pi (r^2 + r^2) \\ L &= 4 \pi r^2 \end{aligned} </math> </div></div> # Buktikan keliling lingkaran <math>K = 2 \pi r</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0) } \\ \text{kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{sehingga } \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{x^2}{r^2 - x^2})}\,dx \\ K &= 4 \int_{0}^{r} \sqrt\frac{r^2}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \sqrt\frac{1}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \frac{1}{\sqrt{r^2 - x^2}}\,dx \\ K &= 4r \left. \arcsin \left (\frac{x}{r} \right) \right|_{0}^{r} \\ K &= 4r \left (\arcsin \left (\frac{r}{r}\right) - \arcsin \left (\frac{0}{r} \right) \right) \\ K &= 4r \left (\arcsin\left (1 \right) - \arcsin \left(0 \right) \right)) \\ K &= 4r \left (\frac{\pi}{2} \right) \\ K &= 2 \pi r \end{aligned} </math> </div></div> # Buktikan luas lingkaran <math>L = \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. } \\ \sin(\theta) &= \frac{x}{r} \\ x &= r \sin(\theta) \\ dx &= r \cos(\theta)\,d\theta \\ \text{dengan turunan di atas } \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - x^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - (r \sin(\theta))^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - r^2 \sin^2(\theta)}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 (1 - \sin^2(\theta))}\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta)\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta) (r \cos(\theta)\,d\theta) \\ L &= 4 \int_{0}^{r} r^2 \cos^2(\theta)\,d\theta \\ L &= 4r^2 \int_{0}^{r} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta \\ L &= 2r^2 \int_{0}^{r} (1 + \cos(2\theta))\,d\theta \\ L &= 2r^2 \left (\theta + \frac{1}{2} \sin(2\theta) \right)|_{0}^{r} \\ L &= 2r^2 (\theta + \sin(\theta) \cos(\theta))|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left (\frac{x}{r} \right) + \left (\frac{x}{r} \right) \left (\frac{r^2 - x^2}{r} \right) \right)|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left(\frac{r}{r} \right) + \left (\frac{r}{r} \right) \left (\frac {r^2 - r^2}{r} \right) \right) - \left(\arcsin \left (\frac{0}{r} \right) + \left (\frac{0}{r}\right) \left (\frac{r^2 - 0^2}{r} \right) \right) \\ L &= 2r^2 (\arcsin(1) + 0 - (\arcsin(0) + 0)) \\ L &= 2r^2 \left (\frac{\pi}{2} \right) \\ L &= \pi r^2 \end{aligned} </math> </div></div> # Buktikan luas elips <math>L = \pi ab</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \frac{b \sqrt{a^2 - x^2}}{a} \text{ serta (-''a'', 0) dan (''a'', 0), } \\ L &= 4 \int_{0}^{r} \frac{b \sqrt{a^2 - x^2}}{a}\,dx \\ L &= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx \\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, } \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{\frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx}{4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx} \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{b}{a} \\ L_\text{elips} &= \frac{b}{a} L_\text{ling} \\ L_\text{elips} &= \frac{b}{a} \pi a^2 \\ L_\text{elips} &= \pi ab \end{aligned} </math> </div></div> # Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ x^-2x &= 4x+7 \\ x^2-6x-7 &= 0 \\ (x+1)(x-7) &= 0 \\ x=-1 &\text{ atau } x=7 \\ y &= 4(-1)+7 = 3 \\ y &= 4(7)+7 = 35 \\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut } \\ L &= \int_{-1}^{7} 4x+7-(x^2-2x)\,dx \\ &= \int_{-1}^{7} -x^2+6x+7\,dx \\ &= -\frac{x^3}{3}+3x^2+7x|_{-1}^{7} \\ &= -\frac{7^3}{3}+3(7)^2+7(7)-(-\frac{(-1)^3}{3}+3(-1)^2+7(-1)) \\ &= \frac{245}{3}-(-\frac{13}{3}) \\ &= \frac{258}{3} \\ &= 86 \\ \end{aligned} </math> </div></div> # Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ 6-1,5x &= x-4 \\ 2,5x &= 10 \\ x &= 4 \\ y &= 4-4 = 0 \\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut } \\ \text{untuk x=0 } \\ y &= 6-1,5(0) = 6 \\ y &= 0-4 = -4 \\ L &= \pi \int_{0}^{4} ((6-1,5x)^2-(x-4)^2)\,dx \\ &= \pi \int_{0}^{4} (36-18x+2,25x^2-(x^2-8x+16))\,dx \\ &= \pi \int_{0}^{4} (1,25x^2-10x+20)\,dx \\ &= \pi (\frac{1,25x^3}{3}-5x^2+20x)|_{0}^{4} \\ &= \pi (\frac{1,25(4)^3}{3}-5(4)^2+20(4)-(\frac{1,25(0)^3}{3}-5(0)^2+20(0))) \\ &= \pi (\frac{30}{3}-0) \\ &= 10\pi \\ \end{aligned} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 1duofh5i500f6zbtbrm5ze9q77q10xy 115029 115028 2026-04-27T02:06:50Z ~2026-25546-23 43043 /* integral pecahan parsial */ 115029 wikitext text/x-wiki == Kaidah umum == :# <math>\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ konstan)}\,\!</math> :# <math>\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx</math> :# <math>\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx</math> :# <math>\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx</math> :# <math>\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\! </math> :# <math>\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C </math> :# <math>\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C </math> :# <math>\int f(x) \, dx = F(x) + C</math> :# <math>\int_a^b f(x) \, dx = F(b) - F(a)</math> :# <math>\int f(x) \circ g(x) dx</math> = f’(g(x)) g’(x) :# <math>\int f^{-1}(x) dx = \frac{1}{f'(f'(x))}</math> == rumus sederhana == :<math>\int \, dx = x + C</math> :<math>\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int {dx \over x} = \ln{\left|x\right|} + C</math> <math>\int {dx \over \sqrt{a^2-b^2 x^2}} = \frac{1}{b} \arcsin \, {bx \over a} + C</math> :<math>\int {dx \over {a^2+b^2x^2}} = {1 \over ab}\arctan \, {bx \over a} + C</math> <math>\int {dx \over x\sqrt{b^2x^2-a^2}} = {1 \over a}\arcsec \, {bx \over a} + C</math> ; Eksponen dan logaritma :<math>\int e^x\,dx = e^x + C</math> :<math>\int a^x\,dx = \frac{a^x}{\ln{a}} + C</math> :<math>\int \ln \, x\,dx = x \cdot \, \ln {x} - x + C</math> :<math>\int \, ^b\!\log \, x\,dx = x \cdot \, ^b\!\log \, x - x \cdot \,^b\!\log e + C</math> ; Trigonometri :<math>\int \sin \, x \, dx = -\cos \, x + C</math> :<math>\int \cos \, x \, dx = \sin \, x + C</math> :<math>\int \tan \, x \, dx = -\ln{\left| \cos \, x \right|} + C</math> :<math>\int \cot \, x \, dx = \ln{\left| \sin \, x \right|} + C</math> :<math>\int \sec \, x \, dx = \ln{\left| \sec \, x + \tan \, x \right|} + C</math> :<math>\int \csc \, x \, dx = -\ln{\left| \csc \, x + \cot \, x \right|} + C</math> ; Hiperbolik :<math>\int \sinh \, x \, dx = \cosh \, x + C</math> :<math>\int \cosh \, x \, dx = \sinh \, x + C</math> :<math>\int \tanh \, x \, dx = \ln{\left| \cosh \, x \right|} + C</math> :<math>\int \coth \, x \, dx = \ln{\left| \sinh \, x \right|} + C</math> :<math>\int \mbox{sech} \, x \, dx = \arctan \,(\sinh \, x) + C</math> :<math>\int \mbox{csch} \, x \, dx = \ln{\left| \tanh \, {x \over 2} \right|} + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : 1. <math>\int x^3 - cos 3x + e^{5x+2} - \frac{1}{2x+5}\,dx</math> : <math>= \frac{1}{3+1}x^{3+1} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : <math>= \frac{x^4}{4} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : 2. <math>\int x(x-5)^4\,dx</math> : <math>= \int x(x^4-4x^3(5)+6x^2(25)-4x(125)+625)\,dx</math> : <math>= \int x(x^4-20x^3+150x^2-500x+625)\,dx</math> : <math>= \int x^5-20x^4+150x^3-500x^2+625x\,dx</math> : <math>= \frac{x^6}{6}-4x^5+\frac{150x^4}{4}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> : <math>= \frac{x^6}{6}-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : 1. <math>\int \frac{\ln \, x}{x}\,dx</math> : <math>u = \ln \, x\, du = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{\ln(x)}{x} dx \\ \int u du \\ \frac{1}{2} u^2 + C \\ \frac{1}{2} ln^2 x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x-5,\, du = dx,\, x = u+5</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4\,dx \\ \int (u+5)u^4\,du \\ \int (u^5+5u^4)\,du \\ \frac{1}{6} u^6+u^5 + C \\ \frac{1}{6} (x-5)^6+(x-5)^5 + C \\ \frac{1}{6} (x^6-6x^5(5)+15x^4(25)-20x^3(125)+15x^2(625)-6x(3125)+15625)+(x^5-5x^4(5)+10x^3(25)-10x^2(125)+5x(625)+3125) + C \\ \frac{1}{6}x^6-5x^5+\frac{125x^4}{2}-\frac{1250x^3}{3}+\frac{3125x^2}{2}-3125x+\frac{15625}{6}+x^5-25x^4+250x^3-1250x^2+3125x+3125 + C \\ \frac{1}{6}x^6-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C \\ \end{aligned} </math> </div></div> === integral parsial === ;Cara 1: Rumus Integral parsial diselesaikan dengan rumus berikut. : <math>\int u\,dv = uv - \int v\,du </math> Berikut contoh penyelesaian cara parsial dengan rumus. : 1. <math>\int x \sin \, x \, dx</math> : <math>u = x, du = 1 \, dx, dv = \sin \, x \, dx, v = -\cos \, x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \,dx \\ (x)(-\cos \, x) - \int (-\cos \, x)(1 dx) \\ -x \cos \, x + \int \cos \, x dx \\ -x \cos \, x + \sin \, x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x,\, du = 1\,dx,\, dv = (x-5)^4\,dx,\, v = \frac{1}{5}(x-5)^5</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - \int (\frac{1}{5}(x-5)^5)(1\,dx) \\ \frac{x}{5}(x-5)^5 - \int \frac{1}{5}(x-5)^5\,dx \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6\,dx \\ \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x\,dx</math> : <math>u = \ln \,x,\, du = \frac{1}{x}\,dx,\, dv = x\,dx,\, v = \frac{x^2}{2}</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x\, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{x^2}{2})(\frac{1}{x}\,dx) \\ \frac{x^2}{2} \ln \,x - \int \frac{x^2}{2}(\frac{1}{x})\,dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \,dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x\,dx</math> ini terjadi dua kali integral parsial : <math>u = \sin \,x,\, du = \cos \,x\,dx,\, dv = e^x\,dx,\, v = e^x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x\, dx \\ (\sin \,x)(e^x) - \int (e^x)(\cos \,x\,dx) \\ e^x \sin \,x - \int e^x \cos \,x\,dx \\ u = \cos \,x,\, du = -\sin \,x\,dx,\, dv = e^x\,dx,\, v = e^x \\ (\cos \,x)(e^x) - \int (e^x)(-\sin \,x\,dx) \\ e^x \cos \,x + \int e^x \sin \,x\,dx \\ e^x \sin \,x - (e^x \cos \,x + \int e^x \sin \,x\,dx) \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x\,dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> ;Cara 2: Tabel Untuk <math display="inline">\int u\,dv</math>, berlaku ketentuan sebagai berikut. {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>u</math> || <math>dv</math> |- | - || <math>\frac{du}{dx}</math> || <math>v</math> |- | + || <math>\frac{d^2u}{dx^2}</math> || <math>\int v\,dx</math> |} Berikut contoh penyelesaian cara parsial dengan tabel. : 1. <math>\int x \sin \, x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>\sin \, x</math> |- | - || <math>1</math> || <math>-\cos \, x</math> |- | + || <math>0</math> || <math>-\sin \, x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \, dx \\ (x)(-\cos \, x) - (1)(-\sin \, x) + C \\ -x \cos \, x + \sin \, x + C \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4 \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>(x-5)^4</math> |- | - || <math>1</math> || <math>\frac{1}{5}(x-5)^5</math> |- | + || <math>0</math> || <math>\frac{1}{30}(x-5)^6</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - (1)(\frac{1}{30}(x-5)^6) + C \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6 + C \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\ln \,x</math> || <math>x</math> |- | - || <math>\frac{1}{x}</math> || <math>\frac{x^2}{2}</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x \, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{1}{x})(\frac{x^2}{2}) \, dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \, dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x \, dx</math> ini terjadi dua kali integral parsial {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\sin \,x</math> || <math>e^x</math> |- | - || <math>\cos \, x</math> || <math>e^x</math> |- | + || <math>-\sin \,x</math> || <math>e^x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x \, dx \\ (\sin \,x)(e^x) - (\cos \,x)(e^x) + \int (-\sin \,x)(e^x) \, dx \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x) \, dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : 1. <math>\int \frac{1}{x^2 - 4}\,dx</math> Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{1}{x^2 - 4} \\ =&\; \frac{1}{(x + 2)(x - 2)} \\ =&\; \frac{A}{x + 2} + \frac{B}{x - 2} \\ =&\; \frac{A(x - 2) + B(x + 2)}{x^2 - 4} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> : 2. <math>\int \frac{x^2+5x+4}{(x-3)^3}\,dx</math> : cara 1 Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^3} \\ \text{untuk mencari C yaitu } (3)^2+5(3)+4=28 \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x - 3)^2 + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x^2 - 6x + 9) + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 - 6Ax + 9A + Bx - 3B}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 + (- 6A + B)x +(9 - 3B)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ \end{aligned} </math> Kita tahu bahwa <math display="inline">A = 1</math> dan <math display="inline">-6A + B = 5</math> dapat diselesaikan, yaitu <math display="inline">A = 1</math> dan {{nowrap|<math display="inline">B = 11</math>.}} : <math> \begin{aligned} &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int (\frac{1}{x - 3} + \frac{11}{(x - 3)^2} + \frac{28}{(x - 3)^3})\,dx \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : cara 2 Pertama, ubah u=x-3 menjadi x=u+3 dan du=dx tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{(u+3)^2+5(u+3)+4}{u^3} \\ =&\; \frac{u^2+11u+28}{u^3} \\ &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int \frac{u^2+11u+28}{u^3}\,du \\ =&\; \int (\frac{u^2}{u^3} + \frac{11u}{u^3} + \frac{28}{u^3})\,du \\ =&\; \int (\frac{1}{u} + \frac{11}{u^2} + \frac{28}{u^3})\,du \\ =&\; \ln u + (-11 u^{-1}) + (-14 u^{-2}) + C \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : 3. <math>\int \frac{x^4+2x^3-13x^2-7x+21}{x^2-4x+3}\,dx</math> Pertama, penyelesaian dengan persamaan polinomial tersebut. : <math> \begin{aligned} =&\; \frac{x^4+2x^3-13x^2-7x+21}{x^2-4x+3} \\ =&\; x^2+6x+8 + \frac{7x-3}{x^2-4x+3} \\ =&\; \frac{1}{3}x^3+3x^2+8x+ \int \frac{7x-3}{x^2-4x+3}\,dx \end{aligned} </math> Kedua, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{7x - 3}{x^2 - 4x + 3} \\ =&\; \frac{7x - 3}{(x - 1)(x - 3)} \\ =&\; \frac{A}{x - 1} + \frac{B}{x - 3} \\ =&\; \frac{A(x - 3) + B(x - 1)}{x^2 - 4x + 3} \\ =&\; \frac{(A + B)x + (-3A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 7</math> dan <math display="inline">-3A - B = -3</math> dapat diselesaikan, yaitu <math display="inline">A = 9</math> dan {{nowrap|<math display="inline">B = -2</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> === integral substitusi trigonometri === {| class="wikitable" |- ! Bentuk !! Substitusi Trigonometri |- | <math>\sqrt{a^2 - b^2 x^2}</math> || <math>x = \frac{a}{b} sin \, \alpha </math> |- | <math>\sqrt{a^2 + b^2 x^2}</math> || <math>x = \frac{a}{b} tan \, \alpha</math> |- | <math>\sqrt{b^2 x^2 - a^2}</math> || <math>x = \frac{a}{b} sec \, \alpha</math> |} Berikut contoh penyelesaian cara substitusi trigonometri. : <math>\int \frac{dx}{x^2 \sqrt{x^2 + 4}}</math> : <math>x = 2 \tan \, A, dx = 2 \sec^2 \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \int \frac{dx}{x^2\sqrt{x^2+4}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(2 \tan \, A)^2 \sqrt{4 + (2 \tan \, A)^2}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 + 4 \tan^2 \, A}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4(1 + \tan^2 \, A)}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 \sec^2 \, A }} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(4 \tan^2 \, A)(2 \sec \, A)} \\ =&\; \int \frac{\sec \, A \,dA}{4 \tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\sec \, A \,dA}{\tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A} \,dA \end{aligned} </math> Substitusi berikut dapat dibuat. : <math>\int \frac{\cos \, A}{\sin^2 \, A} \,dA</math> : <math>t = \sin \, A, dt = \cos \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A}\,dA \\ =&\; \frac{1}{4} \int \frac{dt}{t^2} \\ =&\; \frac{1}{4} \left(-\frac{1}{t}\right) + C \\ =&\; -\frac{1}{4 t} + C \\ =&\; -\frac{1}{4 \sin \, A} + C \end{aligned} </math> Ingat bahwa <math display="inline">\sin \, A = \frac{x}{\sqrt{x^2 + 4}}</math> berlaku. : <math> \begin{aligned} =&\; -\frac{1}{4 \sin \, A} + C \\ =&\; -\frac{\sqrt{x^2 + 4}}{4x} + C \end{aligned} </math> === integral mutlak === : <math>\int |f(x)| \,dA</math> buatlah <math>f(x) \ge 0</math> jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x) : <math>\int_a^c |f(x)| \, dx = \int_a^b f(x) \, dx + \int_b^c -f(x) \, dx = F(b) - F(a) - F(c) + F(b)</math> jika <math>|f(x)| = \begin{cases} f(x), & \mbox{jika } f(x) \ge b, \\ -f(x), & \mbox{jika } f(x) < b. \end{cases}</math>. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif. === integral fungsi ganjil dan integral fungsi genap === : <math>\int_{-a}^a f(x) \,dx</math> dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral: : integral fungsi ganjil : <math>\int_{-a}^a f(x) \,dx</math> = 0 : integral fungsi genap : <math>\int_{-a}^a f(x) \,dx</math> = <math>2 \int_{0}^a f(x) \,dx</math> === integral notasi === : <math>\int_a^b f(x) \,dx = \int_{a-p}^{b-p} f(x) \,dx = \int_{a+p}^{b+p} f(x) \,dx</math> : <math>\int_a^c f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx</math> : <math>\int_a^b f(x \pm k) \,dx = \int_{a \pm k}^{b \pm k} f(x) \,dx</math> === integral terbalik === : <math>\int_a^b f(x) \,dx = -\int_b^a f(x) \,dx</math> == Jenis integral lainnya == === panjang busur === ; Sumbu ''x'' : <math>S = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' : <math>S = \int_{y_1}^{y_2} \sqrt{1 + (f'(y))^2}\,dy</math> === luas daerah === ; Satu kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} f(x)\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} f(y)\,dy</math> ; Dua kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} (f(x_2) - f(x_1))\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} (f(y_2) - f(y_1))\,dy</math> : atau juga <math>L = \frac {D \sqrt{D}}{6 a^2}</math> === luas permukaan benda putar === ; Sumbu ''x'' sebagai poros : <math>L_p = 2 \pi \int_{x_1}^{x_2} f(x)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' sebagai poros : <math>L_p = 2 \pi \int_{y_1}^{y_2} f(y)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(y))^2}\,dy</math> === volume benda putar === ; Satu kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} (f(x))^2\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} (f(y))^2\,dy</math> ; Dua kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} ((f(x_2))^2 - (f(x_1))^2)\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} ((f(y_2))^2 - (f(y_1))^2)\,dy</math> : atau juga <math>V = |\frac {D^2 \sqrt{D}}{30 a^3}|</math> == Integral lipat == Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga. ;contoh # Tentukan hasil dari: * <math>\int (5x-1)(5x^2-2x+7)^4 dx</math> * <math>\int x^2 cos 3x dx</math> * <math>\int sin^2 x cos x dx</math> * <math>\int cos^2 x sin x dx</math> * <math>\int sec x dx</math> * <math>\int csc x dx</math> * <math>\int \frac{1}{2x^2-5x+3} dx</math> * <math>\int \frac{1}{\sqrt{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> * <math>\int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx</math> * <math>\int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx</math> * <math>\int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx</math> * <math>\int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx</math> * <math>\int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx \\ a=1, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-sin^2 x} \\ &= \frac{1}{cos^2 x} \\ &= sec^2 x \\ \int sec^2 x dx &= tan x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx \\ a=1, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-cos^2 x} \\ &= \frac{1}{sin^2 x} \\ &= csc^2 x \\ \int csc^2 x dx &= -cot x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx \\ a=sin x, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{sin x}{1-sin^2 x} \\ &= \frac{sin x}{cos^2 x} \\ &= sec x \cdot tan x \\ \int sec x \cdot tan x dx &= sec x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx \\ a=cos x, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{cos x}{1-cos^2 x} \\ &= \frac{cos x}{sin^2 x} \\ &= csc x \cdot cot x \\ \int csc x \cdot cot x dx &= -csc x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx \\ \text{misalkan } y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y &= \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y^2 &= \frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}} \\ &= \frac{x}{y} \\ y^3 &= x \\ 3y^2 dy &= dx \\ \int y \cdot 3y^2 dy \\ \int 3y^3 dy \\ \frac{3y^4}{4}+C \\ \end{aligned} </math> </div></div> # Tentukan hasil dari: * <math>\int_{1}^{5} 2x-3 dx</math> * <math>\int_{0}^{4} |x-2| dx</math> * <math>\int_{2}^{4} |3-x| dx</math> * <math>\int_{1}^{5} x^2-6x+8 dx</math> * <math>\int_{-3}^{5} |x^2-2x-8| dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} 2x-3 dx &= \left. x^2 - 3x \right|_{1}^{5} \\ &= 10 - (-2) \\ &= 12 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{0}^{4} |x-2| dx \\ \text{ tentukan nilai harga nol } \\ x-2 &= 0 \\ x &= 2 \\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti } \\ |x-2| = \begin{cases} x-2, & \mbox{jika } x \ge 2, \\ -(x-2), & \mbox{jika } x < 2. \end{cases} \\ \int_{0}^{4} |x-2| dx &= \int_{2}^{4} x-2 dx + \int_{0}^{2} -(x-2) dx \\ &= \left. (\frac{x^2}{2}-2x) \right|_{2}^{4} + \left. (\frac{-x^2}{2}+2x) \right|_{0}^{2} \\ &= 0 - (-2) + 2 - 0 \\ &= 4 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{2}^{4} |3-x| dx \\ \text{ tentukan nilai syarat-syarat } \\ 3-x &\ge 0 \\ \text{ harga nol } \\ x &= 3 \\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti } \\ |3-x| = \begin{cases} 3-x, & \mbox{jika } x \le 3, \\ -(3-x), & \mbox{jika } x > 3. \end{cases} \\ \int_{2}^{4} |3-x| dx &= \int_{2}^{3} 3-x dx + \int_{3}^{4} -(3-x) dx \\ &= \left. (3x-\frac{x^2}{2}) \right|_{2}^{3} + \left. (-3x+\frac{x^2}{2}) \right|_{3}^{4} \\ &= 4.5 - 4 - 4 - (-4.5) \\ &= 1 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} x^2-6x+8 dx &= \left. \frac{x^3}{3}-3x^2+8x \right|_{1}^{5} \\ &= \frac{20}{3} - \frac{16}{3} \\ &= \frac{4}{3} \\ &= 1\frac{1}{3} \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{-3}^{5} |x^2-2x-8| dx \\ \text{ tentukan nilai syarat-syarat } \\ x^2-2x-8 &\ge 0 \\ \text{ harga nol } \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x = 4 \\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti } \\ |x^2-2x-8| = \begin{cases} x^2-2x-8, & \mbox{jika } x \le -2, \\ -(x^2-2x-8), & \mbox{jika } -2 < x < 4, \\ x^2-2x-8, & \mbox{jika } x \ge 4 \end{cases} \\ \int_{-3}^{5} |x^2-2x-8| dx &= \int_{-3}^{-2} x^2-2x-8 dx + \int_{-2}^{4} -(x^2-2x-8) dx + \int_{4}^{5} x^2-2x-8 dx \\ &= \left. (\frac{x^3}{3}-x^2-8x) \right|_{-3}^{-2} + \left. (\frac{-x^3}{3}+x^2+8x) \right|_{-2}^{4} + \left. (\frac{x^3}{3}-x^2-8x) \right|_{4}^{5} \\ &= \frac{28}{3} - 6 + \frac{80}{3} - (-\frac{28}{3}) - \frac{70}{3} - (-\frac{80}{3}) \\ &= \frac{128}{3} \\ &= 42\frac{2}{3} \\ \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x\,dx \\ L &= \frac{1}{2} x^2 \\ L &= \frac{1}{2} xy \quad (y = x) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x^2</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x^2\,dx \\ L &= \frac{1}{3} x^3 \\ L &= \frac{1}{3} xy \quad (y = x^2) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = \sqrt{x}</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int \sqrt{x}\,dx \\ L &= \frac{2}{3} x^{\frac{3}{2}} \\ L &= \frac{2}{3} xy \quad (y = \sqrt{x}) \end{aligned} </math> </div></div> # Buktikan luas persegi <math>L = s^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = s \text{ dan titik (''s'', ''s''), } \\ L &= \int_{0}^{s} s\,dx \\ L &= sx |_{0}^{s} \\ L &= ss - 0 \\ L &= s^2 \end{aligned} </math> </div></div> # Buktikan luas persegi panjang <math>L = pl</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = l \text{dan titik (''p'', ''l''), } \\ L &= \int_{0}^{p} l\,dx \\ L &= lx |_{0}^{p} \\ L &= pl - 0 \\ L &= pl \end{aligned} </math> </div></div> * Buktikan luas segitiga <math>L = \frac{at}{2}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{-tx}{a} + t \text { dan titik (''a'', ''t''), } \\ L &= \int_{0}^{a} \left (\frac{-tx}{a} + t \right) \,dx \\ L &= \left. \frac{-tx^2}{2a} + tx \right|_{0}^{a} \\ L &= \frac{-ta^2}{2a} + ta - 0 + 0 \\ L &= \frac{-ta}{2} + ta \\ L &= \frac{at}{2} \end{aligned} </math> </div></div> # Buktikan volume tabung <math>V = \pi r^2t</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ Dengan posisi } y = r \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} r^2\,dx \\ V &= \pi \left. r^2x \right|_{0}^{t} \\ V &= \pi r^2t - 0 \\ V &= \pi r^2t \end{aligned} </math> </div></div> # Buktikan volume kerucut <math>V = \frac{\pi r^2t}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{rx}{t} \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} \left (\frac{rx}{t} \right)^2 \,dx \\ V &= \pi \left. \frac{r^2 x^3}{3t^2} \right|_{0}^{t} \\ V &= \pi \frac{r^2 t^3}{3t^2} - 0 \\ V &= \frac{\pi r^2t}{3} \end{aligned} </math> </div></div> # Buktikan volume bola <math>V = \frac{4 \pi r^3}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), } \\ V &= \pi \int_{-r}^{r} \left(\sqrt {r^2 - x^2}\right)^2\,dx \\ V &= \pi \int_{-r}^{r} r^2 - x^2 dx \\ V &= \pi \left. r^2x - \frac{x^3}{3} \right|_{-r}^{r} \\ V &= \pi \left (r^3 - \frac{r^3}{3} - \left (-r^3 + \frac{r^3}{3} \right) \right) \\ V &= \frac{4 \pi r^3}{3} \end{aligned} </math> </div></div> # Buktikan luas permukaan bola <math>L = 4 \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{selanjutnya } \\ ds &= \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ ds &= \sqrt{1 + \frac{x^2}{r^2 - x^2}}\,dx \\ ds &= \sqrt{\frac{r^2}{r^2 - x^2}}\,dx \\ ds &= \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ \text{sehingga } \\ L &= 2 \pi \int_{-r}^{r} \sqrt{r^2 - x^2} \cdot \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ L &= 2 \pi \int_{-r}^{r} r\,dx \\ L &= 2 \pi rx|_{-r}^{r} \\ L &= 2 \pi (r r - r (-r)) \\ L &= 2 \pi (r^2 + r^2) \\ L &= 4 \pi r^2 \end{aligned} </math> </div></div> # Buktikan keliling lingkaran <math>K = 2 \pi r</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0) } \\ \text{kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{sehingga } \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{x^2}{r^2 - x^2})}\,dx \\ K &= 4 \int_{0}^{r} \sqrt\frac{r^2}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \sqrt\frac{1}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \frac{1}{\sqrt{r^2 - x^2}}\,dx \\ K &= 4r \left. \arcsin \left (\frac{x}{r} \right) \right|_{0}^{r} \\ K &= 4r \left (\arcsin \left (\frac{r}{r}\right) - \arcsin \left (\frac{0}{r} \right) \right) \\ K &= 4r \left (\arcsin\left (1 \right) - \arcsin \left(0 \right) \right)) \\ K &= 4r \left (\frac{\pi}{2} \right) \\ K &= 2 \pi r \end{aligned} </math> </div></div> # Buktikan luas lingkaran <math>L = \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. } \\ \sin(\theta) &= \frac{x}{r} \\ x &= r \sin(\theta) \\ dx &= r \cos(\theta)\,d\theta \\ \text{dengan turunan di atas } \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - x^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - (r \sin(\theta))^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - r^2 \sin^2(\theta)}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 (1 - \sin^2(\theta))}\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta)\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta) (r \cos(\theta)\,d\theta) \\ L &= 4 \int_{0}^{r} r^2 \cos^2(\theta)\,d\theta \\ L &= 4r^2 \int_{0}^{r} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta \\ L &= 2r^2 \int_{0}^{r} (1 + \cos(2\theta))\,d\theta \\ L &= 2r^2 \left (\theta + \frac{1}{2} \sin(2\theta) \right)|_{0}^{r} \\ L &= 2r^2 (\theta + \sin(\theta) \cos(\theta))|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left (\frac{x}{r} \right) + \left (\frac{x}{r} \right) \left (\frac{r^2 - x^2}{r} \right) \right)|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left(\frac{r}{r} \right) + \left (\frac{r}{r} \right) \left (\frac {r^2 - r^2}{r} \right) \right) - \left(\arcsin \left (\frac{0}{r} \right) + \left (\frac{0}{r}\right) \left (\frac{r^2 - 0^2}{r} \right) \right) \\ L &= 2r^2 (\arcsin(1) + 0 - (\arcsin(0) + 0)) \\ L &= 2r^2 \left (\frac{\pi}{2} \right) \\ L &= \pi r^2 \end{aligned} </math> </div></div> # Buktikan luas elips <math>L = \pi ab</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \frac{b \sqrt{a^2 - x^2}}{a} \text{ serta (-''a'', 0) dan (''a'', 0), } \\ L &= 4 \int_{0}^{r} \frac{b \sqrt{a^2 - x^2}}{a}\,dx \\ L &= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx \\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, } \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{\frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx}{4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx} \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{b}{a} \\ L_\text{elips} &= \frac{b}{a} L_\text{ling} \\ L_\text{elips} &= \frac{b}{a} \pi a^2 \\ L_\text{elips} &= \pi ab \end{aligned} </math> </div></div> # Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ x^-2x &= 4x+7 \\ x^2-6x-7 &= 0 \\ (x+1)(x-7) &= 0 \\ x=-1 &\text{ atau } x=7 \\ y &= 4(-1)+7 = 3 \\ y &= 4(7)+7 = 35 \\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut } \\ L &= \int_{-1}^{7} 4x+7-(x^2-2x)\,dx \\ &= \int_{-1}^{7} -x^2+6x+7\,dx \\ &= -\frac{x^3}{3}+3x^2+7x|_{-1}^{7} \\ &= -\frac{7^3}{3}+3(7)^2+7(7)-(-\frac{(-1)^3}{3}+3(-1)^2+7(-1)) \\ &= \frac{245}{3}-(-\frac{13}{3}) \\ &= \frac{258}{3} \\ &= 86 \\ \end{aligned} </math> </div></div> # Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ 6-1,5x &= x-4 \\ 2,5x &= 10 \\ x &= 4 \\ y &= 4-4 = 0 \\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut } \\ \text{untuk x=0 } \\ y &= 6-1,5(0) = 6 \\ y &= 0-4 = -4 \\ L &= \pi \int_{0}^{4} ((6-1,5x)^2-(x-4)^2)\,dx \\ &= \pi \int_{0}^{4} (36-18x+2,25x^2-(x^2-8x+16))\,dx \\ &= \pi \int_{0}^{4} (1,25x^2-10x+20)\,dx \\ &= \pi (\frac{1,25x^3}{3}-5x^2+20x)|_{0}^{4} \\ &= \pi (\frac{1,25(4)^3}{3}-5(4)^2+20(4)-(\frac{1,25(0)^3}{3}-5(0)^2+20(0))) \\ &= \pi (\frac{30}{3}-0) \\ &= 10\pi \\ \end{aligned} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 38ddgi0kyqc3pcyndsu740cnjuy1n03 115030 115029 2026-04-27T02:18:38Z ~2026-25546-23 43043 /* integral pecahan parsial */ 115030 wikitext text/x-wiki == Kaidah umum == :# <math>\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ konstan)}\,\!</math> :# <math>\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx</math> :# <math>\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx</math> :# <math>\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx</math> :# <math>\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\! </math> :# <math>\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C </math> :# <math>\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C </math> :# <math>\int f(x) \, dx = F(x) + C</math> :# <math>\int_a^b f(x) \, dx = F(b) - F(a)</math> :# <math>\int f(x) \circ g(x) dx</math> = f’(g(x)) g’(x) :# <math>\int f^{-1}(x) dx = \frac{1}{f'(f'(x))}</math> == rumus sederhana == :<math>\int \, dx = x + C</math> :<math>\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int {dx \over x} = \ln{\left|x\right|} + C</math> <math>\int {dx \over \sqrt{a^2-b^2 x^2}} = \frac{1}{b} \arcsin \, {bx \over a} + C</math> :<math>\int {dx \over {a^2+b^2x^2}} = {1 \over ab}\arctan \, {bx \over a} + C</math> <math>\int {dx \over x\sqrt{b^2x^2-a^2}} = {1 \over a}\arcsec \, {bx \over a} + C</math> ; Eksponen dan logaritma :<math>\int e^x\,dx = e^x + C</math> :<math>\int a^x\,dx = \frac{a^x}{\ln{a}} + C</math> :<math>\int \ln \, x\,dx = x \cdot \, \ln {x} - x + C</math> :<math>\int \, ^b\!\log \, x\,dx = x \cdot \, ^b\!\log \, x - x \cdot \,^b\!\log e + C</math> ; Trigonometri :<math>\int \sin \, x \, dx = -\cos \, x + C</math> :<math>\int \cos \, x \, dx = \sin \, x + C</math> :<math>\int \tan \, x \, dx = -\ln{\left| \cos \, x \right|} + C</math> :<math>\int \cot \, x \, dx = \ln{\left| \sin \, x \right|} + C</math> :<math>\int \sec \, x \, dx = \ln{\left| \sec \, x + \tan \, x \right|} + C</math> :<math>\int \csc \, x \, dx = -\ln{\left| \csc \, x + \cot \, x \right|} + C</math> ; Hiperbolik :<math>\int \sinh \, x \, dx = \cosh \, x + C</math> :<math>\int \cosh \, x \, dx = \sinh \, x + C</math> :<math>\int \tanh \, x \, dx = \ln{\left| \cosh \, x \right|} + C</math> :<math>\int \coth \, x \, dx = \ln{\left| \sinh \, x \right|} + C</math> :<math>\int \mbox{sech} \, x \, dx = \arctan \,(\sinh \, x) + C</math> :<math>\int \mbox{csch} \, x \, dx = \ln{\left| \tanh \, {x \over 2} \right|} + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : 1. <math>\int x^3 - cos 3x + e^{5x+2} - \frac{1}{2x+5}\,dx</math> : <math>= \frac{1}{3+1}x^{3+1} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : <math>= \frac{x^4}{4} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : 2. <math>\int x(x-5)^4\,dx</math> : <math>= \int x(x^4-4x^3(5)+6x^2(25)-4x(125)+625)\,dx</math> : <math>= \int x(x^4-20x^3+150x^2-500x+625)\,dx</math> : <math>= \int x^5-20x^4+150x^3-500x^2+625x\,dx</math> : <math>= \frac{x^6}{6}-4x^5+\frac{150x^4}{4}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> : <math>= \frac{x^6}{6}-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : 1. <math>\int \frac{\ln \, x}{x}\,dx</math> : <math>u = \ln \, x\, du = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{\ln(x)}{x} dx \\ \int u du \\ \frac{1}{2} u^2 + C \\ \frac{1}{2} ln^2 x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x-5,\, du = dx,\, x = u+5</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4\,dx \\ \int (u+5)u^4\,du \\ \int (u^5+5u^4)\,du \\ \frac{1}{6} u^6+u^5 + C \\ \frac{1}{6} (x-5)^6+(x-5)^5 + C \\ \frac{1}{6} (x^6-6x^5(5)+15x^4(25)-20x^3(125)+15x^2(625)-6x(3125)+15625)+(x^5-5x^4(5)+10x^3(25)-10x^2(125)+5x(625)+3125) + C \\ \frac{1}{6}x^6-5x^5+\frac{125x^4}{2}-\frac{1250x^3}{3}+\frac{3125x^2}{2}-3125x+\frac{15625}{6}+x^5-25x^4+250x^3-1250x^2+3125x+3125 + C \\ \frac{1}{6}x^6-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C \\ \end{aligned} </math> </div></div> === integral parsial === ;Cara 1: Rumus Integral parsial diselesaikan dengan rumus berikut. : <math>\int u\,dv = uv - \int v\,du </math> Berikut contoh penyelesaian cara parsial dengan rumus. : 1. <math>\int x \sin \, x \, dx</math> : <math>u = x, du = 1 \, dx, dv = \sin \, x \, dx, v = -\cos \, x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \,dx \\ (x)(-\cos \, x) - \int (-\cos \, x)(1 dx) \\ -x \cos \, x + \int \cos \, x dx \\ -x \cos \, x + \sin \, x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x,\, du = 1\,dx,\, dv = (x-5)^4\,dx,\, v = \frac{1}{5}(x-5)^5</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - \int (\frac{1}{5}(x-5)^5)(1\,dx) \\ \frac{x}{5}(x-5)^5 - \int \frac{1}{5}(x-5)^5\,dx \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6\,dx \\ \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x\,dx</math> : <math>u = \ln \,x,\, du = \frac{1}{x}\,dx,\, dv = x\,dx,\, v = \frac{x^2}{2}</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x\, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{x^2}{2})(\frac{1}{x}\,dx) \\ \frac{x^2}{2} \ln \,x - \int \frac{x^2}{2}(\frac{1}{x})\,dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \,dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x\,dx</math> ini terjadi dua kali integral parsial : <math>u = \sin \,x,\, du = \cos \,x\,dx,\, dv = e^x\,dx,\, v = e^x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x\, dx \\ (\sin \,x)(e^x) - \int (e^x)(\cos \,x\,dx) \\ e^x \sin \,x - \int e^x \cos \,x\,dx \\ u = \cos \,x,\, du = -\sin \,x\,dx,\, dv = e^x\,dx,\, v = e^x \\ (\cos \,x)(e^x) - \int (e^x)(-\sin \,x\,dx) \\ e^x \cos \,x + \int e^x \sin \,x\,dx \\ e^x \sin \,x - (e^x \cos \,x + \int e^x \sin \,x\,dx) \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x\,dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> ;Cara 2: Tabel Untuk <math display="inline">\int u\,dv</math>, berlaku ketentuan sebagai berikut. {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>u</math> || <math>dv</math> |- | - || <math>\frac{du}{dx}</math> || <math>v</math> |- | + || <math>\frac{d^2u}{dx^2}</math> || <math>\int v\,dx</math> |} Berikut contoh penyelesaian cara parsial dengan tabel. : 1. <math>\int x \sin \, x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>\sin \, x</math> |- | - || <math>1</math> || <math>-\cos \, x</math> |- | + || <math>0</math> || <math>-\sin \, x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \, dx \\ (x)(-\cos \, x) - (1)(-\sin \, x) + C \\ -x \cos \, x + \sin \, x + C \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4 \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>(x-5)^4</math> |- | - || <math>1</math> || <math>\frac{1}{5}(x-5)^5</math> |- | + || <math>0</math> || <math>\frac{1}{30}(x-5)^6</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - (1)(\frac{1}{30}(x-5)^6) + C \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6 + C \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\ln \,x</math> || <math>x</math> |- | - || <math>\frac{1}{x}</math> || <math>\frac{x^2}{2}</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x \, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{1}{x})(\frac{x^2}{2}) \, dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \, dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x \, dx</math> ini terjadi dua kali integral parsial {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\sin \,x</math> || <math>e^x</math> |- | - || <math>\cos \, x</math> || <math>e^x</math> |- | + || <math>-\sin \,x</math> || <math>e^x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x \, dx \\ (\sin \,x)(e^x) - (\cos \,x)(e^x) + \int (-\sin \,x)(e^x) \, dx \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x) \, dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : 1. <math>\int \frac{1}{x^2 - 4}\,dx</math> Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{1}{x^2 - 4} \\ =&\; \frac{1}{(x + 2)(x - 2)} \\ =&\; \frac{A}{x + 2} + \frac{B}{x - 2} \\ =&\; \frac{A(x - 2) + B(x + 2)}{x^2 - 4} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> : 2. <math>\int \frac{x^2+5x+4}{(x-3)^3}\,dx</math> : cara 1 Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^3} \\ \text{untuk mencari C yaitu } (3)^2+5(3)+4=28 \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x - 3)^2 + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x^2 - 6x + 9) + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 - 6Ax + 9A + Bx - 3B}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 + (- 6A + B)x +(9 - 3B)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ \end{aligned} </math> Kita tahu bahwa <math display="inline">A = 1</math> dan <math display="inline">-6A + B = 5</math> dapat diselesaikan, yaitu <math display="inline">A = 1</math> dan {{nowrap|<math display="inline">B = 11</math>.}} : <math> \begin{aligned} &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int (\frac{1}{x - 3} + \frac{11}{(x - 3)^2} + \frac{28}{(x - 3)^3})\,dx \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : cara 2 Pertama, ubah u=x-3 menjadi x=u+3 dan du=dx tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{(u+3)^2+5(u+3)+4}{u^3} \\ =&\; \frac{u^2+11u+28}{u^3} \\ &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int \frac{u^2+11u+28}{u^3}\,du \\ =&\; \int (\frac{u^2}{u^3} + \frac{11u}{u^3} + \frac{28}{u^3})\,du \\ =&\; \int (\frac{1}{u} + \frac{11}{u^2} + \frac{28}{u^3})\,du \\ =&\; \ln u + (-11 u^{-1}) + (-14 u^{-2}) + C \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : 3. <math>\int \frac{x^4+2x^3-13x^2-7x+21}{x^2-4x+3}\,dx</math> Pertama, penyelesaian dengan persamaan polinomial tersebut. : <math> \begin{aligned} =&\; \frac{x^4+2x^3-13x^2-7x+21}{x^2-4x+3} \\ =&\; x^2+6x+8 + \frac{7x-3}{x^2-4x+3} \\ =&\; \frac{1}{3}x^3+3x^2+8x+ \int \frac{7x-3}{x^2-4x+3}\,dx \end{aligned} </math> Kedua, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{7x - 3}{x^2 - 4x + 3} \\ =&\; \frac{7x - 3}{(x - 1)(x - 3)} \\ =&\; \frac{A}{x - 1} + \frac{B}{x - 3} \\ =&\; \frac{A(x - 3) + B(x - 1)}{x^2 - 4x + 3} \\ =&\; \frac{(A + B)x + (-3A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 7</math> dan <math display="inline">-3A - B = -3</math> dapat diselesaikan, yaitu <math display="inline">A = -2</math> dan {{nowrap|<math display="inline">B = 9</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> === integral substitusi trigonometri === {| class="wikitable" |- ! Bentuk !! Substitusi Trigonometri |- | <math>\sqrt{a^2 - b^2 x^2}</math> || <math>x = \frac{a}{b} sin \, \alpha </math> |- | <math>\sqrt{a^2 + b^2 x^2}</math> || <math>x = \frac{a}{b} tan \, \alpha</math> |- | <math>\sqrt{b^2 x^2 - a^2}</math> || <math>x = \frac{a}{b} sec \, \alpha</math> |} Berikut contoh penyelesaian cara substitusi trigonometri. : <math>\int \frac{dx}{x^2 \sqrt{x^2 + 4}}</math> : <math>x = 2 \tan \, A, dx = 2 \sec^2 \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \int \frac{dx}{x^2\sqrt{x^2+4}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(2 \tan \, A)^2 \sqrt{4 + (2 \tan \, A)^2}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 + 4 \tan^2 \, A}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4(1 + \tan^2 \, A)}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 \sec^2 \, A }} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(4 \tan^2 \, A)(2 \sec \, A)} \\ =&\; \int \frac{\sec \, A \,dA}{4 \tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\sec \, A \,dA}{\tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A} \,dA \end{aligned} </math> Substitusi berikut dapat dibuat. : <math>\int \frac{\cos \, A}{\sin^2 \, A} \,dA</math> : <math>t = \sin \, A, dt = \cos \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A}\,dA \\ =&\; \frac{1}{4} \int \frac{dt}{t^2} \\ =&\; \frac{1}{4} \left(-\frac{1}{t}\right) + C \\ =&\; -\frac{1}{4 t} + C \\ =&\; -\frac{1}{4 \sin \, A} + C \end{aligned} </math> Ingat bahwa <math display="inline">\sin \, A = \frac{x}{\sqrt{x^2 + 4}}</math> berlaku. : <math> \begin{aligned} =&\; -\frac{1}{4 \sin \, A} + C \\ =&\; -\frac{\sqrt{x^2 + 4}}{4x} + C \end{aligned} </math> === integral mutlak === : <math>\int |f(x)| \,dA</math> buatlah <math>f(x) \ge 0</math> jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x) : <math>\int_a^c |f(x)| \, dx = \int_a^b f(x) \, dx + \int_b^c -f(x) \, dx = F(b) - F(a) - F(c) + F(b)</math> jika <math>|f(x)| = \begin{cases} f(x), & \mbox{jika } f(x) \ge b, \\ -f(x), & \mbox{jika } f(x) < b. \end{cases}</math>. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif. === integral fungsi ganjil dan integral fungsi genap === : <math>\int_{-a}^a f(x) \,dx</math> dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral: : integral fungsi ganjil : <math>\int_{-a}^a f(x) \,dx</math> = 0 : integral fungsi genap : <math>\int_{-a}^a f(x) \,dx</math> = <math>2 \int_{0}^a f(x) \,dx</math> === integral notasi === : <math>\int_a^b f(x) \,dx = \int_{a-p}^{b-p} f(x) \,dx = \int_{a+p}^{b+p} f(x) \,dx</math> : <math>\int_a^c f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx</math> : <math>\int_a^b f(x \pm k) \,dx = \int_{a \pm k}^{b \pm k} f(x) \,dx</math> === integral terbalik === : <math>\int_a^b f(x) \,dx = -\int_b^a f(x) \,dx</math> == Jenis integral lainnya == === panjang busur === ; Sumbu ''x'' : <math>S = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' : <math>S = \int_{y_1}^{y_2} \sqrt{1 + (f'(y))^2}\,dy</math> === luas daerah === ; Satu kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} f(x)\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} f(y)\,dy</math> ; Dua kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} (f(x_2) - f(x_1))\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} (f(y_2) - f(y_1))\,dy</math> : atau juga <math>L = \frac {D \sqrt{D}}{6 a^2}</math> === luas permukaan benda putar === ; Sumbu ''x'' sebagai poros : <math>L_p = 2 \pi \int_{x_1}^{x_2} f(x)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' sebagai poros : <math>L_p = 2 \pi \int_{y_1}^{y_2} f(y)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(y))^2}\,dy</math> === volume benda putar === ; Satu kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} (f(x))^2\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} (f(y))^2\,dy</math> ; Dua kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} ((f(x_2))^2 - (f(x_1))^2)\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} ((f(y_2))^2 - (f(y_1))^2)\,dy</math> : atau juga <math>V = |\frac {D^2 \sqrt{D}}{30 a^3}|</math> == Integral lipat == Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga. ;contoh # Tentukan hasil dari: * <math>\int (5x-1)(5x^2-2x+7)^4 dx</math> * <math>\int x^2 cos 3x dx</math> * <math>\int sin^2 x cos x dx</math> * <math>\int cos^2 x sin x dx</math> * <math>\int sec x dx</math> * <math>\int csc x dx</math> * <math>\int \frac{1}{2x^2-5x+3} dx</math> * <math>\int \frac{1}{\sqrt{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> * <math>\int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx</math> * <math>\int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx</math> * <math>\int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx</math> * <math>\int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx</math> * <math>\int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx \\ a=1, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-sin^2 x} \\ &= \frac{1}{cos^2 x} \\ &= sec^2 x \\ \int sec^2 x dx &= tan x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx \\ a=1, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-cos^2 x} \\ &= \frac{1}{sin^2 x} \\ &= csc^2 x \\ \int csc^2 x dx &= -cot x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx \\ a=sin x, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{sin x}{1-sin^2 x} \\ &= \frac{sin x}{cos^2 x} \\ &= sec x \cdot tan x \\ \int sec x \cdot tan x dx &= sec x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx \\ a=cos x, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{cos x}{1-cos^2 x} \\ &= \frac{cos x}{sin^2 x} \\ &= csc x \cdot cot x \\ \int csc x \cdot cot x dx &= -csc x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx \\ \text{misalkan } y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y &= \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y^2 &= \frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}} \\ &= \frac{x}{y} \\ y^3 &= x \\ 3y^2 dy &= dx \\ \int y \cdot 3y^2 dy \\ \int 3y^3 dy \\ \frac{3y^4}{4}+C \\ \end{aligned} </math> </div></div> # Tentukan hasil dari: * <math>\int_{1}^{5} 2x-3 dx</math> * <math>\int_{0}^{4} |x-2| dx</math> * <math>\int_{2}^{4} |3-x| dx</math> * <math>\int_{1}^{5} x^2-6x+8 dx</math> * <math>\int_{-3}^{5} |x^2-2x-8| dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} 2x-3 dx &= \left. x^2 - 3x \right|_{1}^{5} \\ &= 10 - (-2) \\ &= 12 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{0}^{4} |x-2| dx \\ \text{ tentukan nilai harga nol } \\ x-2 &= 0 \\ x &= 2 \\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti } \\ |x-2| = \begin{cases} x-2, & \mbox{jika } x \ge 2, \\ -(x-2), & \mbox{jika } x < 2. \end{cases} \\ \int_{0}^{4} |x-2| dx &= \int_{2}^{4} x-2 dx + \int_{0}^{2} -(x-2) dx \\ &= \left. (\frac{x^2}{2}-2x) \right|_{2}^{4} + \left. (\frac{-x^2}{2}+2x) \right|_{0}^{2} \\ &= 0 - (-2) + 2 - 0 \\ &= 4 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{2}^{4} |3-x| dx \\ \text{ tentukan nilai syarat-syarat } \\ 3-x &\ge 0 \\ \text{ harga nol } \\ x &= 3 \\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti } \\ |3-x| = \begin{cases} 3-x, & \mbox{jika } x \le 3, \\ -(3-x), & \mbox{jika } x > 3. \end{cases} \\ \int_{2}^{4} |3-x| dx &= \int_{2}^{3} 3-x dx + \int_{3}^{4} -(3-x) dx \\ &= \left. (3x-\frac{x^2}{2}) \right|_{2}^{3} + \left. (-3x+\frac{x^2}{2}) \right|_{3}^{4} \\ &= 4.5 - 4 - 4 - (-4.5) \\ &= 1 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} x^2-6x+8 dx &= \left. \frac{x^3}{3}-3x^2+8x \right|_{1}^{5} \\ &= \frac{20}{3} - \frac{16}{3} \\ &= \frac{4}{3} \\ &= 1\frac{1}{3} \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{-3}^{5} |x^2-2x-8| dx \\ \text{ tentukan nilai syarat-syarat } \\ x^2-2x-8 &\ge 0 \\ \text{ harga nol } \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x = 4 \\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti } \\ |x^2-2x-8| = \begin{cases} x^2-2x-8, & \mbox{jika } x \le -2, \\ -(x^2-2x-8), & \mbox{jika } -2 < x < 4, \\ x^2-2x-8, & \mbox{jika } x \ge 4 \end{cases} \\ \int_{-3}^{5} |x^2-2x-8| dx &= \int_{-3}^{-2} x^2-2x-8 dx + \int_{-2}^{4} -(x^2-2x-8) dx + \int_{4}^{5} x^2-2x-8 dx \\ &= \left. (\frac{x^3}{3}-x^2-8x) \right|_{-3}^{-2} + \left. (\frac{-x^3}{3}+x^2+8x) \right|_{-2}^{4} + \left. (\frac{x^3}{3}-x^2-8x) \right|_{4}^{5} \\ &= \frac{28}{3} - 6 + \frac{80}{3} - (-\frac{28}{3}) - \frac{70}{3} - (-\frac{80}{3}) \\ &= \frac{128}{3} \\ &= 42\frac{2}{3} \\ \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x\,dx \\ L &= \frac{1}{2} x^2 \\ L &= \frac{1}{2} xy \quad (y = x) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x^2</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x^2\,dx \\ L &= \frac{1}{3} x^3 \\ L &= \frac{1}{3} xy \quad (y = x^2) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = \sqrt{x}</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int \sqrt{x}\,dx \\ L &= \frac{2}{3} x^{\frac{3}{2}} \\ L &= \frac{2}{3} xy \quad (y = \sqrt{x}) \end{aligned} </math> </div></div> # Buktikan luas persegi <math>L = s^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = s \text{ dan titik (''s'', ''s''), } \\ L &= \int_{0}^{s} s\,dx \\ L &= sx |_{0}^{s} \\ L &= ss - 0 \\ L &= s^2 \end{aligned} </math> </div></div> # Buktikan luas persegi panjang <math>L = pl</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = l \text{dan titik (''p'', ''l''), } \\ L &= \int_{0}^{p} l\,dx \\ L &= lx |_{0}^{p} \\ L &= pl - 0 \\ L &= pl \end{aligned} </math> </div></div> * Buktikan luas segitiga <math>L = \frac{at}{2}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{-tx}{a} + t \text { dan titik (''a'', ''t''), } \\ L &= \int_{0}^{a} \left (\frac{-tx}{a} + t \right) \,dx \\ L &= \left. \frac{-tx^2}{2a} + tx \right|_{0}^{a} \\ L &= \frac{-ta^2}{2a} + ta - 0 + 0 \\ L &= \frac{-ta}{2} + ta \\ L &= \frac{at}{2} \end{aligned} </math> </div></div> # Buktikan volume tabung <math>V = \pi r^2t</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ Dengan posisi } y = r \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} r^2\,dx \\ V &= \pi \left. r^2x \right|_{0}^{t} \\ V &= \pi r^2t - 0 \\ V &= \pi r^2t \end{aligned} </math> </div></div> # Buktikan volume kerucut <math>V = \frac{\pi r^2t}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{rx}{t} \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} \left (\frac{rx}{t} \right)^2 \,dx \\ V &= \pi \left. \frac{r^2 x^3}{3t^2} \right|_{0}^{t} \\ V &= \pi \frac{r^2 t^3}{3t^2} - 0 \\ V &= \frac{\pi r^2t}{3} \end{aligned} </math> </div></div> # Buktikan volume bola <math>V = \frac{4 \pi r^3}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), } \\ V &= \pi \int_{-r}^{r} \left(\sqrt {r^2 - x^2}\right)^2\,dx \\ V &= \pi \int_{-r}^{r} r^2 - x^2 dx \\ V &= \pi \left. r^2x - \frac{x^3}{3} \right|_{-r}^{r} \\ V &= \pi \left (r^3 - \frac{r^3}{3} - \left (-r^3 + \frac{r^3}{3} \right) \right) \\ V &= \frac{4 \pi r^3}{3} \end{aligned} </math> </div></div> # Buktikan luas permukaan bola <math>L = 4 \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{selanjutnya } \\ ds &= \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ ds &= \sqrt{1 + \frac{x^2}{r^2 - x^2}}\,dx \\ ds &= \sqrt{\frac{r^2}{r^2 - x^2}}\,dx \\ ds &= \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ \text{sehingga } \\ L &= 2 \pi \int_{-r}^{r} \sqrt{r^2 - x^2} \cdot \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ L &= 2 \pi \int_{-r}^{r} r\,dx \\ L &= 2 \pi rx|_{-r}^{r} \\ L &= 2 \pi (r r - r (-r)) \\ L &= 2 \pi (r^2 + r^2) \\ L &= 4 \pi r^2 \end{aligned} </math> </div></div> # Buktikan keliling lingkaran <math>K = 2 \pi r</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0) } \\ \text{kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{sehingga } \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{x^2}{r^2 - x^2})}\,dx \\ K &= 4 \int_{0}^{r} \sqrt\frac{r^2}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \sqrt\frac{1}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \frac{1}{\sqrt{r^2 - x^2}}\,dx \\ K &= 4r \left. \arcsin \left (\frac{x}{r} \right) \right|_{0}^{r} \\ K &= 4r \left (\arcsin \left (\frac{r}{r}\right) - \arcsin \left (\frac{0}{r} \right) \right) \\ K &= 4r \left (\arcsin\left (1 \right) - \arcsin \left(0 \right) \right)) \\ K &= 4r \left (\frac{\pi}{2} \right) \\ K &= 2 \pi r \end{aligned} </math> </div></div> # Buktikan luas lingkaran <math>L = \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. } \\ \sin(\theta) &= \frac{x}{r} \\ x &= r \sin(\theta) \\ dx &= r \cos(\theta)\,d\theta \\ \text{dengan turunan di atas } \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - x^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - (r \sin(\theta))^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - r^2 \sin^2(\theta)}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 (1 - \sin^2(\theta))}\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta)\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta) (r \cos(\theta)\,d\theta) \\ L &= 4 \int_{0}^{r} r^2 \cos^2(\theta)\,d\theta \\ L &= 4r^2 \int_{0}^{r} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta \\ L &= 2r^2 \int_{0}^{r} (1 + \cos(2\theta))\,d\theta \\ L &= 2r^2 \left (\theta + \frac{1}{2} \sin(2\theta) \right)|_{0}^{r} \\ L &= 2r^2 (\theta + \sin(\theta) \cos(\theta))|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left (\frac{x}{r} \right) + \left (\frac{x}{r} \right) \left (\frac{r^2 - x^2}{r} \right) \right)|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left(\frac{r}{r} \right) + \left (\frac{r}{r} \right) \left (\frac {r^2 - r^2}{r} \right) \right) - \left(\arcsin \left (\frac{0}{r} \right) + \left (\frac{0}{r}\right) \left (\frac{r^2 - 0^2}{r} \right) \right) \\ L &= 2r^2 (\arcsin(1) + 0 - (\arcsin(0) + 0)) \\ L &= 2r^2 \left (\frac{\pi}{2} \right) \\ L &= \pi r^2 \end{aligned} </math> </div></div> # Buktikan luas elips <math>L = \pi ab</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \frac{b \sqrt{a^2 - x^2}}{a} \text{ serta (-''a'', 0) dan (''a'', 0), } \\ L &= 4 \int_{0}^{r} \frac{b \sqrt{a^2 - x^2}}{a}\,dx \\ L &= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx \\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, } \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{\frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx}{4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx} \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{b}{a} \\ L_\text{elips} &= \frac{b}{a} L_\text{ling} \\ L_\text{elips} &= \frac{b}{a} \pi a^2 \\ L_\text{elips} &= \pi ab \end{aligned} </math> </div></div> # Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ x^-2x &= 4x+7 \\ x^2-6x-7 &= 0 \\ (x+1)(x-7) &= 0 \\ x=-1 &\text{ atau } x=7 \\ y &= 4(-1)+7 = 3 \\ y &= 4(7)+7 = 35 \\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut } \\ L &= \int_{-1}^{7} 4x+7-(x^2-2x)\,dx \\ &= \int_{-1}^{7} -x^2+6x+7\,dx \\ &= -\frac{x^3}{3}+3x^2+7x|_{-1}^{7} \\ &= -\frac{7^3}{3}+3(7)^2+7(7)-(-\frac{(-1)^3}{3}+3(-1)^2+7(-1)) \\ &= \frac{245}{3}-(-\frac{13}{3}) \\ &= \frac{258}{3} \\ &= 86 \\ \end{aligned} </math> </div></div> # Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ 6-1,5x &= x-4 \\ 2,5x &= 10 \\ x &= 4 \\ y &= 4-4 = 0 \\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut } \\ \text{untuk x=0 } \\ y &= 6-1,5(0) = 6 \\ y &= 0-4 = -4 \\ L &= \pi \int_{0}^{4} ((6-1,5x)^2-(x-4)^2)\,dx \\ &= \pi \int_{0}^{4} (36-18x+2,25x^2-(x^2-8x+16))\,dx \\ &= \pi \int_{0}^{4} (1,25x^2-10x+20)\,dx \\ &= \pi (\frac{1,25x^3}{3}-5x^2+20x)|_{0}^{4} \\ &= \pi (\frac{1,25(4)^3}{3}-5(4)^2+20(4)-(\frac{1,25(0)^3}{3}-5(0)^2+20(0))) \\ &= \pi (\frac{30}{3}-0) \\ &= 10\pi \\ \end{aligned} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 5f1rpw7v10hffyme5u83bt8wfumj91s 115031 115030 2026-04-27T02:23:27Z ~2026-25546-23 43043 /* integral pecahan parsial */ 115031 wikitext text/x-wiki == Kaidah umum == :# <math>\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ konstan)}\,\!</math> :# <math>\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx</math> :# <math>\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx</math> :# <math>\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx</math> :# <math>\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\! </math> :# <math>\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C </math> :# <math>\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C </math> :# <math>\int f(x) \, dx = F(x) + C</math> :# <math>\int_a^b f(x) \, dx = F(b) - F(a)</math> :# <math>\int f(x) \circ g(x) dx</math> = f’(g(x)) g’(x) :# <math>\int f^{-1}(x) dx = \frac{1}{f'(f'(x))}</math> == rumus sederhana == :<math>\int \, dx = x + C</math> :<math>\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int {dx \over x} = \ln{\left|x\right|} + C</math> <math>\int {dx \over \sqrt{a^2-b^2 x^2}} = \frac{1}{b} \arcsin \, {bx \over a} + C</math> :<math>\int {dx \over {a^2+b^2x^2}} = {1 \over ab}\arctan \, {bx \over a} + C</math> <math>\int {dx \over x\sqrt{b^2x^2-a^2}} = {1 \over a}\arcsec \, {bx \over a} + C</math> ; Eksponen dan logaritma :<math>\int e^x\,dx = e^x + C</math> :<math>\int a^x\,dx = \frac{a^x}{\ln{a}} + C</math> :<math>\int \ln \, x\,dx = x \cdot \, \ln {x} - x + C</math> :<math>\int \, ^b\!\log \, x\,dx = x \cdot \, ^b\!\log \, x - x \cdot \,^b\!\log e + C</math> ; Trigonometri :<math>\int \sin \, x \, dx = -\cos \, x + C</math> :<math>\int \cos \, x \, dx = \sin \, x + C</math> :<math>\int \tan \, x \, dx = -\ln{\left| \cos \, x \right|} + C</math> :<math>\int \cot \, x \, dx = \ln{\left| \sin \, x \right|} + C</math> :<math>\int \sec \, x \, dx = \ln{\left| \sec \, x + \tan \, x \right|} + C</math> :<math>\int \csc \, x \, dx = -\ln{\left| \csc \, x + \cot \, x \right|} + C</math> ; Hiperbolik :<math>\int \sinh \, x \, dx = \cosh \, x + C</math> :<math>\int \cosh \, x \, dx = \sinh \, x + C</math> :<math>\int \tanh \, x \, dx = \ln{\left| \cosh \, x \right|} + C</math> :<math>\int \coth \, x \, dx = \ln{\left| \sinh \, x \right|} + C</math> :<math>\int \mbox{sech} \, x \, dx = \arctan \,(\sinh \, x) + C</math> :<math>\int \mbox{csch} \, x \, dx = \ln{\left| \tanh \, {x \over 2} \right|} + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : 1. <math>\int x^3 - cos 3x + e^{5x+2} - \frac{1}{2x+5}\,dx</math> : <math>= \frac{1}{3+1}x^{3+1} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : <math>= \frac{x^4}{4} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : 2. <math>\int x(x-5)^4\,dx</math> : <math>= \int x(x^4-4x^3(5)+6x^2(25)-4x(125)+625)\,dx</math> : <math>= \int x(x^4-20x^3+150x^2-500x+625)\,dx</math> : <math>= \int x^5-20x^4+150x^3-500x^2+625x\,dx</math> : <math>= \frac{x^6}{6}-4x^5+\frac{150x^4}{4}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> : <math>= \frac{x^6}{6}-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : 1. <math>\int \frac{\ln \, x}{x}\,dx</math> : <math>u = \ln \, x\, du = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{\ln(x)}{x} dx \\ \int u du \\ \frac{1}{2} u^2 + C \\ \frac{1}{2} ln^2 x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x-5,\, du = dx,\, x = u+5</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4\,dx \\ \int (u+5)u^4\,du \\ \int (u^5+5u^4)\,du \\ \frac{1}{6} u^6+u^5 + C \\ \frac{1}{6} (x-5)^6+(x-5)^5 + C \\ \frac{1}{6} (x^6-6x^5(5)+15x^4(25)-20x^3(125)+15x^2(625)-6x(3125)+15625)+(x^5-5x^4(5)+10x^3(25)-10x^2(125)+5x(625)+3125) + C \\ \frac{1}{6}x^6-5x^5+\frac{125x^4}{2}-\frac{1250x^3}{3}+\frac{3125x^2}{2}-3125x+\frac{15625}{6}+x^5-25x^4+250x^3-1250x^2+3125x+3125 + C \\ \frac{1}{6}x^6-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C \\ \end{aligned} </math> </div></div> === integral parsial === ;Cara 1: Rumus Integral parsial diselesaikan dengan rumus berikut. : <math>\int u\,dv = uv - \int v\,du </math> Berikut contoh penyelesaian cara parsial dengan rumus. : 1. <math>\int x \sin \, x \, dx</math> : <math>u = x, du = 1 \, dx, dv = \sin \, x \, dx, v = -\cos \, x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \,dx \\ (x)(-\cos \, x) - \int (-\cos \, x)(1 dx) \\ -x \cos \, x + \int \cos \, x dx \\ -x \cos \, x + \sin \, x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x,\, du = 1\,dx,\, dv = (x-5)^4\,dx,\, v = \frac{1}{5}(x-5)^5</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - \int (\frac{1}{5}(x-5)^5)(1\,dx) \\ \frac{x}{5}(x-5)^5 - \int \frac{1}{5}(x-5)^5\,dx \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6\,dx \\ \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x\,dx</math> : <math>u = \ln \,x,\, du = \frac{1}{x}\,dx,\, dv = x\,dx,\, v = \frac{x^2}{2}</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x\, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{x^2}{2})(\frac{1}{x}\,dx) \\ \frac{x^2}{2} \ln \,x - \int \frac{x^2}{2}(\frac{1}{x})\,dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \,dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x\,dx</math> ini terjadi dua kali integral parsial : <math>u = \sin \,x,\, du = \cos \,x\,dx,\, dv = e^x\,dx,\, v = e^x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x\, dx \\ (\sin \,x)(e^x) - \int (e^x)(\cos \,x\,dx) \\ e^x \sin \,x - \int e^x \cos \,x\,dx \\ u = \cos \,x,\, du = -\sin \,x\,dx,\, dv = e^x\,dx,\, v = e^x \\ (\cos \,x)(e^x) - \int (e^x)(-\sin \,x\,dx) \\ e^x \cos \,x + \int e^x \sin \,x\,dx \\ e^x \sin \,x - (e^x \cos \,x + \int e^x \sin \,x\,dx) \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x\,dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> ;Cara 2: Tabel Untuk <math display="inline">\int u\,dv</math>, berlaku ketentuan sebagai berikut. {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>u</math> || <math>dv</math> |- | - || <math>\frac{du}{dx}</math> || <math>v</math> |- | + || <math>\frac{d^2u}{dx^2}</math> || <math>\int v\,dx</math> |} Berikut contoh penyelesaian cara parsial dengan tabel. : 1. <math>\int x \sin \, x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>\sin \, x</math> |- | - || <math>1</math> || <math>-\cos \, x</math> |- | + || <math>0</math> || <math>-\sin \, x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \, dx \\ (x)(-\cos \, x) - (1)(-\sin \, x) + C \\ -x \cos \, x + \sin \, x + C \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4 \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>(x-5)^4</math> |- | - || <math>1</math> || <math>\frac{1}{5}(x-5)^5</math> |- | + || <math>0</math> || <math>\frac{1}{30}(x-5)^6</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - (1)(\frac{1}{30}(x-5)^6) + C \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6 + C \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\ln \,x</math> || <math>x</math> |- | - || <math>\frac{1}{x}</math> || <math>\frac{x^2}{2}</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x \, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{1}{x})(\frac{x^2}{2}) \, dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \, dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x \, dx</math> ini terjadi dua kali integral parsial {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\sin \,x</math> || <math>e^x</math> |- | - || <math>\cos \, x</math> || <math>e^x</math> |- | + || <math>-\sin \,x</math> || <math>e^x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x \, dx \\ (\sin \,x)(e^x) - (\cos \,x)(e^x) + \int (-\sin \,x)(e^x) \, dx \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x) \, dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : 1. <math>\int \frac{1}{x^2 - 4}\,dx</math> Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{1}{x^2 - 4} \\ =&\; \frac{1}{(x + 2)(x - 2)} \\ =&\; \frac{A}{x + 2} + \frac{B}{x - 2} \\ =&\; \frac{A(x - 2) + B(x + 2)}{x^2 - 4} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> : 2. <math>\int \frac{x^2+5x+4}{(x-3)^3}\,dx</math> : cara 1 Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^3} \\ &\; \text{untuk mencari C yaitu } (3)^2+5(3)+4=28 \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x - 3)^2 + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x^2 - 6x + 9) + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 - 6Ax + 9A + Bx - 3B}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 + (- 6A + B)x +(9 - 3B)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ \end{aligned} </math> Kita tahu bahwa <math display="inline">A = 1</math> dan <math display="inline">-6A + B = 5</math> dapat diselesaikan, yaitu <math display="inline">A = 1</math> dan {{nowrap|<math display="inline">B = 11</math>.}} : <math> \begin{aligned} &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int (\frac{1}{x - 3} + \frac{11}{(x - 3)^2} + \frac{28}{(x - 3)^3})\,dx \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : cara 2 Pertama, ubah u=x-3 menjadi x=u+3 dan du=dx tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{(u+3)^2+5(u+3)+4}{u^3} \\ =&\; \frac{u^2+11u+28}{u^3} \\ &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int \frac{u^2+11u+28}{u^3}\,du \\ =&\; \int (\frac{u^2}{u^3} + \frac{11u}{u^3} + \frac{28}{u^3})\,du \\ =&\; \int (\frac{1}{u} + \frac{11}{u^2} + \frac{28}{u^3})\,du \\ =&\; \ln u + (-11 u^{-1}) + (-14 u^{-2}) + C \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : 3. <math>\int \frac{x^4+2x^3-13x^2-7x+21}{x^2-4x+3}\,dx</math> Pertama, penyelesaian dengan persamaan polinomial tersebut. : <math> \begin{aligned} =&\; \frac{x^4+2x^3-13x^2-7x+21}{x^2-4x+3} \\ =&\; x^2+6x+8 + \frac{7x-3}{x^2-4x+3} \\ =&\; \frac{1}{3}x^3+3x^2+8x+ \int \frac{7x-3}{x^2-4x+3}\,dx \end{aligned} </math> Kedua, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{7x - 3}{x^2 - 4x + 3} \\ =&\; \frac{7x - 3}{(x - 1)(x - 3)} \\ =&\; \frac{A}{x - 1} + \frac{B}{x - 3} \\ =&\; \frac{A(x - 3) + B(x - 1)}{x^2 - 4x + 3} \\ =&\; \frac{(A + B)x + (-3A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 7</math> dan <math display="inline">-3A - B = -3</math> dapat diselesaikan, yaitu <math display="inline">A = -2</math> dan {{nowrap|<math display="inline">B = 9</math>.}} : <math> \begin{aligned} &\; \frac{1}{3}x^3+3x^2+8x+ \int \frac{7x-3}{x^2-4x+3}\,dx \\ =&\; \frac{1}{3}x^3+3x^2+8x+ \int (-\frac{2}{(x - 1)} + \frac{9}{x - 3})\,dx \\ =&\; \frac{1}{3}x^3+3x^2+8x- 2 \ln|x - 3| + 9 \ln|x - 3| \,dx + C \\ \end{aligned}</math> === integral substitusi trigonometri === {| class="wikitable" |- ! Bentuk !! Substitusi Trigonometri |- | <math>\sqrt{a^2 - b^2 x^2}</math> || <math>x = \frac{a}{b} sin \, \alpha </math> |- | <math>\sqrt{a^2 + b^2 x^2}</math> || <math>x = \frac{a}{b} tan \, \alpha</math> |- | <math>\sqrt{b^2 x^2 - a^2}</math> || <math>x = \frac{a}{b} sec \, \alpha</math> |} Berikut contoh penyelesaian cara substitusi trigonometri. : <math>\int \frac{dx}{x^2 \sqrt{x^2 + 4}}</math> : <math>x = 2 \tan \, A, dx = 2 \sec^2 \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \int \frac{dx}{x^2\sqrt{x^2+4}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(2 \tan \, A)^2 \sqrt{4 + (2 \tan \, A)^2}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 + 4 \tan^2 \, A}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4(1 + \tan^2 \, A)}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 \sec^2 \, A }} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(4 \tan^2 \, A)(2 \sec \, A)} \\ =&\; \int \frac{\sec \, A \,dA}{4 \tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\sec \, A \,dA}{\tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A} \,dA \end{aligned} </math> Substitusi berikut dapat dibuat. : <math>\int \frac{\cos \, A}{\sin^2 \, A} \,dA</math> : <math>t = \sin \, A, dt = \cos \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A}\,dA \\ =&\; \frac{1}{4} \int \frac{dt}{t^2} \\ =&\; \frac{1}{4} \left(-\frac{1}{t}\right) + C \\ =&\; -\frac{1}{4 t} + C \\ =&\; -\frac{1}{4 \sin \, A} + C \end{aligned} </math> Ingat bahwa <math display="inline">\sin \, A = \frac{x}{\sqrt{x^2 + 4}}</math> berlaku. : <math> \begin{aligned} =&\; -\frac{1}{4 \sin \, A} + C \\ =&\; -\frac{\sqrt{x^2 + 4}}{4x} + C \end{aligned} </math> === integral mutlak === : <math>\int |f(x)| \,dA</math> buatlah <math>f(x) \ge 0</math> jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x) : <math>\int_a^c |f(x)| \, dx = \int_a^b f(x) \, dx + \int_b^c -f(x) \, dx = F(b) - F(a) - F(c) + F(b)</math> jika <math>|f(x)| = \begin{cases} f(x), & \mbox{jika } f(x) \ge b, \\ -f(x), & \mbox{jika } f(x) < b. \end{cases}</math>. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif. === integral fungsi ganjil dan integral fungsi genap === : <math>\int_{-a}^a f(x) \,dx</math> dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral: : integral fungsi ganjil : <math>\int_{-a}^a f(x) \,dx</math> = 0 : integral fungsi genap : <math>\int_{-a}^a f(x) \,dx</math> = <math>2 \int_{0}^a f(x) \,dx</math> === integral notasi === : <math>\int_a^b f(x) \,dx = \int_{a-p}^{b-p} f(x) \,dx = \int_{a+p}^{b+p} f(x) \,dx</math> : <math>\int_a^c f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx</math> : <math>\int_a^b f(x \pm k) \,dx = \int_{a \pm k}^{b \pm k} f(x) \,dx</math> === integral terbalik === : <math>\int_a^b f(x) \,dx = -\int_b^a f(x) \,dx</math> == Jenis integral lainnya == === panjang busur === ; Sumbu ''x'' : <math>S = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' : <math>S = \int_{y_1}^{y_2} \sqrt{1 + (f'(y))^2}\,dy</math> === luas daerah === ; Satu kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} f(x)\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} f(y)\,dy</math> ; Dua kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} (f(x_2) - f(x_1))\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} (f(y_2) - f(y_1))\,dy</math> : atau juga <math>L = \frac {D \sqrt{D}}{6 a^2}</math> === luas permukaan benda putar === ; Sumbu ''x'' sebagai poros : <math>L_p = 2 \pi \int_{x_1}^{x_2} f(x)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' sebagai poros : <math>L_p = 2 \pi \int_{y_1}^{y_2} f(y)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(y))^2}\,dy</math> === volume benda putar === ; Satu kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} (f(x))^2\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} (f(y))^2\,dy</math> ; Dua kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} ((f(x_2))^2 - (f(x_1))^2)\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} ((f(y_2))^2 - (f(y_1))^2)\,dy</math> : atau juga <math>V = |\frac {D^2 \sqrt{D}}{30 a^3}|</math> == Integral lipat == Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga. ;contoh # Tentukan hasil dari: * <math>\int (5x-1)(5x^2-2x+7)^4 dx</math> * <math>\int x^2 cos 3x dx</math> * <math>\int sin^2 x cos x dx</math> * <math>\int cos^2 x sin x dx</math> * <math>\int sec x dx</math> * <math>\int csc x dx</math> * <math>\int \frac{1}{2x^2-5x+3} dx</math> * <math>\int \frac{1}{\sqrt{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> * <math>\int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx</math> * <math>\int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx</math> * <math>\int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx</math> * <math>\int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx</math> * <math>\int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx \\ a=1, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-sin^2 x} \\ &= \frac{1}{cos^2 x} \\ &= sec^2 x \\ \int sec^2 x dx &= tan x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx \\ a=1, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-cos^2 x} \\ &= \frac{1}{sin^2 x} \\ &= csc^2 x \\ \int csc^2 x dx &= -cot x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx \\ a=sin x, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{sin x}{1-sin^2 x} \\ &= \frac{sin x}{cos^2 x} \\ &= sec x \cdot tan x \\ \int sec x \cdot tan x dx &= sec x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx \\ a=cos x, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{cos x}{1-cos^2 x} \\ &= \frac{cos x}{sin^2 x} \\ &= csc x \cdot cot x \\ \int csc x \cdot cot x dx &= -csc x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx \\ \text{misalkan } y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y &= \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y^2 &= \frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}} \\ &= \frac{x}{y} \\ y^3 &= x \\ 3y^2 dy &= dx \\ \int y \cdot 3y^2 dy \\ \int 3y^3 dy \\ \frac{3y^4}{4}+C \\ \end{aligned} </math> </div></div> # Tentukan hasil dari: * <math>\int_{1}^{5} 2x-3 dx</math> * <math>\int_{0}^{4} |x-2| dx</math> * <math>\int_{2}^{4} |3-x| dx</math> * <math>\int_{1}^{5} x^2-6x+8 dx</math> * <math>\int_{-3}^{5} |x^2-2x-8| dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} 2x-3 dx &= \left. x^2 - 3x \right|_{1}^{5} \\ &= 10 - (-2) \\ &= 12 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{0}^{4} |x-2| dx \\ \text{ tentukan nilai harga nol } \\ x-2 &= 0 \\ x &= 2 \\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti } \\ |x-2| = \begin{cases} x-2, & \mbox{jika } x \ge 2, \\ -(x-2), & \mbox{jika } x < 2. \end{cases} \\ \int_{0}^{4} |x-2| dx &= \int_{2}^{4} x-2 dx + \int_{0}^{2} -(x-2) dx \\ &= \left. (\frac{x^2}{2}-2x) \right|_{2}^{4} + \left. (\frac{-x^2}{2}+2x) \right|_{0}^{2} \\ &= 0 - (-2) + 2 - 0 \\ &= 4 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{2}^{4} |3-x| dx \\ \text{ tentukan nilai syarat-syarat } \\ 3-x &\ge 0 \\ \text{ harga nol } \\ x &= 3 \\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti } \\ |3-x| = \begin{cases} 3-x, & \mbox{jika } x \le 3, \\ -(3-x), & \mbox{jika } x > 3. \end{cases} \\ \int_{2}^{4} |3-x| dx &= \int_{2}^{3} 3-x dx + \int_{3}^{4} -(3-x) dx \\ &= \left. (3x-\frac{x^2}{2}) \right|_{2}^{3} + \left. (-3x+\frac{x^2}{2}) \right|_{3}^{4} \\ &= 4.5 - 4 - 4 - (-4.5) \\ &= 1 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} x^2-6x+8 dx &= \left. \frac{x^3}{3}-3x^2+8x \right|_{1}^{5} \\ &= \frac{20}{3} - \frac{16}{3} \\ &= \frac{4}{3} \\ &= 1\frac{1}{3} \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{-3}^{5} |x^2-2x-8| dx \\ \text{ tentukan nilai syarat-syarat } \\ x^2-2x-8 &\ge 0 \\ \text{ harga nol } \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x = 4 \\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti } \\ |x^2-2x-8| = \begin{cases} x^2-2x-8, & \mbox{jika } x \le -2, \\ -(x^2-2x-8), & \mbox{jika } -2 < x < 4, \\ x^2-2x-8, & \mbox{jika } x \ge 4 \end{cases} \\ \int_{-3}^{5} |x^2-2x-8| dx &= \int_{-3}^{-2} x^2-2x-8 dx + \int_{-2}^{4} -(x^2-2x-8) dx + \int_{4}^{5} x^2-2x-8 dx \\ &= \left. (\frac{x^3}{3}-x^2-8x) \right|_{-3}^{-2} + \left. (\frac{-x^3}{3}+x^2+8x) \right|_{-2}^{4} + \left. (\frac{x^3}{3}-x^2-8x) \right|_{4}^{5} \\ &= \frac{28}{3} - 6 + \frac{80}{3} - (-\frac{28}{3}) - \frac{70}{3} - (-\frac{80}{3}) \\ &= \frac{128}{3} \\ &= 42\frac{2}{3} \\ \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x\,dx \\ L &= \frac{1}{2} x^2 \\ L &= \frac{1}{2} xy \quad (y = x) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x^2</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x^2\,dx \\ L &= \frac{1}{3} x^3 \\ L &= \frac{1}{3} xy \quad (y = x^2) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = \sqrt{x}</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int \sqrt{x}\,dx \\ L &= \frac{2}{3} x^{\frac{3}{2}} \\ L &= \frac{2}{3} xy \quad (y = \sqrt{x}) \end{aligned} </math> </div></div> # Buktikan luas persegi <math>L = s^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = s \text{ dan titik (''s'', ''s''), } \\ L &= \int_{0}^{s} s\,dx \\ L &= sx |_{0}^{s} \\ L &= ss - 0 \\ L &= s^2 \end{aligned} </math> </div></div> # Buktikan luas persegi panjang <math>L = pl</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = l \text{dan titik (''p'', ''l''), } \\ L &= \int_{0}^{p} l\,dx \\ L &= lx |_{0}^{p} \\ L &= pl - 0 \\ L &= pl \end{aligned} </math> </div></div> * Buktikan luas segitiga <math>L = \frac{at}{2}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{-tx}{a} + t \text { dan titik (''a'', ''t''), } \\ L &= \int_{0}^{a} \left (\frac{-tx}{a} + t \right) \,dx \\ L &= \left. \frac{-tx^2}{2a} + tx \right|_{0}^{a} \\ L &= \frac{-ta^2}{2a} + ta - 0 + 0 \\ L &= \frac{-ta}{2} + ta \\ L &= \frac{at}{2} \end{aligned} </math> </div></div> # Buktikan volume tabung <math>V = \pi r^2t</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ Dengan posisi } y = r \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} r^2\,dx \\ V &= \pi \left. r^2x \right|_{0}^{t} \\ V &= \pi r^2t - 0 \\ V &= \pi r^2t \end{aligned} </math> </div></div> # Buktikan volume kerucut <math>V = \frac{\pi r^2t}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{rx}{t} \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} \left (\frac{rx}{t} \right)^2 \,dx \\ V &= \pi \left. \frac{r^2 x^3}{3t^2} \right|_{0}^{t} \\ V &= \pi \frac{r^2 t^3}{3t^2} - 0 \\ V &= \frac{\pi r^2t}{3} \end{aligned} </math> </div></div> # Buktikan volume bola <math>V = \frac{4 \pi r^3}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), } \\ V &= \pi \int_{-r}^{r} \left(\sqrt {r^2 - x^2}\right)^2\,dx \\ V &= \pi \int_{-r}^{r} r^2 - x^2 dx \\ V &= \pi \left. r^2x - \frac{x^3}{3} \right|_{-r}^{r} \\ V &= \pi \left (r^3 - \frac{r^3}{3} - \left (-r^3 + \frac{r^3}{3} \right) \right) \\ V &= \frac{4 \pi r^3}{3} \end{aligned} </math> </div></div> # Buktikan luas permukaan bola <math>L = 4 \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{selanjutnya } \\ ds &= \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ ds &= \sqrt{1 + \frac{x^2}{r^2 - x^2}}\,dx \\ ds &= \sqrt{\frac{r^2}{r^2 - x^2}}\,dx \\ ds &= \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ \text{sehingga } \\ L &= 2 \pi \int_{-r}^{r} \sqrt{r^2 - x^2} \cdot \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ L &= 2 \pi \int_{-r}^{r} r\,dx \\ L &= 2 \pi rx|_{-r}^{r} \\ L &= 2 \pi (r r - r (-r)) \\ L &= 2 \pi (r^2 + r^2) \\ L &= 4 \pi r^2 \end{aligned} </math> </div></div> # Buktikan keliling lingkaran <math>K = 2 \pi r</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0) } \\ \text{kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{sehingga } \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{x^2}{r^2 - x^2})}\,dx \\ K &= 4 \int_{0}^{r} \sqrt\frac{r^2}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \sqrt\frac{1}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \frac{1}{\sqrt{r^2 - x^2}}\,dx \\ K &= 4r \left. \arcsin \left (\frac{x}{r} \right) \right|_{0}^{r} \\ K &= 4r \left (\arcsin \left (\frac{r}{r}\right) - \arcsin \left (\frac{0}{r} \right) \right) \\ K &= 4r \left (\arcsin\left (1 \right) - \arcsin \left(0 \right) \right)) \\ K &= 4r \left (\frac{\pi}{2} \right) \\ K &= 2 \pi r \end{aligned} </math> </div></div> # Buktikan luas lingkaran <math>L = \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. } \\ \sin(\theta) &= \frac{x}{r} \\ x &= r \sin(\theta) \\ dx &= r \cos(\theta)\,d\theta \\ \text{dengan turunan di atas } \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - x^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - (r \sin(\theta))^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - r^2 \sin^2(\theta)}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 (1 - \sin^2(\theta))}\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta)\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta) (r \cos(\theta)\,d\theta) \\ L &= 4 \int_{0}^{r} r^2 \cos^2(\theta)\,d\theta \\ L &= 4r^2 \int_{0}^{r} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta \\ L &= 2r^2 \int_{0}^{r} (1 + \cos(2\theta))\,d\theta \\ L &= 2r^2 \left (\theta + \frac{1}{2} \sin(2\theta) \right)|_{0}^{r} \\ L &= 2r^2 (\theta + \sin(\theta) \cos(\theta))|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left (\frac{x}{r} \right) + \left (\frac{x}{r} \right) \left (\frac{r^2 - x^2}{r} \right) \right)|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left(\frac{r}{r} \right) + \left (\frac{r}{r} \right) \left (\frac {r^2 - r^2}{r} \right) \right) - \left(\arcsin \left (\frac{0}{r} \right) + \left (\frac{0}{r}\right) \left (\frac{r^2 - 0^2}{r} \right) \right) \\ L &= 2r^2 (\arcsin(1) + 0 - (\arcsin(0) + 0)) \\ L &= 2r^2 \left (\frac{\pi}{2} \right) \\ L &= \pi r^2 \end{aligned} </math> </div></div> # Buktikan luas elips <math>L = \pi ab</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \frac{b \sqrt{a^2 - x^2}}{a} \text{ serta (-''a'', 0) dan (''a'', 0), } \\ L &= 4 \int_{0}^{r} \frac{b \sqrt{a^2 - x^2}}{a}\,dx \\ L &= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx \\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, } \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{\frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx}{4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx} \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{b}{a} \\ L_\text{elips} &= \frac{b}{a} L_\text{ling} \\ L_\text{elips} &= \frac{b}{a} \pi a^2 \\ L_\text{elips} &= \pi ab \end{aligned} </math> </div></div> # Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ x^-2x &= 4x+7 \\ x^2-6x-7 &= 0 \\ (x+1)(x-7) &= 0 \\ x=-1 &\text{ atau } x=7 \\ y &= 4(-1)+7 = 3 \\ y &= 4(7)+7 = 35 \\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut } \\ L &= \int_{-1}^{7} 4x+7-(x^2-2x)\,dx \\ &= \int_{-1}^{7} -x^2+6x+7\,dx \\ &= -\frac{x^3}{3}+3x^2+7x|_{-1}^{7} \\ &= -\frac{7^3}{3}+3(7)^2+7(7)-(-\frac{(-1)^3}{3}+3(-1)^2+7(-1)) \\ &= \frac{245}{3}-(-\frac{13}{3}) \\ &= \frac{258}{3} \\ &= 86 \\ \end{aligned} </math> </div></div> # Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ 6-1,5x &= x-4 \\ 2,5x &= 10 \\ x &= 4 \\ y &= 4-4 = 0 \\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut } \\ \text{untuk x=0 } \\ y &= 6-1,5(0) = 6 \\ y &= 0-4 = -4 \\ L &= \pi \int_{0}^{4} ((6-1,5x)^2-(x-4)^2)\,dx \\ &= \pi \int_{0}^{4} (36-18x+2,25x^2-(x^2-8x+16))\,dx \\ &= \pi \int_{0}^{4} (1,25x^2-10x+20)\,dx \\ &= \pi (\frac{1,25x^3}{3}-5x^2+20x)|_{0}^{4} \\ &= \pi (\frac{1,25(4)^3}{3}-5(4)^2+20(4)-(\frac{1,25(0)^3}{3}-5(0)^2+20(0))) \\ &= \pi (\frac{30}{3}-0) \\ &= 10\pi \\ \end{aligned} </math> </div></div> [[Kategori:Soal-Soal Matematika]] pgx16th07rtbl9h94sbxisy1nfbqsvo 115032 115031 2026-04-27T02:25:03Z ~2026-25546-23 43043 /* integral pecahan parsial */ 115032 wikitext text/x-wiki == Kaidah umum == :# <math>\int af(x)\,dx = a\int f(x)\,dx \qquad\mbox{(}a \mbox{ konstan)}\,\!</math> :# <math>\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx</math> :# <math>\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx</math> :# <math>\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left[f'(x) \left(\int g(x)\,dx\right)\right]\,dx</math> :# <math>\int [f(x)]^n f'(x)\,dx = {[f(x)]^{n+1} \over n+1} + C \qquad\mbox{(untuk } n\neq -1\mbox{)}\,\! </math> :# <math>\int {f'(x)\over f(x)}\,dx= \ln{\left|f(x)\right|} + C </math> :# <math>\int {f'(x) f(x)}\,dx= {1 \over 2} [ f(x) ]^2 + C </math> :# <math>\int f(x) \, dx = F(x) + C</math> :# <math>\int_a^b f(x) \, dx = F(b) - F(a)</math> :# <math>\int f(x) \circ g(x) dx</math> = f’(g(x)) g’(x) :# <math>\int f^{-1}(x) dx = \frac{1}{f'(f'(x))}</math> == rumus sederhana == :<math>\int \, dx = x + C</math> :<math>\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C\qquad\mbox{ jika }n \ne -1</math> :<math>\int {dx \over x} = \ln{\left|x\right|} + C</math> <math>\int {dx \over \sqrt{a^2-b^2 x^2}} = \frac{1}{b} \arcsin \, {bx \over a} + C</math> :<math>\int {dx \over {a^2+b^2x^2}} = {1 \over ab}\arctan \, {bx \over a} + C</math> <math>\int {dx \over x\sqrt{b^2x^2-a^2}} = {1 \over a}\arcsec \, {bx \over a} + C</math> ; Eksponen dan logaritma :<math>\int e^x\,dx = e^x + C</math> :<math>\int a^x\,dx = \frac{a^x}{\ln{a}} + C</math> :<math>\int \ln \, x\,dx = x \cdot \, \ln {x} - x + C</math> :<math>\int \, ^b\!\log \, x\,dx = x \cdot \, ^b\!\log \, x - x \cdot \,^b\!\log e + C</math> ; Trigonometri :<math>\int \sin \, x \, dx = -\cos \, x + C</math> :<math>\int \cos \, x \, dx = \sin \, x + C</math> :<math>\int \tan \, x \, dx = -\ln{\left| \cos \, x \right|} + C</math> :<math>\int \cot \, x \, dx = \ln{\left| \sin \, x \right|} + C</math> :<math>\int \sec \, x \, dx = \ln{\left| \sec \, x + \tan \, x \right|} + C</math> :<math>\int \csc \, x \, dx = -\ln{\left| \csc \, x + \cot \, x \right|} + C</math> ; Hiperbolik :<math>\int \sinh \, x \, dx = \cosh \, x + C</math> :<math>\int \cosh \, x \, dx = \sinh \, x + C</math> :<math>\int \tanh \, x \, dx = \ln{\left| \cosh \, x \right|} + C</math> :<math>\int \coth \, x \, dx = \ln{\left| \sinh \, x \right|} + C</math> :<math>\int \mbox{sech} \, x \, dx = \arctan \,(\sinh \, x) + C</math> :<math>\int \mbox{csch} \, x \, dx = \ln{\left| \tanh \, {x \over 2} \right|} + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : 1. <math>\int x^3 - cos 3x + e^{5x+2} - \frac{1}{2x+5}\,dx</math> : <math>= \frac{1}{3+1}x^{3+1} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : <math>= \frac{x^4}{4} - \frac{sin 3x}{3} + \frac{e^{5x+2}}{5} - \frac{ln (2x+5)}{2} + C</math> : 2. <math>\int x(x-5)^4\,dx</math> : <math>= \int x(x^4-4x^3(5)+6x^2(25)-4x(125)+625)\,dx</math> : <math>= \int x(x^4-20x^3+150x^2-500x+625)\,dx</math> : <math>= \int x^5-20x^4+150x^3-500x^2+625x\,dx</math> : <math>= \frac{x^6}{6}-4x^5+\frac{150x^4}{4}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> : <math>= \frac{x^6}{6}-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C</math> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : 1. <math>\int \frac{\ln \, x}{x}\,dx</math> : <math>u = \ln \, x\, du = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{\ln(x)}{x} dx \\ \int u du \\ \frac{1}{2} u^2 + C \\ \frac{1}{2} ln^2 x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x-5,\, du = dx,\, x = u+5</math> Dengan menggunakan rumus di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4\,dx \\ \int (u+5)u^4\,du \\ \int (u^5+5u^4)\,du \\ \frac{1}{6} u^6+u^5 + C \\ \frac{1}{6} (x-5)^6+(x-5)^5 + C \\ \frac{1}{6} (x^6-6x^5(5)+15x^4(25)-20x^3(125)+15x^2(625)-6x(3125)+15625)+(x^5-5x^4(5)+10x^3(25)-10x^2(125)+5x(625)+3125) + C \\ \frac{1}{6}x^6-5x^5+\frac{125x^4}{2}-\frac{1250x^3}{3}+\frac{3125x^2}{2}-3125x+\frac{15625}{6}+x^5-25x^4+250x^3-1250x^2+3125x+3125 + C \\ \frac{1}{6}x^6-4x^5+\frac{75x^4}{2}-\frac{500x^3}{3}+\frac{625x^2}{2} + C \\ \end{aligned} </math> </div></div> === integral parsial === ;Cara 1: Rumus Integral parsial diselesaikan dengan rumus berikut. : <math>\int u\,dv = uv - \int v\,du </math> Berikut contoh penyelesaian cara parsial dengan rumus. : 1. <math>\int x \sin \, x \, dx</math> : <math>u = x, du = 1 \, dx, dv = \sin \, x \, dx, v = -\cos \, x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \,dx \\ (x)(-\cos \, x) - \int (-\cos \, x)(1 dx) \\ -x \cos \, x + \int \cos \, x dx \\ -x \cos \, x + \sin \, x + C \\ \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4\,dx</math> : <math>u = x,\, du = 1\,dx,\, dv = (x-5)^4\,dx,\, v = \frac{1}{5}(x-5)^5</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - \int (\frac{1}{5}(x-5)^5)(1\,dx) \\ \frac{x}{5}(x-5)^5 - \int \frac{1}{5}(x-5)^5\,dx \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6\,dx \\ \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x\,dx</math> : <math>u = \ln \,x,\, du = \frac{1}{x}\,dx,\, dv = x\,dx,\, v = \frac{x^2}{2}</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x\, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{x^2}{2})(\frac{1}{x}\,dx) \\ \frac{x^2}{2} \ln \,x - \int \frac{x^2}{2}(\frac{1}{x})\,dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \,dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x\,dx</math> ini terjadi dua kali integral parsial : <math>u = \sin \,x,\, du = \cos \,x\,dx,\, dv = e^x\,dx,\, v = e^x</math> Dengan menggunakan rumus di atas, : <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x\, dx \\ (\sin \,x)(e^x) - \int (e^x)(\cos \,x\,dx) \\ e^x \sin \,x - \int e^x \cos \,x\,dx \\ u = \cos \,x,\, du = -\sin \,x\,dx,\, dv = e^x\,dx,\, v = e^x \\ (\cos \,x)(e^x) - \int (e^x)(-\sin \,x\,dx) \\ e^x \cos \,x + \int e^x \sin \,x\,dx \\ e^x \sin \,x - (e^x \cos \,x + \int e^x \sin \,x\,dx) \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x\,dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> ;Cara 2: Tabel Untuk <math display="inline">\int u\,dv</math>, berlaku ketentuan sebagai berikut. {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>u</math> || <math>dv</math> |- | - || <math>\frac{du}{dx}</math> || <math>v</math> |- | + || <math>\frac{d^2u}{dx^2}</math> || <math>\int v\,dx</math> |} Berikut contoh penyelesaian cara parsial dengan tabel. : 1. <math>\int x \sin \, x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>\sin \, x</math> |- | - || <math>1</math> || <math>-\cos \, x</math> |- | + || <math>0</math> || <math>-\sin \, x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \sin \, x \, dx \\ (x)(-\cos \, x) - (1)(-\sin \, x) + C \\ -x \cos \, x + \sin \, x + C \end{aligned} </math> </div></div> : 2. <math>\int x(x-5)^4 \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>x</math> || <math>(x-5)^4</math> |- | - || <math>1</math> || <math>\frac{1}{5}(x-5)^5</math> |- | + || <math>0</math> || <math>\frac{1}{30}(x-5)^6</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x(x-5)^4 \, dx \\ (x)(\frac{1}{5}(x-5)^5) - (1)(\frac{1}{30}(x-5)^6) + C \\ \frac{x}{5}(x-5)^5 - \frac{1}{30}(x-5)^6 + C \end{aligned} </math> </div></div> : 3. <math>\int x \ln \,x \, dx</math> {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\ln \,x</math> || <math>x</math> |- | - || <math>\frac{1}{x}</math> || <math>\frac{x^2}{2}</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int x \ln \,x \, dx \\ (\ln \,x)(\frac{x^2}{2}) - \int (\frac{1}{x})(\frac{x^2}{2}) \, dx \\ \frac{x^2}{2} \ln \,x - \int \frac{x}{2} \, dx \\ \frac{x^2}{2} \ln \,x - \frac{x^2}{4} + C \\ \end{aligned} </math> </div></div> : 4. <math>\int e^x \sin \,x \, dx</math> ini terjadi dua kali integral parsial {| class="wikitable" |- ! Tanda !! Turunan !! Integral |- | + || <math>\sin \,x</math> || <math>e^x</math> |- | - || <math>\cos \, x</math> || <math>e^x</math> |- | + || <math>-\sin \,x</math> || <math>e^x</math> |} Dengan tabel di atas, <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int e^x \sin \,x \, dx \\ (\sin \,x)(e^x) - (\cos \,x)(e^x) + \int (-\sin \,x)(e^x) \, dx \\ e^x \sin \,x - e^x \cos \,x - \int e^x \sin \,x) \, dx \\ \text{misalkan } I = \int e^x \sin \,x \, dx \\ I &= e^x \sin \,x - e^x \cos \,x - I \\ 2I &= e^x \sin \,x - e^x \cos \,x \\ I &= \frac{e^x \sin \,x - e^x \cos \,x}{2} \\ &= \frac{e^x \sin \,x}{2} - \frac{e^x \cos \,x}{2} + C \\ \end{aligned} </math> </div></div> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : 1. <math>\int \frac{1}{x^2 - 4}\,dx</math> Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{1}{x^2 - 4} \\ =&\; \frac{1}{(x + 2)(x - 2)} \\ =&\; \frac{A}{x + 2} + \frac{B}{x - 2} \\ =&\; \frac{A(x - 2) + B(x + 2)}{x^2 - 4} \\ =&\; \frac{(A + B)x - 2(A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 0</math> dan <math display="inline">A - B = -\frac{1}{2}</math> dapat diselesaikan, yaitu <math display="inline">A = -\frac{1}{4}</math> dan {{nowrap|<math display="inline">B = \frac{1}{4}</math>.}} : <math> \begin{aligned} &\; \int \frac{1}{x^2 - 4}\,dx \\ =&\; \int (-\frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)})\,dx \\ =&\; \frac{1}{4} \int (\frac{1}{x - 2} - \frac{1}{x + 2})\,dx \\ =&\; \frac{1}{4} (\ln|x - 2| - \ln|x + 2|) + C \\ =&\; \frac{1}{4} \ln|\frac{x - 2}{x + 2}| + C \end{aligned}</math> : 2. <math>\int \frac{x^2+5x+4}{(x-3)^3}\,dx</math> : cara 1 Pertama, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^3} \\ &\; \text{untuk mencari C yaitu } (3)^2+5(3)+4=28 \\ =&\; \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x - 3)^2 + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{A(x^2 - 6x + 9) + B(x - 3)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 - 6Ax + 9A + Bx - 3B}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ =&\; \frac{Ax^2 + (- 6A + B)x +(9 - 3B)}{(x - 3)^3} + \frac{28}{(x - 3)^3} \\ \end{aligned} </math> Kita tahu bahwa <math display="inline">A = 1</math> dan <math display="inline">-6A + B = 5</math> dapat diselesaikan, yaitu <math display="inline">A = 1</math> dan {{nowrap|<math display="inline">B = 11</math>.}} : <math> \begin{aligned} &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int (\frac{1}{x - 3} + \frac{11}{(x - 3)^2} + \frac{28}{(x - 3)^3})\,dx \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : cara 2 Pertama, ubah u=x-3 menjadi x=u+3 dan du=dx tersebut. : <math> \begin{aligned} =&\; \frac{x^2+5x+4}{(x-3)^3} \\ =&\; \frac{(u+3)^2+5(u+3)+4}{u^3} \\ =&\; \frac{u^2+11u+28}{u^3} \\ &\; \int \frac{x^2+5x+4}{(x-3)^3}\,dx \\ =&\; \int \frac{u^2+11u+28}{u^3}\,du \\ =&\; \int (\frac{u^2}{u^3} + \frac{11u}{u^3} + \frac{28}{u^3})\,du \\ =&\; \int (\frac{1}{u} + \frac{11}{u^2} + \frac{28}{u^3})\,du \\ =&\; \ln u + (-11 u^{-1}) + (-14 u^{-2}) + C \\ =&\; \ln|x - 3| + (-11 (x - 3)^{-1}) + (-14 (x - 3)^{-2}) + C \\ =&\; \ln|x - 3| - \frac{11}{x - 3} - \frac{14}{(x - 3)^2} + C \end{aligned}</math> : 3. <math>\int \frac{x^4+2x^3-13x^2-7x+21}{x^2-4x+3}\,dx</math> Pertama, penyelesaian dengan persamaan polinomial tersebut. : <math> \begin{aligned} =&\; \frac{x^4+2x^3-13x^2-7x+21}{x^2-4x+3} \\ =&\; x^2+6x+8 + \frac{7x-3}{x^2-4x+3} \\ =&\; \frac{1}{3}x^3+3x^2+8x+ \int \frac{7x-3}{x^2-4x+3}\,dx \end{aligned} </math> Kedua, pisahkan pecahan tersebut. : <math> \begin{aligned} =&\; \frac{7x - 3}{x^2 - 4x + 3} \\ =&\; \frac{7x - 3}{(x - 1)(x - 3)} \\ =&\; \frac{A}{x - 1} + \frac{B}{x - 3} \\ =&\; \frac{A(x - 3) + B(x - 1)}{x^2 - 4x + 3} \\ =&\; \frac{(A + B)x + (-3A - B)}{x^2 - 4} \end{aligned} </math> Kita tahu bahwa <math display="inline">A + B = 7</math> dan <math display="inline">-3A - B = -3</math> dapat diselesaikan, yaitu <math display="inline">A = -2</math> dan {{nowrap|<math display="inline">B = 9</math>.}} : <math> \begin{aligned} &\; \frac{1}{3}x^3+3x^2+8x+ \int \frac{7x-3}{x^2-4x+3}\,dx \\ =&\; \frac{1}{3}x^3+3x^2+8x+ \int (-\frac{2}{(x - 1)} + \frac{9}{x - 3})\,dx \\ =&\; \frac{1}{3}x^3+3x^2+8x- 2 \ln|x - 1| + 9 \ln|x - 3| \,dx + C \\ \end{aligned}</math> === integral substitusi trigonometri === {| class="wikitable" |- ! Bentuk !! Substitusi Trigonometri |- | <math>\sqrt{a^2 - b^2 x^2}</math> || <math>x = \frac{a}{b} sin \, \alpha </math> |- | <math>\sqrt{a^2 + b^2 x^2}</math> || <math>x = \frac{a}{b} tan \, \alpha</math> |- | <math>\sqrt{b^2 x^2 - a^2}</math> || <math>x = \frac{a}{b} sec \, \alpha</math> |} Berikut contoh penyelesaian cara substitusi trigonometri. : <math>\int \frac{dx}{x^2 \sqrt{x^2 + 4}}</math> : <math>x = 2 \tan \, A, dx = 2 \sec^2 \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \int \frac{dx}{x^2\sqrt{x^2+4}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(2 \tan \, A)^2 \sqrt{4 + (2 \tan \, A)^2}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 + 4 \tan^2 \, A}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4(1 + \tan^2 \, A)}} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{4 \tan^2 \, A \sqrt{4 \sec^2 \, A }} \\ =&\; \int \frac{2 \sec^2 \, A \,dA}{(4 \tan^2 \, A)(2 \sec \, A)} \\ =&\; \int \frac{\sec \, A \,dA}{4 \tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\sec \, A \,dA}{\tan^2 \, A} \\ =&\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A} \,dA \end{aligned} </math> Substitusi berikut dapat dibuat. : <math>\int \frac{\cos \, A}{\sin^2 \, A} \,dA</math> : <math>t = \sin \, A, dt = \cos \, A \,dA</math> Dengan substitusi di atas, : <math> \begin{aligned} &\; \frac{1}{4} \int \frac{\cos \, A}{\sin^2 \, A}\,dA \\ =&\; \frac{1}{4} \int \frac{dt}{t^2} \\ =&\; \frac{1}{4} \left(-\frac{1}{t}\right) + C \\ =&\; -\frac{1}{4 t} + C \\ =&\; -\frac{1}{4 \sin \, A} + C \end{aligned} </math> Ingat bahwa <math display="inline">\sin \, A = \frac{x}{\sqrt{x^2 + 4}}</math> berlaku. : <math> \begin{aligned} =&\; -\frac{1}{4 \sin \, A} + C \\ =&\; -\frac{\sqrt{x^2 + 4}}{4x} + C \end{aligned} </math> === integral mutlak === : <math>\int |f(x)| \,dA</math> buatlah <math>f(x) \ge 0</math> jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x) : <math>\int_a^c |f(x)| \, dx = \int_a^b f(x) \, dx + \int_b^c -f(x) \, dx = F(b) - F(a) - F(c) + F(b)</math> jika <math>|f(x)| = \begin{cases} f(x), & \mbox{jika } f(x) \ge b, \\ -f(x), & \mbox{jika } f(x) < b. \end{cases}</math>. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif. === integral fungsi ganjil dan integral fungsi genap === : <math>\int_{-a}^a f(x) \,dx</math> dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral: : integral fungsi ganjil : <math>\int_{-a}^a f(x) \,dx</math> = 0 : integral fungsi genap : <math>\int_{-a}^a f(x) \,dx</math> = <math>2 \int_{0}^a f(x) \,dx</math> === integral notasi === : <math>\int_a^b f(x) \,dx = \int_{a-p}^{b-p} f(x) \,dx = \int_{a+p}^{b+p} f(x) \,dx</math> : <math>\int_a^c f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx</math> : <math>\int_a^b f(x \pm k) \,dx = \int_{a \pm k}^{b \pm k} f(x) \,dx</math> === integral terbalik === : <math>\int_a^b f(x) \,dx = -\int_b^a f(x) \,dx</math> == Jenis integral lainnya == === panjang busur === ; Sumbu ''x'' : <math>S = \int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' : <math>S = \int_{y_1}^{y_2} \sqrt{1 + (f'(y))^2}\,dy</math> === luas daerah === ; Satu kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} f(x)\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} f(y)\,dy</math> ; Dua kurva ; Sumbu ''x'' : <math>L = \int_{x_1}^{x_2} (f(x_2) - f(x_1))\,dx</math> ; Sumbu ''y'' : <math>L = \int_{y_1}^{y_2} (f(y_2) - f(y_1))\,dy</math> : atau juga <math>L = \frac {D \sqrt{D}}{6 a^2}</math> === luas permukaan benda putar === ; Sumbu ''x'' sebagai poros : <math>L_p = 2 \pi \int_{x_1}^{x_2} f(x)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(x))^2}\,dx</math> ; Sumbu ''y'' sebagai poros : <math>L_p = 2 \pi \int_{y_1}^{y_2} f(y)\,ds</math> dengan : <math>ds = \sqrt {1 + (f'(y))^2}\,dy</math> === volume benda putar === ; Satu kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} (f(x))^2\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} (f(y))^2\,dy</math> ; Dua kurva ; Sumbu ''x'' sebagai poros : <math>V = \pi \int_{x_1}^{x_2} ((f(x_2))^2 - (f(x_1))^2)\,dx</math> ; Sumbu ''y'' sebagai poros : <math>V = \pi \int_{y_1}^{y_2} ((f(y_2))^2 - (f(y_1))^2)\,dy</math> : atau juga <math>V = |\frac {D^2 \sqrt{D}}{30 a^3}|</math> == Integral lipat == Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga. ;contoh # Tentukan hasil dari: * <math>\int (5x-1)(5x^2-2x+7)^4 dx</math> * <math>\int x^2 cos 3x dx</math> * <math>\int sin^2 x cos x dx</math> * <math>\int cos^2 x sin x dx</math> * <math>\int sec x dx</math> * <math>\int csc x dx</math> * <math>\int \frac{1}{2x^2-5x+3} dx</math> * <math>\int \frac{1}{\sqrt{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> * <math>\int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx</math> * <math>\int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx</math> * <math>\int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx</math> * <math>\int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx</math> * <math>\int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+sin^2 x+sin^4 x+sin^6 x+ \dots) dx \\ a=1, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-sin^2 x} \\ &= \frac{1}{cos^2 x} \\ &= sec^2 x \\ \int sec^2 x dx &= tan x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (1+cos^2 x+cos^4 x+cos^6 x+ \dots) dx \\ a=1, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{1}{1-cos^2 x} \\ &= \frac{1}{sin^2 x} \\ &= csc^2 x \\ \int csc^2 x dx &= -cot x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (sin x+sin^3 x+sin^5 x+sin^7 x+ \dots) dx \\ a=sin x, r=sin^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{sin x}{1-sin^2 x} \\ &= \frac{sin x}{cos^2 x} \\ &= sec x \cdot tan x \\ \int sec x \cdot tan x dx &= sec x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int (cos x+cos^3 x+cos^5 x+cos^7 x+ \dots) dx \\ a=cos x, r=cos^2 x \\ S_{\infty} &= \frac{a}{1-r} \\ &= \frac{cos x}{1-cos^2 x} \\ &= \frac{cos x}{sin^2 x} \\ &= csc x \cdot cot x \\ \int csc x \cdot cot x dx &= -csc x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} dx \\ \text{misalkan } y = \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y &= \sqrt{\frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}}} \\ y^2 &= \frac{x}{\sqrt{\frac{x}{\sqrt{\frac{x}{\dots}}}}} \\ &= \frac{x}{y} \\ y^3 &= x \\ 3y^2 dy &= dx \\ \int y \cdot 3y^2 dy \\ \int 3y^3 dy \\ \frac{3y^4}{4}+C \\ \end{aligned} </math> </div></div> # Tentukan hasil dari: * <math>\int_{1}^{5} 2x-3 dx</math> * <math>\int_{0}^{4} |x-2| dx</math> * <math>\int_{2}^{4} |3-x| dx</math> * <math>\int_{1}^{5} x^2-6x+8 dx</math> * <math>\int_{-3}^{5} |x^2-2x-8| dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} 2x-3 dx &= \left. x^2 - 3x \right|_{1}^{5} \\ &= 10 - (-2) \\ &= 12 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{0}^{4} |x-2| dx \\ \text{ tentukan nilai harga nol } \\ x-2 &= 0 \\ x &= 2 \\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti } \\ |x-2| = \begin{cases} x-2, & \mbox{jika } x \ge 2, \\ -(x-2), & \mbox{jika } x < 2. \end{cases} \\ \int_{0}^{4} |x-2| dx &= \int_{2}^{4} x-2 dx + \int_{0}^{2} -(x-2) dx \\ &= \left. (\frac{x^2}{2}-2x) \right|_{2}^{4} + \left. (\frac{-x^2}{2}+2x) \right|_{0}^{2} \\ &= 0 - (-2) + 2 - 0 \\ &= 4 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{2}^{4} |3-x| dx \\ \text{ tentukan nilai syarat-syarat } \\ 3-x &\ge 0 \\ \text{ harga nol } \\ x &= 3 \\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti } \\ |3-x| = \begin{cases} 3-x, & \mbox{jika } x \le 3, \\ -(3-x), & \mbox{jika } x > 3. \end{cases} \\ \int_{2}^{4} |3-x| dx &= \int_{2}^{3} 3-x dx + \int_{3}^{4} -(3-x) dx \\ &= \left. (3x-\frac{x^2}{2}) \right|_{2}^{3} + \left. (-3x+\frac{x^2}{2}) \right|_{3}^{4} \\ &= 4.5 - 4 - 4 - (-4.5) \\ &= 1 \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{1}^{5} x^2-6x+8 dx &= \left. \frac{x^3}{3}-3x^2+8x \right|_{1}^{5} \\ &= \frac{20}{3} - \frac{16}{3} \\ &= \frac{4}{3} \\ &= 1\frac{1}{3} \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int_{-3}^{5} |x^2-2x-8| dx \\ \text{ tentukan nilai syarat-syarat } \\ x^2-2x-8 &\ge 0 \\ \text{ harga nol } \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x = 4 \\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti } \\ |x^2-2x-8| = \begin{cases} x^2-2x-8, & \mbox{jika } x \le -2, \\ -(x^2-2x-8), & \mbox{jika } -2 < x < 4, \\ x^2-2x-8, & \mbox{jika } x \ge 4 \end{cases} \\ \int_{-3}^{5} |x^2-2x-8| dx &= \int_{-3}^{-2} x^2-2x-8 dx + \int_{-2}^{4} -(x^2-2x-8) dx + \int_{4}^{5} x^2-2x-8 dx \\ &= \left. (\frac{x^3}{3}-x^2-8x) \right|_{-3}^{-2} + \left. (\frac{-x^3}{3}+x^2+8x) \right|_{-2}^{4} + \left. (\frac{x^3}{3}-x^2-8x) \right|_{4}^{5} \\ &= \frac{28}{3} - 6 + \frac{80}{3} - (-\frac{28}{3}) - \frac{70}{3} - (-\frac{80}{3}) \\ &= \frac{128}{3} \\ &= 42\frac{2}{3} \\ \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x\,dx \\ L &= \frac{1}{2} x^2 \\ L &= \frac{1}{2} xy \quad (y = x) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = x^2</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int x^2\,dx \\ L &= \frac{1}{3} x^3 \\ L &= \frac{1}{3} xy \quad (y = x^2) \end{aligned} </math> </div></div> # Tentukan luas (tak tentu) dengan persamaan garis <math>y = \sqrt{x}</math> dan batas-batas sumbu ''y'' dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} L &= \int \sqrt{x}\,dx \\ L &= \frac{2}{3} x^{\frac{3}{2}} \\ L &= \frac{2}{3} xy \quad (y = \sqrt{x}) \end{aligned} </math> </div></div> # Buktikan luas persegi <math>L = s^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = s \text{ dan titik (''s'', ''s''), } \\ L &= \int_{0}^{s} s\,dx \\ L &= sx |_{0}^{s} \\ L &= ss - 0 \\ L &= s^2 \end{aligned} </math> </div></div> # Buktikan luas persegi panjang <math>L = pl</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = l \text{dan titik (''p'', ''l''), } \\ L &= \int_{0}^{p} l\,dx \\ L &= lx |_{0}^{p} \\ L &= pl - 0 \\ L &= pl \end{aligned} </math> </div></div> * Buktikan luas segitiga <math>L = \frac{at}{2}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{-tx}{a} + t \text { dan titik (''a'', ''t''), } \\ L &= \int_{0}^{a} \left (\frac{-tx}{a} + t \right) \,dx \\ L &= \left. \frac{-tx^2}{2a} + tx \right|_{0}^{a} \\ L &= \frac{-ta^2}{2a} + ta - 0 + 0 \\ L &= \frac{-ta}{2} + ta \\ L &= \frac{at}{2} \end{aligned} </math> </div></div> # Buktikan volume tabung <math>V = \pi r^2t</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ Dengan posisi } y = r \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} r^2\,dx \\ V &= \pi \left. r^2x \right|_{0}^{t} \\ V &= \pi r^2t - 0 \\ V &= \pi r^2t \end{aligned} </math> </div></div> # Buktikan volume kerucut <math>V = \frac{\pi r^2t}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \frac{rx}{t} \text{ dan titik (''t'', ''r''), } \\ V &= \pi \int_{0}^{t} \left (\frac{rx}{t} \right)^2 \,dx \\ V &= \pi \left. \frac{r^2 x^3}{3t^2} \right|_{0}^{t} \\ V &= \pi \frac{r^2 t^3}{3t^2} - 0 \\ V &= \frac{\pi r^2t}{3} \end{aligned} </math> </div></div> # Buktikan volume bola <math>V = \frac{4 \pi r^3}{3}</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), } \\ V &= \pi \int_{-r}^{r} \left(\sqrt {r^2 - x^2}\right)^2\,dx \\ V &= \pi \int_{-r}^{r} r^2 - x^2 dx \\ V &= \pi \left. r^2x - \frac{x^3}{3} \right|_{-r}^{r} \\ V &= \pi \left (r^3 - \frac{r^3}{3} - \left (-r^3 + \frac{r^3}{3} \right) \right) \\ V &= \frac{4 \pi r^3}{3} \end{aligned} </math> </div></div> # Buktikan luas permukaan bola <math>L = 4 \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{selanjutnya } \\ ds &= \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ ds &= \sqrt{1 + \frac{x^2}{r^2 - x^2}}\,dx \\ ds &= \sqrt{\frac{r^2}{r^2 - x^2}}\,dx \\ ds &= \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ \text{sehingga } \\ L &= 2 \pi \int_{-r}^{r} \sqrt{r^2 - x^2} \cdot \frac{r}{\sqrt{r^2 - x^2}}\,dx \\ L &= 2 \pi \int_{-r}^{r} r\,dx \\ L &= 2 \pi rx|_{-r}^{r} \\ L &= 2 \pi (r r - r (-r)) \\ L &= 2 \pi (r^2 + r^2) \\ L &= 4 \pi r^2 \end{aligned} </math> </div></div> # Buktikan keliling lingkaran <math>K = 2 \pi r</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0) } \\ \text{kita tahu bahwa turunannya adalah } \\ y' &= \frac{-x}{\sqrt{r^2 - x^2}} \\ \text{sehingga } \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{-x}{\sqrt{r^2 - x^2}})^2}\,dx \\ K &= 4 \int_{0}^{r} \sqrt{1 + (\frac{x^2}{r^2 - x^2})}\,dx \\ K &= 4 \int_{0}^{r} \sqrt\frac{r^2}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \sqrt\frac{1}{r^2 - x^2}\,dx \\ K &= 4r \int_{0}^{r} \frac{1}{\sqrt{r^2 - x^2}}\,dx \\ K &= 4r \left. \arcsin \left (\frac{x}{r} \right) \right|_{0}^{r} \\ K &= 4r \left (\arcsin \left (\frac{r}{r}\right) - \arcsin \left (\frac{0}{r} \right) \right) \\ K &= 4r \left (\arcsin\left (1 \right) - \arcsin \left(0 \right) \right)) \\ K &= 4r \left (\frac{\pi}{2} \right) \\ K &= 2 \pi r \end{aligned} </math> </div></div> # Buktikan luas lingkaran <math>L = \pi r^2</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \sqrt{r^2 - x^2} \text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. } \\ \sin(\theta) &= \frac{x}{r} \\ x &= r \sin(\theta) \\ dx &= r \cos(\theta)\,d\theta \\ \text{dengan turunan di atas } \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - x^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - (r \sin(\theta))^2}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 - r^2 \sin^2(\theta)}\,dx \\ L &= 4 \int_{0}^{r} \sqrt{r^2 (1 - \sin^2(\theta))}\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta)\,dx \\ L &= 4 \int_{0}^{r} r \cos(\theta) (r \cos(\theta)\,d\theta) \\ L &= 4 \int_{0}^{r} r^2 \cos^2(\theta)\,d\theta \\ L &= 4r^2 \int_{0}^{r} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta \\ L &= 2r^2 \int_{0}^{r} (1 + \cos(2\theta))\,d\theta \\ L &= 2r^2 \left (\theta + \frac{1}{2} \sin(2\theta) \right)|_{0}^{r} \\ L &= 2r^2 (\theta + \sin(\theta) \cos(\theta))|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left (\frac{x}{r} \right) + \left (\frac{x}{r} \right) \left (\frac{r^2 - x^2}{r} \right) \right)|_{0}^{r} \\ L &= 2r^2 \left (\arcsin \left(\frac{r}{r} \right) + \left (\frac{r}{r} \right) \left (\frac {r^2 - r^2}{r} \right) \right) - \left(\arcsin \left (\frac{0}{r} \right) + \left (\frac{0}{r}\right) \left (\frac{r^2 - 0^2}{r} \right) \right) \\ L &= 2r^2 (\arcsin(1) + 0 - (\arcsin(0) + 0)) \\ L &= 2r^2 \left (\frac{\pi}{2} \right) \\ L &= \pi r^2 \end{aligned} </math> </div></div> # Buktikan luas elips <math>L = \pi ab</math> dengan cara integral! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{Dengan posisi } y = \frac{b \sqrt{a^2 - x^2}}{a} \text{ serta (-''a'', 0) dan (''a'', 0), } \\ L &= 4 \int_{0}^{r} \frac{b \sqrt{a^2 - x^2}}{a}\,dx \\ L &= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx \\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, } \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{\frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2}\,dx}{4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx} \\ \frac{L_\text{elips}}{L_\text{ling}} &= \frac{b}{a} \\ L_\text{elips} &= \frac{b}{a} L_\text{ling} \\ L_\text{elips} &= \frac{b}{a} \pi a^2 \\ L_\text{elips} &= \pi ab \end{aligned} </math> </div></div> # Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ x^-2x &= 4x+7 \\ x^2-6x-7 &= 0 \\ (x+1)(x-7) &= 0 \\ x=-1 &\text{ atau } x=7 \\ y &= 4(-1)+7 = 3 \\ y &= 4(7)+7 = 35 \\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut } \\ L &= \int_{-1}^{7} 4x+7-(x^2-2x)\,dx \\ &= \int_{-1}^{7} -x^2+6x+7\,dx \\ &= -\frac{x^3}{3}+3x^2+7x|_{-1}^{7} \\ &= -\frac{7^3}{3}+3(7)^2+7(7)-(-\frac{(-1)^3}{3}+3(-1)^2+7(-1)) \\ &= \frac{245}{3}-(-\frac{13}{3}) \\ &= \frac{258}{3} \\ &= 86 \\ \end{aligned} </math> </div></div> # Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{ cari titik potong dari kedua persamaan tersebut } \\ y_1 &= y_2 \\ 6-1,5x &= x-4 \\ 2,5x &= 10 \\ x &= 4 \\ y &= 4-4 = 0 \\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut } \\ \text{untuk x=0 } \\ y &= 6-1,5(0) = 6 \\ y &= 0-4 = -4 \\ L &= \pi \int_{0}^{4} ((6-1,5x)^2-(x-4)^2)\,dx \\ &= \pi \int_{0}^{4} (36-18x+2,25x^2-(x^2-8x+16))\,dx \\ &= \pi \int_{0}^{4} (1,25x^2-10x+20)\,dx \\ &= \pi (\frac{1,25x^3}{3}-5x^2+20x)|_{0}^{4} \\ &= \pi (\frac{1,25(4)^3}{3}-5(4)^2+20(4)-(\frac{1,25(0)^3}{3}-5(0)^2+20(0))) \\ &= \pi (\frac{30}{3}-0) \\ &= 10\pi \\ \end{aligned} </math> </div></div> [[Kategori:Soal-Soal Matematika]] 0fxmv3iqfbwi413gxpjr1gx36n5mpfn 0 23568 115011 114990 2026-04-26T12:26:31Z ~2026-25546-23 43043 115011 wikitext text/x-wiki contoh soal # Berapa hasil dari <math>\sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Misalkan 2020 = p} \\ \sqrt{2015 \cdot 2017 \cdot 2023 \cdot 2025 + 64} &= \sqrt{(2020-5) \cdot (2020-3) \cdot (2020+3) \cdot (2020+5) + 64} \\ &= \sqrt{(p-5) \cdot (p-3) \cdot (p+3) \cdot (p+5) + 64} \\ &= \sqrt{(p-5) \cdot (p+5) \cdot (p-3) \cdot (p+3) + 64} \\ &= \sqrt{(p^2-25) \cdot (p^2-9) + 64} \\ &= \sqrt{p^4-34p^2+ 225 + 64} \\ &= \sqrt{p^4-34p^2+ 289} \\ &= \sqrt{(p^2-17)^2} \\ &= p^2-17 \\ &= 2020^2-17 \\ &= (2000+20)^2-17 \\ &= 4.000.000+80.000+400-17 \\ &= 4.080.383 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} = \frac{9}{10}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}}}{\sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}}} &= \frac{9}{10} \\ \text{misalkan untuk } \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} = p \\ \sqrt{x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}}} &= p \\ x^2-x-\sqrt{x^2-x-\sqrt{x^2-x-\sqrt{\dots}}} &= p^2 \\ x^2-x-p &= p^2 \\ x^2-2x+1+x-1 &= p^2+p \\ (x-1)^2+(x-1) &= p^2+p \\ x-1 &= p \\ \text{misalkan untuk } \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ \sqrt[3]{x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}}} &= q \\ x^2\sqrt[3]{x^2\sqrt[3]{x^2 \dots}} &= q^3 \\ x^2 q &= q^3 \\ x^2 &= q^2 \\ x &= q \\ \frac{x-1}{x} &= \frac{9}{10} \\ x &= 10 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>(\frac{x}{x+10})^{x+10}=\frac{1}{1024}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{x+10}{x})^{-(x+10)} &= (1024)^{-1} \\ (\frac{x+10}{x})^{x+10} &= 1024 \\ (\frac{x+10}{x})^{x+10} &= 2^{10} \\ (\frac{x+10}{x})^{\frac{x+10}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= 2 \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (\frac{1}{2})^{-1} \\ (1+\frac{10}{x})^{1+\frac{x}{10}} &= (1+(-\frac{1}{2}))^{(1+(-\frac{2}{1}))} \\ \frac{10}{x} &= -\frac{1}{2} \\ x &= -20 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{2}+\sqrt{x+\frac{1}{4}} &= (\sqrt{x+\frac{1}{4}})^2+2 \cdot \sqrt{x+\frac{1}{4}} \cdot \frac{1}{2}+(\frac{1}{2})^2 \\ &= (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 \\ x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}} &= 4 \\ x+\sqrt{(\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2} &= 4 \\ x+\sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 4 \\ (\sqrt{x+\frac{1}{4}}+\frac{1}{2})^2 &= 4 \\ \sqrt{x+\frac{1}{4}}+\frac{1}{2} &= 2 \\ \sqrt{x+\frac{1}{4}} &= \frac{3}{2} \\ x+\frac{1}{4} &= \frac{9}{4} \\ x &= 2 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{x^3}{\sqrt{8-x^2}}+x^2-8=0</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^3}{\sqrt{8-x^2}}+x^2-8 &= 0 \\ \frac{x^3}{\sqrt{8-x^2}} &= 8-x^2 \\ x^3 &= (8-x^2)^{\frac{3}{2}} \\ x &= (8-x^2)^{\frac{1}{2}} \\ x^2 &= 8-x^2 \\ 2x^2-8 &= 0 \\ x^2-4 &= 0 \\ (x-2)(x+2) &= 0 \\ \text{membuktikan } \\ x=2 \text{ maka hasilnya 0 } \\ x=-2 \text{ maka hasilnya -8 } \\ \text{jadi } x=2 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\sqrt{3x+5+\sqrt{4x+5}} = x</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{3x+5+\sqrt{4x+5}} &= x \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \text{misalkan } \sqrt{4x+5}=y \text{ dan } 4x+5=y^2 \\ \sqrt{4x+5+\sqrt{4x+5}-x} &= x \\ \sqrt{y^2+y-x} &= x \\ y^2+y &= x^2+x \\ y=x \\ 4x+5 &= y^2 \\ 4x+5 &= x^2 \\ x^2-4x-5 &= 0 \\ (x-5)(x+1) &= 0 \\ x=5 &\text{ atau } x=-1 \text{ (TM) } \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\sqrt{1+\sqrt{1+x}} = \sqrt[3]{x}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{1+\sqrt{1+x}} &= \sqrt[3]{x} \\ \sqrt[3]{x} &= n \\ x &= n^3 \\ \sqrt{1+\sqrt{1+n^3}} &= n \\ 1+\sqrt{1+n^3} &= n^2 \\ \sqrt{1+n^3} &= n^2-1 \\ 1+n^3 &= n^4-2n^2+1 \\ n^4-n^3-2n^2 &= 0 \\ n^2(n^2-n-2) &= 0 \\ n^2(n-2)(n+1) &= 0 \\ n=0, n=2 \text{ atau } n=-1 \\ n &= 0 \\ x &= 0^3 \\ &= 0 \\ n &= 2 \\ x &= 2^3 \\ &= 8 \\ n &= -1 \\ x &= (-1)^3 \\ &= -1 \\ \text{yang paling mungkin untuk nilai x adalah } 8 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}}{\sqrt{x}}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}} &= \frac{\sqrt{1+x}}{\sqrt{x}} \\ \sqrt{x}(\sqrt{1+x}+\sqrt{x}) &= (\sqrt{1+x}-\sqrt{x})\sqrt{1+x} \\ \sqrt{x(1+x)}+x &= 1+x-\sqrt{x(1+x)} \\ 2\sqrt{x(1+x)} &= 1 \\ \sqrt{x(1+x)} &= \frac{1}{2} \\ x(1+x) &= \frac{1}{4} \\ x^2+x &= \frac{1}{4} \\ 4x^2+4x &= 1 \\ 4x^2+4x-1 &= 0 \\ x &= \frac{-4 \pm \sqrt{4^2-4(4)(-1)}}{2(4)} \\ &= \frac{-4 \pm \sqrt{32}}{8} \\ &= \frac{-4 \pm 4\sqrt{2}}{8} \\ &= \frac{-1 \pm \sqrt{2}}{2} \\ \text{karena akar x harus minimal nol jadi } x = \frac{-1+\sqrt{2}}{2} \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\frac{11}{19}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x-\sqrt{x+1}}{x+\sqrt{x+1}} &= \frac{11}{19} \\ \text{misalkan } \sqrt{x+1}=y \text{ dan } x=y^2-1 \\ \frac{y^2-1-y}{y^2-1+y} &= \frac{11}{19} \\ 19(y^2-y-1) &= 11(y^2+y-1) \\ 19y^2-19y-19 &= 11y^2+11y-11 \\ 8y^2-30y-8 &= 0 \\ 4y^2-15y-4 &= 0 \\ (4y+1)(y-4) &= 0 \\ y=-\frac{1}{4} \text{ (TM) atau } & y=4 \\ x &= 4^2-1 \\ &= 15 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=98</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}} &= 98 \\ \text{misalkan } \sqrt{x^2-1}=y \\ \frac{x+y}{x-y}+\frac{x-y}{x+y} &= 98 \\ \frac{(x+y)^2+(x-y)^2}{(x-y)(x+y)} &= 98 \\ \frac{x^2+2xy+y^2+x^2-2xy+y^2}{x^2-y^2} &= 98 \\ \frac{2(x^2+y^2)}{x^2-y^2} &= 98 \\ \frac{x^2+y^2}{x^2-y^2} &= 49 \\ x^2+y^2 &= 49(x^2-y^2) \\ x^2+y^2 &= 49x^2-49y^2 \\ 48x^2 &= 50y^2 \\ 24x^2 &= 25y^2 \\ 24x^2 &= 25(\sqrt{x^2-1})^2 \\ 24x^2 &= 25(x^2-1) \\ 24x^2 &= 25x^2-25 \\ x^2 &= 25 \\ x &= \pm 5 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \sqrt{x}=y \text{ dan } x=y^2 \\ \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} &= \frac{5}{4}\sqrt{\frac{x}{x+\sqrt{x}}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\sqrt{\frac{y^2}{y^2+y}} \\ \sqrt{y^2+y}-\sqrt{y^2-y} &= \frac{5}{4}\frac{y}{\sqrt{y^2+y}} \\ y^2+y-\sqrt{(y^2+y)(y^2-y)} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^4-y^2} &= \frac{5}{4}y \\ y^2+y-\sqrt{y^2(y^2-1)} &= \frac{5}{4}y \\ y(y+1)-y\sqrt{y^2-1} &= \frac{5}{4}y \\ y+1-\sqrt{y^2-1} &= \frac{5}{4} \\ -\sqrt{y^2-1} &= \frac{1}{4}-y \\ y^2-1 &= (\frac{1}{4}-y)^2 \\ y^2-1 &= \frac{1}{16}-\frac{1}{2}y+y^2 \\ -1 &= \frac{1}{16}-\frac{1}{2}y \\ \frac{1}{2}y &= \frac{1}{16}+1 \\ \frac{1}{2}y &= \frac{17}{16} \\ y &= \frac{17}{8} \\ x &= (\frac{17}{8})^2 \\ &= \frac{289}{64} \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\sqrt[4]{62+x}+\sqrt[4]{275-x}=7</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ misalkan } \sqrt[4]{62+x}=a, 62+x=a^4, \sqrt[4]{275-x}=b \text{ dan } 275-x=b^4 \\ a+b &= 7 \\ (a+b)^2 &= 49 \\ a^2+b^2+2ab &= 49 \\ a^2+b^2 &= 49-2ab \\ a^4+b^4 &= 62+x+275-x \\ (a^2+b^2)^2-2(ab)^2 &= 337 \\ (49-2ab)^2-2(ab)^2 &= 337 \\ 2401-196ab+4(ab)^2-2(ab)^2 &= 337 \\ 2(ab)^2-196ab+2064 &= 0 \\ (ab)^2-98ab+1032 &= 0 \\ (ab-12)(ab-86) &= 0 \\ ab = 12 \text{ atau } & ab = 86 \text{ (TM) karena hasil kali maksimum yaitu 12 } \\ ab =12 \text{ dan } a+b=7 \\ a+b &= 7 \\ b &= 7-a \\ ab &= 12 \\ a(7-a) &= 12 \\ -a^2+7a &= 12 \\ a^2-7a+12 &= 0 \\ (a-3)(a-4) &= 0 \\ a=3 \text{ atau } & a=4 \\ a=3, b=4 \\ 62+x &= a^4 \\ 62+x &= (3)^4 \\ 62+x &= 81 \\ x &= 19 \\ a=4, b=3 \\ 62+x &= a^4 \\ 62+x &= (4)^4 \\ 62+x &= 256 \\ x &= 194 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2}=7</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{(8+x)^2}-\sqrt[3]{(8+x)(27-x)}+\sqrt[3]{(27-x)^2} &= 7 \\ (\sqrt[3]{8+x})^2-\sqrt[3]{8+x} \sqrt[3]{27-x}+(\sqrt[3]{27-x})^2 &= 7 \\ \text{misalkan } \sqrt[3]{8+x}=a, 8+x=a^3, \sqrt[3]{27-x}=b \text{ dan } 27-x=b^3 \\ a^2-ab+b^2 &= 7 \\ a^3+b^3 &= 8+x+27-x \\ &= 35 \\ a^3+b^3 &= (a+b)(a^2-ab+b^2) \\ 35 &= (a+b)(7) \\ a+b &= 5 \\ b &= 5-a \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ 5^3 &= 35+3ab(5) \\ 125 &= 35+15ab \\ 80 &= 15ab \\ ab &= 6 \\ a(5-a) &= 6 \\ 5a-a^2 &= 6 \\ a^2-5a+6 &= 6 \\ (a-2)(a-3) &= 6 \\ a=2 &\text{ atau } a=3 \\ a=2, b=3 \text{ dan } a=3,b=2 \\ 8+x &= a^3 \\ &= 2^3 \\ &= 8 \\ x &= 0 \\ 8+x &= a^3 \\ &= 3^3 \\ &= 27 \\ x &= 19 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>3^x+5^x-9^x+15^x-25^x=1</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3^x+5^x-9^x+15^x-25^x &= 1 \\ 3^x+5^x-(3^2)^x+(3 \cdot 5)^x-(5^2)^x &= 1 \\ 3^x+5^x-(3^x)^2+(3^x \cdot 5^x)-(5^x)^2 &= 1 \\ \text{misalkan } 3^x=a \text{ dan } 5^x=b \\ a+b-a^2+ab-b^2 &= 1 \\ a^2-ab+b^2-a-b+1 &= 0 \\ 2a^2-2ab+2b^2-2a-2b+2 &= 0 \\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1 &= 0 \\ (a-b)^2+(a-1)^2+(b-1)^2 &= 0 \\ a-b=0; a-1=0; b-1 &= 0 \\ a=b &= 1 \\ 3^x &= 1 \\ x &= 0 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>^6log x^2+^{6x}log \frac{6}{x}=1</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ \text{misalkan } 6x=a \text{ maka } x=\frac{a}{6} \\ ^6log x^2+^{6x}log \frac{6}{x} &= 1 \\ ^6log (\frac{a}{6})^2+^{6 \frac{a}{6}}log \frac{6}{\frac{a}{6}} &= 1 \\ ^6log \frac{a^2}{6^2}+^alog \frac{6^2}{a} &= 1 \\ ^6log a^2-^6log 6^2+^alog 6^2-^alog a &= 1 \\ 2 ^6log a-2 ^6log 6+2 ^alog 6-^alog a &= 1 \\ 2 ^6log a-2+2 \frac{1}{^6log a}-1 &= 1 \\ 2 ^6log a+2 \frac{1}{^6log a}-4 &= 0 \\ 2 ^6log^2 a-4 ^6log a+2 &= 0 \\ ^6log^2 a-2 ^6log a+1 &= 0 \\ (^6log a-1)^2 &= 0 \\ ^6log a &= 1 \\ a &= 6 \\ x &= \frac{a}{6} \\ &= \frac{6}{6} \\ &= 1 \\ \end{align} </math> </div></div> # Berapa nilai x dari (x+500)³+x=20? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+500)^3+x &= 20 \\ \text{misalkan } a=x+500 \text{ maka } x=a-500 \\ a^3+a-500 &= 20 \\ a^3+a &= 520 \\ a(a^2+1) &= 8 \cdot 65 \\ a(a^2+1) &= 8(64+1) \\ a(a^2+1) &= 8(8^2+1) \\ a &= 8 \\ x &= 8-500 \\ &= -492 \\ \end{align} </math> </div></div> # Berapa nilai x dari <math>\sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2}=0</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}}-\frac{1}{2} &= 0 \\ \sqrt[n]{\frac{x^n+4^n}{x^n+16^n}} &= \frac{1}{2} \\ \frac{x^n+4^n}{x^n+16^n} &= (\frac{1}{2})^n \\ \frac{x^n+4^n}{x^n+16^n} &= \frac{1}{2^n} \\ 2^n(x^n+4^n) &= x^n+16^n \\ 2^n(x^n+2^{2n}) &= x^n+2^{4n} \\ 2^n \cdot x^n+2^{3n} &= x^n+2^{4n} \\ 2^n \cdot x^n-x^n &= 2^{4n}-2^{3n} \\ x^n(2^n-1) &= 2^{3n}(2^n-1) \\ x^n &= 2^{3n} \\ x^n &= (2^3)^n \\ x^n &= 8^n \\ x &= 8 \\ \end{align} </math> </div></div> # Berapa hasil dari <math>\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } x=\frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ x &= \frac{\sqrt{30}+\sqrt{25}+\sqrt{24}+\sqrt{20}}{\sqrt{20}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5 \cdot 6}+\sqrt{5 \cdot 5}+\sqrt{6 \cdot 4}+\sqrt{5 \cdot 4}}{\sqrt{5 \cdot 4}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{5} \cdot \sqrt{6}+\sqrt{5} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{4}}{2 \cdot \sqrt{5}+\sqrt{6}+\sqrt{4}} \\ &= \frac{\sqrt{6} \cdot \sqrt{5}+\sqrt{6} \cdot \sqrt{4}+\sqrt{5} \cdot \sqrt{5}+\sqrt{5} \cdot \sqrt{4}}{\sqrt{5}+\sqrt{6}+\sqrt{5}+\sqrt{4}} \\ &= \frac{\sqrt{6}(\sqrt{5}+\sqrt{4})+\sqrt{5}(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ &= \frac{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}} \\ \frac{1}{x} &= \frac{\sqrt{6}+\sqrt{5}+\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})}+\frac{\sqrt{5}+\sqrt{4}}{(\sqrt{6}+\sqrt{5})(\sqrt{5}+\sqrt{4})} \\ &= \frac{1}{\sqrt{5}+\sqrt{4}}+\frac{1}{\sqrt{6}+\sqrt{5}} \\ &= \frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5} \\ &= \frac{\sqrt{5}-\sqrt{4}}{1}+\frac{\sqrt{6}-\sqrt{5}}{1} \\ &= \sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5} \\ &= \sqrt{6}-\sqrt{4} \\ &= \sqrt{6}-2 \\ x &= \frac{1}{\sqrt{6}-2} \\ &= \frac{\sqrt{6}+2}{6-4} \\ &= \frac{\sqrt{6}+2}{2} \\ &= 1+\frac{\sqrt{6}}{2} \\ \end{align} </math> </div></div> # Berapa hasil dari <math>(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}})^5 \\ \text{misalkan } x=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ x &= \frac{\sqrt{6}+\sqrt{2}}{\sqrt{32}} \\ &= \frac{\sqrt{2}(\sqrt{3}+1)}{4\sqrt{2}} \\ &= \frac{\sqrt{3}+1}{4} \\ 4x &= \sqrt{3}+1 \\ 4x-1 &= \sqrt{3} \\ (4x-1)^2 &= 3 \\ 16x^2-8x+1 &= 3 \\ 16x^2 &= 8x+2 \\ 8x^2 &= 4x+1 \\ x^2 &= \frac{4x+1}{8} \\ \text{cara 1 } \\ x^3 &= x \cdot x^2 \\ &= x(\frac{4x+1}{8}) \\ &= \frac{4x^2+x}{8} \\ &= \frac{4x^2}{8}+\frac{x}{8} \\ &= \frac{4(\frac{4x+1}{8})}{8}+\frac{x}{8} \\ &= \frac{16x+4}{64}+\frac{x}{8} \\ &= \frac{4x+1}{16}+\frac{x}{8} \\ &= \frac{4x+1+2x}{16} \\ &= \frac{6x+1}{16} \\ x^5 &= x^2 \cdot x^3 \\ &= (\frac{4x+1}{8})(\frac{6x+1}{16}) \\ &= \frac{24x^2+10x+1}{128} \\ &= \frac{24x^2}{128}+\frac{10x+1}{128} \\ &= \frac{24(\frac{4x+1}{8})}{128}+\frac{10x+1}{128} \\ &= \frac{96x+24}{1024}+\frac{10x+1}{128} \\ &= \frac{96x+24+80x+8}{1024} \\ &= \frac{176x+32}{1024} \\ &= \frac{176x}{1024}+\frac{32}{1024} \\ &= \frac{176}{1024}(\frac{\sqrt{3}+1}{4})+\frac{32}{1024} \\ &= \frac{44(\sqrt{3}+1)}{1024}+\frac{32}{1024} \\ &= \frac{44\sqrt{3}+44}{1024}+\frac{32}{1024} \\ &= \frac{76+44\sqrt{3}}{1024} \\ &= \frac{19+11\sqrt{3}}{256} \\ \text{cara 2 } \\ x^4 &= (x^2)^2 \\ &= (\frac{4x+1}{8})^2 \\ &= \frac{16x^2+8x+1}{64} \\ &= \frac{16x^2}{64}+\frac{8x}{64}+\frac{1}{64} \\ &= \frac{x^2}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{\frac{4x+1}{8}}{4}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{4x}{32}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{8}+\frac{1}{32}+\frac{x}{8}+\frac{1}{64} \\ &= \frac{x}{4}+\frac{3}{64} \\ x^5 &= x \cdot x^4 \\ &= (\frac{\sqrt{3}+1}{4})(\frac{x}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\frac{\sqrt{3}+1}{4}}{4}+\frac{3}{64}) \\ &= (\frac{\sqrt{3}+1}{4})(\frac{\sqrt{3}+1}{16}+\frac{3}{64}) \\ &= \frac{(\sqrt{3}+1)^2}{64}+(\frac{\sqrt{3}+1}{4})\frac{3}{64} \\ &= \frac{3+2\sqrt{3}+1}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{4+2\sqrt{3}}{64}+\frac{3(\sqrt{3}+1)}{256} \\ &= \frac{16+8\sqrt{3}}{256}+\frac{3\sqrt{3}+3}{256} \\ &= \frac{19+11\sqrt{3}}{256} \\ \end{align} </math> </div></div> # Berapa hasil dari <math>\frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{1}{4}+\frac{5}{16}+\frac{9}{64}+\frac{13}{256}+\dots \\ \frac{x}{4} &= \frac{1}{16}+\frac{5}{64}+\frac{9}{256}+\frac{13}{1.024}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+\frac{4}{16}+\frac{4}{64}+\frac{4}{256}+\dots \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots) \\ \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\dots &= \frac{1}{1-\frac{1}{4}} \\ &= \frac{4}{3} \\ \frac{3x}{4} &= \frac{1}{4}+4(\frac{4}{3}) \\ &= \frac{1}{4}+\frac{16}{3} \\ &= \frac{67}{12} \\ x &= \frac{67}{9} \\ \end{align} </math> </div></div> # Berapa nilai y-x jika <math>\frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} = \frac{x}{y}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1+2+3+4+ \dots + 106}{4+5+6+7+ \dots + 109} &= \frac{x}{y} \\ \frac{\frac{106 \times 107}{2}}{\frac{106}{2}(4+109)} &= \frac{x}{y} \\ \frac{53 \times 107}{53 \times 113} &= \frac{x}{y} \\ y-x &= 113-107 = 6 \\ \end{align} </math> </div></div> # Berapa angka satuan dari hasil 17²⁰²⁴? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan angka satuannya} \\ 17^1 &= 7 \\ 17^2 &= 9 \\ 17^3 &= 3 \\ 17^4 &= 1 \\ 17^5 &= 7 \\ 17^6 &= 9 \\ 17^7 &= 3 \\ 17^8 &= 1 \\ \text{Ini berarti berulang sebanyak 4 kali. Jadi 2024 dibagi 4 bersisa 0 maka angka satuannya yaitu 1} \end{align} </math> </div></div> # Berapa angka satuan dari hasil 1! + 2! + 3! + 4! + …. + 2024!? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ 1! + 2! + 3! + 4! + \dots + 2024! &= 1 + (1x2) + (1x2x3) + (1x2x3x4) + \dots + 2024! \\ &= 1 + 2 + 6 + 24 + 120 + 720 + \dots + 2024! \\ \text{Karena perkalian dikalikan 4,5,6, dst pasti angka satuan nya 0 maka } 1+2+6+24 = 33 \text{ jadi angka satuannya adalah } 3 \end{align} </math> </div></div> # Berapa hasil sisa jika 1! + 2! + 3! + 4! + ….. + 2024! dibagi 12? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{Perhatikan} \\ \frac{1! + 2! + 3! + 4! + \dots + 2024!}{12} &= \frac{1 + 1x2 + 1x2x3 + 1x2x3x4 + \dots + 2024!}{12} \\ &= \frac{1 + 2 + 6 + 24 + \dots + 2024!}{12} \\ \text{karena 4! + 5! + …. + 2024! dapat habis dibagi 12 yang berasal dari 3x4 jadi } 1+2+6 = 9 \end{align} </math> </div></div> # Penjumlahan bilangan 1 masing-masing seperti 1+1+1+1+… sebanyak 88 buah ditambah x dan y maka hasilnya A dan perkalian bilangan 1 masing-masing 1x1x1x… sebanyak 88 buah dikali x dan y maka hasilnya A maka berapa nilai A? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{penjumlahan} \\ 1+1+1+1+ \dots \text{ (sebanyak 88 buah) }+x+y &= A \\ 88+x+y &= A \\ \text{perkalian} \\ 1 \times 1 \times 1 \times \dots \text{ (sebanyak 88 buah) }\times x \times y &= A \\ x \times y &= A \\ 88+x+y &= xy \\ xy-y &= 88+x \\ y(x-1) &= 88+x \\ y &= \frac{88+x}{x-1} \\ \text{uji selidiki untuk x=2} \\ y &= \frac{88+2}{2-1} \\ &= 90 \\ \text{buktikan} \\ 88+x+y &= xy \\ 88+2+90 &= 2(90) \\ 180 &= 180 \\ \text{terbukti} \\ \text{nilai A adalah } 180 \\ \end{align} </math> </div></div> # Berapakah nilai x, y dan z dari <math>x+y-z=1, x^2+y^2-z^2=-5 \text{ dan } x^3+y^3-z^3=-53</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+y-z &= 1 \\ x+y &= z+1 \\ x^2+2xy+y^2 &= z^2+2z+1 \\ x^2+y^2-z^2 &= 2z+1-2xy \\ -5 &= 2z+1-2xy \\ 2xy &= 2z+6 \\ xy &= z+3 \\ x^2+y^2-z^2 &= -5 \\ x^2+y^2 &= z^2-5 \\ x^3+y^3-z^3 &= -53 \\ (x+y)(x^2-xy+y^2)-z^3+53 &= 0 \\ (x+y)(x^2+y^2-xy)-z^3+53 &= 0 \\ (z+1)(z^2-5-(z+3))-z^3+53 &= 0 \\ (z+1)(z^2-z-8)-z^3+53 &= 0 \\ z^3-z^2-8z+z^2-z-8-z^3+53 &= 0 \\ -9z+45 &= 0 \\ -9z &= -45 \\ z &= 5 \\ x+y &= 5+1 \\ x+y &= 6 \\ x &= 6-y \\ xy &= 5+3 \\ xy &= 8 \\ (6-y)y &= 8 \\ 6y-y^2 &= 8 \\ y^2-6y+8 &= 0 \\ (y-4)(y-2) &= 0 \\ y=4 \text{ atau } y=2 \\ \text{jika } y=4 \\ x+y &= z+1 \\ x+4 &= 5+1 \\ x &= 2 \\ \text{jika } y=2 \\ x+y &= z+1 \\ x+2 &= 5+1 \\ x &= 4 \\ \end{align} </math> </div></div> # Berapakah nilai titik koordinat (x,y) dari <math>\sqrt{x+y}+\sqrt{x-y}=\sqrt{\frac{432x}{13y}}</math> dan <math>\sqrt{x+y}-\sqrt{x-y}=\sqrt{\frac{52y}{3x}}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x+y}+\sqrt{x-y} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+y}-\sqrt{x-y} &= \sqrt{\frac{52y}{3x}} \\ (\sqrt{x+y}+\sqrt{x-y})(\sqrt{x+y}-\sqrt{x-y}) &= \sqrt{\frac{432x}{13y}} \cdot \sqrt{\frac{52y}{3x}} \\ x+y-x+y &= \sqrt{\frac{432x \cdot 52y}{13y \cdot 3x}} \\ 2y &= \sqrt{144 \cdot 4} \\ 2y &= \sqrt{576} \\ 2y &= 24 \\ y &= 12 \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13y}} \\ \sqrt{x+12}+\sqrt{x-12} &= \sqrt{\frac{432x}{13(12)}} \\ x+12+x-12+2 \cdot \sqrt{x+12} \cdot \sqrt{x-12} &= \frac{36x}{13} \\ 2x+2 \sqrt{x^2-144} &= \frac{36x}{13} \\ 2(x+\sqrt{x^2-144}) &= \frac{36x}{13} \\ x+\sqrt{x^2-144} &= \frac{18x}{13} \\ \sqrt{x^2-144} &= \frac{5x}{13} \\ x^2-144 &= \frac{25x^2}{169} \\ \frac{144x^2}{169}-144 &= 0 \\ \frac{x^2}{169}-1 &= 0 \\ x^2-169 &= 0 \\ (x-13)(x+13) &= 0 \\ x_1=13 &\text{ atau } x_2=-13 \text{ (TM) karena } x>y \\ \end{align} </math> jadi titik koordinat (13,12) </div></div> # Berapakah nilai dari <math>x^2-7x</math> jika <math>(x-2)^2+\frac{1}{(x-2)^2} = 11</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x-2)^2+\frac{1}{(x-2)^2} &= 11 \\ (x-2)^2-2(x-2)\frac{1}{(x-2)}+\frac{1}{(x-2)^2} &= 11-2 \\ (x-2-\frac{1}{x-2})^2 &= 9 \\ x-2-\frac{1}{x-2} &= 3 \\ (x-2)^2-1 &= 3(x-2) \\ x^2-4x+4-1 &= 3x-6 \\ x^2-7x &= -9 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>\frac{(x+y)^2(x+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)}</math> jika xy+yz+xz=1? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy+yz+xz &= 1 \\ x^2+xy+yz+xz &= x^2+1 \\ x(x+y)+z(x+y) &= x^2+1 \\ (x+y)(x+z) &= x^2+1 \\ \text{dengan pola yang sama } \\ (y+x)(y+z) &= y^2+1 \\ (x+z)(y+z) &= z^2+1 \\ \frac{(x+y)^2(y+z)^2(x+z)^2}{(x^2+1)(y^2+1)(z^2+1)} &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)(x+z)(y+x)(y+z)(x+z)(y+z)} \\ &= \frac{(x+y)^2(y+z)^2(x+z)^2}{(x+y)^2(y+z)^2(x+z)^2} \\ &= 1 \\ \end{align} </math> </div></div> # Berapakah nilai dari w+x+y+z jika w+5=x+4=y+3=z+2=w+x+y+z+5? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} w+5 &= w+x+y+z+5 \\ x+4 &= w+x+y+z+5 \\ y+3 &= w+x+y+z+5 \\ z+2 &= w+x+y+z+5 \\ \text{jumlahkan keempat persamaan } \\ w+x+y+z+14 &= 4(w+x+y+z+5) \\ w+x+y+z+14 &= 4(w+x+y+z)+20 \\ 3(w+x+y+z) &= -6 \\ w+x+y+z &= -2 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>\frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2}</math> jika <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3</math> dan x+y+z=xyz? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2y^2+y^2z^2+x^2z^2}{x^2y^2z^2} &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2} \\ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}) \\ 3^2 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{z+x+y}{xyz}) \\ 9 &= \frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}+2(\frac{xyz}{xyz}) \\ &= \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2 \\ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} &= 7 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>\frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z}</math> jika <math>x^2+y^2+z^2 = -2(ab+bc+ac)</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+z^2 &= -2(xy+yz+xz) \\ x^2+y^2+z^2+2(xy+yz+xz) &= 0 \\ (x+y+z)^2 &= 0 \\ x+y+z &= 0 \\ x+y &= -z \\ x+z &= -y \\ y+z &= -x \\ \frac{2z}{x+y}-\frac{5y}{x+z}-\frac{7x}{y+z} &= \frac{2z}{-z}-\frac{5y}{-y}-\frac{7x}{-x} \\ &= -2-(-5)-(-7) \\ &= 10 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>\frac{20xyz}{xy+yz+xz}</math> jika <math>16^x = 256^y = 625^z = 40</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 16^x = 256^y = 625^z &= 40 \\ 2^{4x} = 4^{4y} = 5^{4z} &= 40 \\ 2^{4x} &= 40 \\ 2 &= 40^{\frac{1}{4x}} \\ 4^{4y} &= 40 \\ 4 &= 40^{\frac{1}{4y}} \\ 5^{4z} &= 40 \\ 5 &= 40^{\frac{1}{4z}} \\ 2 \cdot 4 \cdot 5 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x}} \cdot 40^{\frac{1}{4y}} \cdot 40^{\frac{1}{4z}} \\ 40 &= 40^{\frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z}} \\ 1 &= \frac{1}{4x} + \frac{1}{4y} + \frac{1}{4z} \\ 4 &= \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\ \frac{20xyz}{xy+yz+xz} &= 20 \cdot \frac{xyz}{xy+yz+xz} \\ &= 20 \cdot (\frac{xy+yz+xz}{xyz})^{-1} \\ &= 20 \cdot (\frac{1}{z} + \frac{1}{x} + \frac{1}{y})^{-1} \\ &= 20 \cdot (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{-1} \\ &= 20 \cdot (4)^{-1} \\ &= 20 \cdot \frac{1}{4} \\ &= 5 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>\frac{x^2}{x^4+3x^2+1}</math> jika <math>6x^2+25x+6=0</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 6x^2+25x+6 &= 0 \\ 6x+25+\frac{6}{x} &= 0 \\ 6(x+\frac{1}{x}) &= -25 \\ x+\frac{1}{x} &= \frac{-25}{6} \\ (c+\frac{1}{x})^2 &= (\frac{-25}{6})^2 \\ x^2+2+\frac{1}{x^2} &= \frac{625}{36} \\ x^2+\frac{1}{x^2} &= \frac{625}{36}-2 \\ x^2+\frac{1}{x^2} &= \frac{553}{36} \\ \frac{x^2}{x^4+3x^2+1} &= \frac{1}{x^2+3+\frac{1}{x^2}} \\ &= \frac{1}{a^2+\frac{1}{x^2}+3} \\ &= \frac{1}{\frac{553}{36}+3} \\ &= \frac{1}{\frac{661}{36}} \\ &= \frac{36}{661} \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>\frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{(9+4\sqrt{5})^{1013}}{(38+17\sqrt{5})^{675}}+6-\sqrt{5} &= \frac{(9+2\sqrt{20})^{1013}}{((2)^3+3(2)^2(\sqrt{5})+3(2)(\sqrt{5})^2+(\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{((2+\sqrt{5})^2)^{1013}}{((2+\sqrt{5})^3)^{675}}+6-\sqrt{5} \\ &= \frac{(2+\sqrt{5})^{2026}}{(2+\sqrt{5})^{2025}}+6-\sqrt{5} \\ &= 2+\sqrt{5}+6-\sqrt{5} \\ &= 8 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>27x^3+\frac{8}{x^3}</math> jika <math>3x+\frac{2}{x}=6</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 3x+\frac{2}{x} &= 6 \\ (3x+\frac{2}{x})^3 &= 6^3 \\ 27x^3+3(3x)(\frac{2}{x})(3x+\frac{2}{x})+\frac{8}{x^3} &= 216 \\ 27x^3+18(6)+\frac{8}{x^3} &= 216 \\ 27x^3+108+\frac{8}{x^3} &= 216 \\ 27x^3+\frac{8}{x^3} &= 108 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>x^6+\frac{8}{x^3}</math> jika <math>x^3+\frac{1}{x^3}=8</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^3+\frac{1}{x^3} &= 8 \\ x^3 &= 8-\frac{1}{x^3} \\ x^6 &= 8x^3-1 \\ x^6+\frac{8}{x^3} &= 8x^3-1+\frac{8}{x^3} \\ &= 8x^3+\frac{8}{x^3}-1 \\ &= 8(x^3+\frac{1}{x^3})-1 \\ &= 8(8)-1 \\ &= 63 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>4x+\frac{25}{x}</math> jika <math>2\sqrt{x}+\frac{5}{\sqrt{x}}=4x-\frac{25}{x}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 2\sqrt{x}+\frac{5}{\sqrt{x}} &= 4x-\frac{25}{x} \\ 2\sqrt{x}+\frac{5}{\sqrt{x}} &= (2\sqrt{x}+\frac{5}{\sqrt{x}})(2\sqrt{x}-\frac{5}{\sqrt{x}}) \\ 1 &= 2\sqrt{x}-\frac{5}{\sqrt{x}} \\ 1^2 &= (2\sqrt{x}-\frac{5}{\sqrt{x}})^2 \\ 1 &= 4x-20+\frac{25}{x} \\ 4x+\frac{25}{x} &= 21 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\frac{x^2-x+1}{x^2+x+1}=\frac{5}{6}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2-x+1}{x^2+x+1} &= \frac{5}{6} \\ \frac{x^2+1-x}{x^2+1+x} &= \frac{5}{6} \\ \frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1} &= \frac{5}{6} \\ \text{ misalkan } x+\frac{1}{x} &= y \\ \frac{y-1}{y+1} &= \frac{5}{6} \\ 6(y-1) &= 5(y+1) \\ 6y-6 &= 5y+5 \\ y &= 11 \\ x+\frac{1}{x} &= 11 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt{x}+x=1</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt{x}+x &= 1 \\ x-1 &= -\sqrt{x} \\ (x-1)^2 &= (-\sqrt{x})^2 \\ x^2-2x+1 &= x \\ x^2-3x+1 &= 0 \\ x-3+\frac{1}{x} &= 0 \\ x+\frac{1}{x} &= 3 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>x+\frac{1}{x}</math> jika <math>\sqrt[3]{x}-\sqrt[3]{x-36}=3</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x}-\sqrt[3]{x-36} &= 3 \\ (\sqrt[3]{x}-\sqrt[3]{x-36})^3 &= 3^3 \\ x-(x-36)-3 \sqrt[3]{x(x-36)}(\sqrt[3]{x}-\sqrt[3]{x-36}) &= 27 \\ 36-3 \sqrt[3]{x(x-36)}3 &= 27 \\ -9 \sqrt[3]{x(x-36)} &= -9 \\ \sqrt[3]{x(x-36)} &= 1 \\ x(x-36) &= 1 \\ x^2-36x-1 &= 0 \\ x-36-\frac{1}{x} &= 0 \\ x-\frac{1}{x} &= 36 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>x+\frac{16}{x}</math> jika <math>x-3\sqrt{x}=4</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x-3\sqrt{x} &= 4 \\ x-4 &= 3\sqrt{x} \\ x^2-8x+16 &= 9x \\ x^2-17x+16 &= 0 \\ x-17+\frac{16}{x} &= 0 \\ x+\frac{16}{x} &= 17 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>\frac{x^2}{x^4+4}</math> jika <math>x^2-7x+2=0</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2-7x+2 &= 0 \\ x^2+2 &= 7x \\ x+\frac{2}{x} &= 7 \\ x^2+4+\frac{4}{x^2} &= 49 \\ x^2+\frac{4}{x^2} &= 45 \\ \frac{x^4+4}{x^2} &= 45 \\ \frac{x^2}{x^4+4} &= \frac{1}{45} \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1}</math> jika <math>x^{\frac{1}{4}}+x^{-\frac{1}{4}}=5</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{1}{2}}+2+x^{-\frac{1}{2}} &= 25 \\ x^{\frac{1}{2}}+x^{-\frac{1}{2}} &= 23 \\ x+2+x^{-1} &= 529 \\ x+x^{-1} &= 527 \\ x^{\frac{1}{4}}+x^{-\frac{1}{4}} &= 5 \\ x^{\frac{3}{4}}+3(x^{\frac{1}{4}}+x^{-\frac{1}{4}})+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+3(5)+x^{-\frac{3}{4}} &= 125 \\ x^{\frac{3}{4}}+x^{-\frac{3}{4}} &= 110 \\ x+x^{\frac{3}{4}}+x^{-\frac{3}{4}}+x^{-1} &= x+x^{-1}+x^{\frac{3}{4}}+x^{-\frac{3}{4}} \\ &= 527+110 \\ &= 637 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>\sqrt{8x^6+x^5+x^4+5x^3+1}</math> jika <math>\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}=0</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5} &= 0 \\ \frac{x^2+x+1}{x^5} &= 0 \\ x^2+x+1 &= 0 \\ x^2+x+1 &= 0 \\ (x-1)(x^2+x+1) &= 0(x-1) \\ x^3-1 &= 0 \\ x^3 &= 1 \\ x &= 1 \\ \sqrt{8x^6+x^5+x^4+5x^3+1} &= \sqrt{(2x^3)^2+x^3x^2+x^3x+5x^3+1} \\ &= \sqrt{(2(1))^2+(1)x^2+(1)x+5(1)+1} \\ &= \sqrt{(2)^2+x^2+x+5+1} \\ &= \sqrt{4+x^2+x+1+5} \\ &= \sqrt{4+0+5} \\ &= \sqrt{9} \\ &= 3 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>f(1)+f(2)+f(3)+ \dots + f(99)</math> jika <math>f(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{1}{\sqrt{x+1}+\sqrt{x}} \\ &= \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} \\ &= \sqrt{x+1}-\sqrt{x} \\ f(1)+f(2)+f(3)+ \dots + f(98)+f(99) &= \sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\sqrt{3+1}-\sqrt{3}+ \cdot + \sqrt{98+1}-\sqrt{98}+\sqrt{99+1}-\sqrt{99} \\ &= \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+ \cdot + \sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99} \\ &= \sqrt{100}-\sqrt{1} \\ &= 10-1 \\ &= 9 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots + \frac{2024}{2025})</math> jika <math>h(x)=\frac{3}{3+9^x}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} h(x) &= \frac{3}{3+9^x} \\ h(1-x) &= \frac{3}{3+9^{1-x}} \\ &= \frac{3}{3+\frac{9}{9^x}} \\ &= \frac{9^x}{3+9^x} \\ h(x)+h(1-x) &= \frac{3}{3+9^x}+\frac{9^x}{3+9^x} \\ &= \frac{3+9^x}{3+9^x} \\ &= 1 \\ & 5(\frac{1}{2025}+\frac{2}{2025}+\frac{3}{2025}+ \dots +(1-\frac{2}{2025})+(1-\frac{1}{2025})) \\ & 5(1+1+1+ \dots +1+1) \text{ sebanyak 1012 kali } \\ & 5(1012) \\ & 5060 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>\frac{7^{2025} - 7^{2023} + 432}{7^{2024} + 7^{2023} + 72}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7^{2025}-7^{2023}+432}{7^{2024}+7^{2023}+72} &= \frac{7^{2023}7^{2}-7^{2023} + 48 \times 9}{7^{2023}7^1+7^{2023}+8 \times 9} \\ &= \frac{7^{2023}(7^{2}-1)+48 \times 9}{7^{2023}(7^1+1)+8 \times 9} \\ &= \frac{7^{2023}(49-1)+48 \times 9}{7^{2023}(7+1) + 8 \times 9} \\ &= \frac{7^{2023} \times 48+48 \times 9}{7^{2023} \times 8+8 \times 9} \\ &= \frac{48(7^{2023}+9)}{8(7^{2023}+9)} \\ &= \frac{48}{8} \\ &= 6 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>tan (x+\frac{\pi}{4})</math> jika <math>\frac{1}{cos x}-tan x = \frac{4}{5}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{cos x}-tan x &= \frac{4}{5} \\ sec x-tan x &= \frac{4}{5} \\ sec^2 x-tan^2 x &= 1 \\ (sec x+tan x)(sec x-tan x) &= 1 \\ (sec x+tan x)\frac{4}{5} &= 1 \\ sec x+tan x &= \frac{5}{4} \\ \text{kedua persamaan dengan cara metode eliminasi } \\ 2 tan x &= \frac{5}{4}-\frac{4}{5} \\ 2 tan x &= \frac{9}{20} \\ tan x &= \frac{9}{40} \\ tan (x+\frac{\pi}{4}) &= \frac{tan x+tan \frac{\pi}{4}}{1-tan x \cdot tan \frac{\pi}{4}} \\ &= \frac{\frac{9}{40}+1}{1-\frac{9}{40} \cdot 1} \\ &= \frac{\frac{49}{40}}{\frac{31}{40}} \\ &= \frac{49}{31} \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>sin^3 x+csc^3 x</math> jika <math>sin x-csc x = 8</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ Dengan menggunakan rumus: } (a-b)^3 &= a^3-b^3-3ab(a-b) \\ (sin x-csc x)^3 &= sin^3 x-csc^3 x-3sin x csc x(sin x-csc x) \\ 8^3 &= sin^3 x-csc^3 x-3sin x (\frac{1}{sin x})(8) \\ 512 &= sin^3 x-csc^3 x-24 \\ sin^3 x-csc^3 x &= 512+24 \\ sin^3 x-csc^3 x &= 536 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>(sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2</math> jika <math>\frac{1}{sin x}+\frac{1}{cos x} = 10</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{sin x}+\frac{1}{cos x} &= 10 \\ \frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} &= 100 \\ (sin x+\frac{1}{cos x})^2+(cos x+\frac{1}{sin x})^2 &= sin^2 x+\frac{2sin x}{cos x}+\frac{1}{cos^2 x}+cos^2 x+\frac{2cos x}{sin x}+\frac{1}{sin^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2(sin^2 x+cos^2 x)}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+\frac{1}{sin^2 x}+\frac{2}{sin x \cdot cos x}+\frac{1}{cos^2 x} \\ &= 1+100 \\ &= 101 \\ \end{align} </math> </div></div> # Berapakah nilai dari (x-1)⁶ jika <math>x=\frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} sin 80^\circ &= cos 10^\circ \\ sin 80^\circ-cos 10^\circ &= 0 \\ x &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+sin 40^\circ}{sin 80^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+2 sin 20^\circ cos 20^\circ}{cos 10^\circ} \\ &= \frac{4 cos 55^\circ cos 25^\circ cos 10^\circ+4 sin 10^\circ cos 10^\circ cos 20^\circ}{cos 10^\circ} \\ &= 4 cos 55^\circ cos 25^\circ+4 sin 10^\circ cos 20^\circ \\ &= 2(2 cos 55^\circ cos 25^\circ+2 sin 10^\circ cos 20^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ+sin (-10)^\circ) \\ &= 2(cos 80^\circ+cos 30^\circ+sin 30^\circ-sin 10^\circ) \\ &= 2(cos 80^\circ-sin 10^\circ+cos 30^\circ+sin 30^\circ) \\ &= 2(cos 80^\circ-sin (90^\circ-80^\circ)+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(cos 80^\circ-cos 80^\circ+\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= 2(\frac{\sqrt{3}}{2}+\frac{1}{2}) \\ &= \sqrt{3}+1 \\ x-1 &= \sqrt{3} \\ (x-1)^6 &= (\sqrt{3})^6 \\ &= 27 \\ \end{align} </math> </div></div> # Berapakah nilai dari x jika <math>x=\frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \frac{x sin 20^\circ-x^2 sin 10^\circ}{2 sin 20^\circ-sin 40 ^\circ} \\ 2x sin 20^\circ-x sin 40 ^\circ &= x sin 20^\circ-x^2 sin 10^\circ \\ x^2 sin 10^\circ+x sin 20^\circ-x sin 40 ^\circ &= 0 \\ x(x sin 10^\circ+sin 20^\circ-sin 40 ^\circ) &= 0 \\ x = 0 &\text{ atau } x sin 10^\circ+sin 20^\circ-sin 40 ^\circ = 0 \\ x sin 10^\circ+sin 20^\circ-sin 40 ^\circ &= 0 \\ x sin 10^\circ &= sin 40 ^\circ-sin 20^\circ \\ x &= \frac{sin 40 ^\circ-sin 20^\circ}{sin 10^\circ} \\ &= \frac{2 cos 30 ^\circ sin 10^\circ}{sin 10^\circ} \\ &= 2 cos 30 ^\circ \\ &= \frac{2 \sqrt{3}}{2} \\ &= \sqrt{3} \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>\frac{x}{y}</math> jika <math>\frac{x^2}{x^2-16y^2} = \frac{625}{49}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{x^2}{x^2-16y^2} &= \frac{625}{49} \\ \frac{x^2-16y^2}{x^2} &= \frac{49}{625} \text{ (terbalik posisinya)} \\ 1-\frac{16y^2}{x^2} &= \frac{49}{625} \\ \frac{16y^2}{x^2} &= 1 - \frac{49}{625} \\ (\frac{4y}{x})^2 &= \frac{576}{625} \\ (\frac{4y}{x})^2 &= (\frac{24}{25})^2 \\ \frac{4y}{x} &= \frac{24}{25} \\ \frac{y}{x} &= \frac{6}{25} \\ \frac{x}{y} &= \frac{25}{6} \\ \end{align} </math> </div></div> # Berapakah nilai dari xy jika <math>x^4+y^4+x^2y^2=15 \text{ dan } x^2+y^2+xy=5</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+y^2+xy &= 5 \\ x^2+y^2 &= 5-xy \\ x^4+y^4+x^2y^2 &= 15 \\ (x^2)^2+(y^2)^2+2x^2y^2-x^2y^2 &= 15 \\ (x^2+y^2)^2-x^2y^2 &= 15 \\ (5-xy)^2-x^2y^2 &= 15 \\ 25-10xy+x^2y^2-x^2y^2 &= 15 \\ 25-10xy &= 15 \\ 10xy &= 10 \\ xy &= 1 \\ \end{align} </math> </div></div> # Berapakah nilai dari x jika <math>4^x = 63(4^3+1)(4^6+1)(4^{12}+1)+1</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 4^x &= 63(4^3+1)(4^6+1)(4^{12}+1)+1 \\ 4^x-1 &= 63(4^3+1)(4^6+1)(4^{12}+1) \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{4^3-1} \\ &= 63(4^3+1)(4^6+1)(4^{12}+1) \frac{4^3-1}{63} \\ &= (4^3+1)(4^6+1)(4^{12}+1)(4^3-1) \\ &= (4^3-1)(4^3+1)(4^6+1)(4^{12}+1) \\ &= (4^6-1)(4^6+1)(4^{12}+1) \\ &= (4^{12}-1)(4^{12}+1) \\ &= 4^{24}-1 \\ 4^x &= 4^{24} \\ x &= 24 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>\frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1}</math> jika <math>x=\sqrt{9+4\sqrt{5}}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x &= \sqrt{9+4\sqrt{5}} \\ x &= 2+\sqrt{5} \\ x^2 &= 9+4\sqrt{5} \\ x^2-4x &= 9+4\sqrt{5}-4(2+\sqrt{5}) \\ x^2-4x &= 1 \\ x^2 &= 4x+1 \\ x^3 &= x \cdot x^2 \\ &= x(4x+1) \\ &= 4x^2+x \\ &= 4(4x+1)+x \\ &= 16x+4+x \\ &= 17x+4 \\ x^4 &= x \cdot x^3 \\ &= x(17x+4) \\ &= 17x^2+4x \\ &= 17(4x+1)+4x \\ &= 68x+17+4x \\ &= 72x+17 \\ \frac{x^4-5x^3+2x^2+5x+3}{x^2-4x+1} &= \frac{72x+17-5(17x+4)+2(4x+1)+5x+3}{1+1} \\ &= \frac{72x+17-85x-20+8x+2+5x+3}{2} \\ &= \frac{2}{2} \\ &= 1 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>\sqrt{\frac{x^3+1}{x^5-x^4-x^3+x^2}}</math> jika 2x-1=<math>\sqrt{61}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan } \frac{x^3+1}{x^5-x^4-x^3+x^2} = p \\ p &= \frac{x^3+1}{x^5-x^4-x^3+x^2} \\ &= \frac{x^3+1}{x^5-x^4-(x^3-x^2)} \\ &= \frac{x^3+1}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{x^4(x-1)-x^2(x-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)(x^4-x^2)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x^2-1)} \\ &= \frac{(x+1)(x^2-x+1)}{(x-1)x^2(x-1)(x+1)} \\ &= \frac{x^2-x+1}{x^2(x-1)^2} \\ &= \frac{x^2-x+1}{(x(x-1))^2} \\ &= \frac{x(x-1)+1}{(x(x-1))^2} \\ 2x-1 &= \sqrt{61} \\ x &= \frac{\sqrt{61}+1}{2} \\ x-1 &= \frac{\sqrt{61}-1}{2} \\ x(x-1) &= (\frac{\sqrt{61}+1}{2})(\frac{\sqrt{61}-1}{2}) \\ &= \frac{61-1}{4} \\ &= \frac{60}{4} \\ &= 15 \\ p &= \frac{x(x-1)+1}{(x(x-1))^2} \\ &= \frac{15+1}{15^2} \\ &= \frac{16}{15^2} \\ \sqrt{p} &= \sqrt{\frac{16}{15^2}} \\ &= \frac{4}{15} \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>(\frac{x-3}{x})^{25}</math> jika <math>x+\sqrt[5]{8}+\sqrt[5]{2}=1+\sqrt[5]{16}+\sqrt[5]{4}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\sqrt[5]{8}+\sqrt[5]{2} &= 1+\sqrt[5]{16}+\sqrt[5]{4} \\ x+(\sqrt[5]{2})^3+\sqrt[5]{2} &= 1+(\sqrt[5]{2})^4+(\sqrt[5]{2})^2 \\ x &= (\sqrt[5]{2})^4-(\sqrt[5]{2})^3+(\sqrt[5]{2})^2-\sqrt[5]{2}+1 \\ \text{misalkan } \sqrt[5]{2} = p \\ x &= p^4-p^3+p^2-p+1 \\ x &= \frac{p^5+1}{p+1} \\ (\frac{x-3}{x})^{25} &= (1-\frac{3}{x})^{25} \\ &= (1-\frac{3}{\frac{p^5+1}{p+1}})^{25} \\ &= (1-\frac{3(p+1)}{p^5+1})^{25} \\ &= (1-\frac{3(\sqrt[5]{2}+1)}{(\sqrt[5]{2})^5+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{2+1})^{25} \\ &= (1-\frac{(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-(3\sqrt[5]{2}+3)}{3})^{25} \\ &= (\frac{3-3\sqrt[5]{2}-3)}{3})^{25} \\ &= (-\sqrt[5]{2})^{25} \\ &= (-2)^5 \\ &= -32 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>x^{50}+x^{49}+x^{48}+x^{47}+x^{46}</math> jika <math>x^2+x+1=0</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x^2+x+1 &= 0 \\ x^2+x &= -1 \\ \frac{x^3-1}{x-1} &= 0 \\ x^3 &= 1 \\ x &= 1 \\ x^{50}+x^{49}+x^{48}+x^{47}+x^{46} &= x^{48}(x^2+x+1)+x^{45}(x^2+x) \\ &= x^{48}(0)+(x^3)^{15}(-1) \\ &= 0+(1)^{15}(-1) \\ &= -1 \\ \end{align} </math> </div></div> # Berapakah 2<supz>24</sup> dari <math>8^7+8^6+8^5+8^4+8^3+8^2+8+1=A</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= A \\ 8(8^7+8^6+8^5+8^4+8^3+8^2+8+1) &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8 &= 8A \\ 8^8+8^7+8^6+8^5+8^4+8^3+8^2+8+1 &= 8A+1 \\ 8^8+A &= 8A+1 \\ 8^8 &= 7A+1 \\ (2^3)^8 &= 7A+1 \\ 2^{24} &= 7A+1 \\ \end{align} </math> </div></div> # Berapakah nilai dari <math>x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1</math> jika <math>x+\frac{1}{x}=\sqrt{3}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} x+\frac{1}{x} &= \sqrt{3} \\ x^2+2+\frac{1}{x^2} &= 3 \\ x^2-1+\frac{1}{x^2} &= 0 \\ x^2(x^2-1+\frac{1}{x^2}) &= x^2(0) \\ x^4-x^2+1 &= 0 \\ (x^2+1)(x^4-x^2+1) &= (x^2+1)0 \\ x^6-x^4+x^2+x^4-x^2+1 &= 0 \\ x^6+1 &= 0 \\ x^6 &= -1 \\ x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^6+1 &= {x^6}^7+{x^6}^6+{x^6}^5+{x^6}^4+{x^6}^3+{x^6}^2+x^6+1 \\ &= (-1)^7+(-1)^6+(-1)^5+(-1)^4+(-1)^3+(-1)^2-1+1 \\ &= -1+1-1+1-1+1-1+1 \\ &= 0 \\ \end{align} </math> </div></div> # Diberikan fungsi kuadrat f(x)=ax²+bx+c yang memenuhi f(2) = 4 dan f(7) = 49. Jika a ≠ 1 maka berapa nilai dari <math>\frac{c-b}{a-1}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= ax^2+bx+c \\ f(2) &= a(2)^2+2b+c = 4 \\ &= 4a+2b+c = 4 \\ f(7) &= a(7)^2+7b+c = 49 \\ &= 49a+7b+c = 49 \\ 49a+7b+c &= 49 \\ 4a+2b+c &= 4 \\ 45a+5b &= 45 \text{ (f(7) dikurangi f(2)) } \\ 9a+b &= 9 \\ b &= -9a+9 \\ 4a+2b+c &= 4 \\ 4a+2(-9a+9)+c &= 4 \\ 4a-18a+18+c &= 4 \\ -14a+18+c &= 4 \\ c &= 14a-14 \\ \frac{c-b}{a-1} &= \frac{14a-14-(-9a+9)}{a-1} \\ &= \frac{14(a-1)+9(a-1)}{a-1} \\ &= \frac{(14+9)(a-1)}{a-1} \\ &= 23 \\ \end{align} </math> </div></div> # Jika x³+y³ = 242 dan x+y = 11 maka berapa hasil dari (x-y)²? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} (x+y)^3 &= x^3+y^3+3xy(x+y) \\ 11^3 &= 242+3xy(11) \text{ (dibagi 11)} \\ 11^2 &= 22+3xy \\ 121 &= 22+3xy \\ 99 &= 3xy \\ xy &= 33 \\ (x-y)^2 &= x^2+y^2-2xy \\ &= ((x+y)^2-2xy)-2xy \\ &= (x+y)^2-4xy \\ &= 11^2-4(33) \\ &= 121-132 \\ &= -11 \\ \end{align} </math> </div></div> # Berapa f(1)+f(-1) jika <math>f(\frac{ax-b}{bx-a})</math>=x²-5x+6? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{ jika} f(1) = f(\frac{ax-b}{bx-a}) \\ 1 &= \frac{ax-b}{bx-a} \\ bx-a &= ax-b \\ (b-a)x &= -b+a \\ &= -(b-a) \\ &= -1 \\ f(1) &= x^2-5x+6 \\ &= (-1)^2-5(-1)+6 \\ &= 12 \\ \text{ jika} f(-1) = f(\frac{ax-b}{bx-a}) \\ -1 &= \frac{ax-b}{bx-a} \\ -(bx-a) &= ax-b \\ -bx+a &= ax-b \\ (-b-a)x &= -b-a \\ &= 1 \\ f(-1) &= x^2-5x+6 \\ &= (1)^2-5(1)+6 \\ &= 2 \\ f(1)+f(-1) &= 12+2 \\ &= 14 \\ \end{align} </math> </div></div> # berapa f(200) jika f(0)=1 serta f(x)-x=f(x-1)? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)-x &= f(x-1) \\ f(x)-f(x-1) &= x \\ x=1 ; f(1)-f(0) &= 1 \\ x=2 ; f(2)-f(1) &= 2 \\ x=3 ; f(3)-f(2) &= 3 \\ x=4 ; f(4)-f(3) &= 4 \\ \dots \\ x=200 ; f(200)-f(199) &= 200 \\ \text{ jumlahkan tersebut menjadi } \\ f(200)-f(0) &= 1+2+3+4+\dots+200 \\ &= \frac{200 \cdot 201}{2} \\ &= 20.100 \\ f(200)-1 &= 20.100 \\ &= 20.101 \\ \end{align} </math> </div></div> # Misalkan f(x) adalah fungsi rekursif yang berlaku ∀x ∈ R sebagai berikut: : f(x)+f(15-x) = 2024 : f(15+x) = f(x)+2020 maka tentukan nilai dari 2f(2025)+2f(-2025)! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x)+f(15-x) &= 2024 \\ f(15+x) &= f(x)+2020 \\ *\text{cara 1 } \\ \text{ganti x dengan 15+x } \\ f(15+x)+f(-x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) = 8 \\ \text{maka } 2f(2025)+2f(-2025) = 8 \\ *\text{cara 2 } \\ \text{ganti x dengan -x } \\ f(-x)+f(15+x) &= 2024 \\ f(15+x)-f(x) &= 2020 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ f(x)+f(-x) &= 4 \\ \text{lalu dikalikan 2 masing-masing menjadi } \\ 2f(x)+2f(-x) = 8 \\ \text{maka } 2f(2025)+2f(-2025) = 8 \\ \end{align} </math> </div></div> # Misalkan f suatu fungsi rekursif yang memenuhi <math>f(\frac{1}{x}) + \frac{1}{x}f(-x) = 3x</math> untuk setiap bilangan riil x ≠ 0. Tentukan nilai f(3)! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(\frac{1}{x})+\frac{1}{x}f(-x) &= 3x \\ \text{ganti x dengan 1/3 } \\ f(3)+3f(-\frac{1}{3}) &= 1 \\ \text{ganti x dengan -3 } \\ f(-\frac{1}{3}) - \frac{1}{3}f(3) = -9 \\ \text{dikalikan 3 } \\ 3f(-\frac{1}{3})-f(3) &= -27 \\ \text{persamaan 1 dan 2 dihasilkan sebagai berikut } \\ 2f(3) &= 28 \\ f(3) &= 14 \\ \end{align} </math> </div></div> # Diketahui polinom <math>f(7^b-1)=7^{3b}-10</math>. tentukan nilai f(5)! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * cara 1 \\ f(5) &= f(7^b-1) \\ 5 &= 7^b-1 \\ 7^b &= 6 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(6-1) &= 6^3-10 \\ f(5) &= 216-10 \\ &= 206 \\ * cara 2 \\ \text{misalkan } 7^b-1=a \text{ maka } 7^b=a+1 \\ f(7^b-1) &= 7^{3b}-10 \\ &= (7^b)^3-10 \\ f(a) &= (a+1)^3-10 \\ f(5) &= (5+1)^3-10 \\ &= 6^3-10 \\ &= 216-10 \\ &= 206 \\ \end{align} </math> </div></div> # Diketahui polinom <math>f(6^b-7)=6^{3b}-2 \cdot 6^{2b}-4</math>. tentukan nilai f(-2)! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * cara 1 \\ f(-2) &= f(6^b-7) \\ -2 &= 6^b-7 \\ 6^b &= 5 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(5-7) &= 5^3-2 \cdot 5^2-4 \\ f(-2) &= 125-50-4 \\ &= 71 \\ * cara 2 \\ \text{misalkan } 6^b-7=a \text{ maka } 6^b=a+7 \\ f(6^b-7) &= 6^{3b}-2 \cdot 6^{2b}-4 \\ &= (6^b)^3-2 \cdot (6^b)^2-4 \\ f(a) &= (a+7)^3-2(a+7)^2-4 \\ f(-2) &= (-2+7)^3-2(-2+7)^2-4 \\ &= 5^3-2(5)^2-4 \\ &= 125-50-4 \\ &= 71 \\ \end{align} </math> </div></div> # Jika <math>f(xy)=\frac{f(x)}{y}</math> dengan y ≠ 0 serta f(10)=7 maka tentukan nilai f(2)! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(10) &= 7 \\ f(2 \cdot 5) &= 7 \\ f(xy) &= \frac{f(x)}{y} \\ f(2 \cdot 5) &= \frac{f(2)}{5} \\ 7 &= \frac{f(2)}{5} \\ f(2) &= 35 \\ \end{align} </math> </div></div> # Jika <math>f(xy)=\frac{f(x+y)}{xy}</math> dengan f(xy) ≠ 0 serta f(15)=16 maka tentukan nilai f(8)! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(15) &= 16 \\ f(3 \cdot 5) &= 16 \\ f(xy) &= \frac{f(x+y)}{xy} \\ f(3 \cdot 5) &= \frac{f(3+5)}{3 \cdot 5} \\ f(15) &= \frac{f(8)}{15} \\ 16 &= \frac{f(8)}{15} \\ f(8) &= 240 \\ \end{align} </math> </div></div> # Jika <math>f(x+\frac{1}{x}+6)=x^2+\frac{1}{x^2}+15</math> maka tentukan nilai f(16)! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x+\frac{1}{x}+6) &= x^2+\frac{1}{x^2}+15 \\ &= (x+\frac{1}{x})^2-2+15 \\ &= (x+\frac{1}{x})^2+13 \\ \text{misalkan } x+\frac{1}{x} &= p \\ f(x+\frac{1}{x}+6) &= (x+\frac{1}{x})^2+13 \\ f(p+6) &= p^2+13 \\ \text{jika f(16) maka p adalah 10 sebelum ditambahkan 6 } \\ f(p+6) &= p^2+13 \\ f(10+6) &= 10^2+13 \\ f(16) &= 100+13 \\ &= 113 \\ \end{align} </math> </div></div> # tentukan nilai x jika <math>f(x)=\frac{4}{4-x}</math> dan <math>f(x \cdot f(x))^{\frac{f(4x)}{f(x)}}=256</math>! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{4}{4-x} \\ f(4x) &= \frac{4}{4-4x} \\ \frac{f(4x)}{f(x)} &= \frac{\frac{4}{4-4x}}{\frac{4}{4-x}} \\ &= \frac{4-x}{4-4x} \\ f(x \cdot f(x)) &= f(x(\frac{4}{4-x})) \\ &= f(\frac{4x}{4-x}) \\ &= \frac{4}{4-(\frac{4x}{4-x})} \\ &= \frac{4}{\frac{16-4x-4x}{4-x}} \\ &= \frac{4}{\frac{16-8x}{4-x}} \\ &= \frac{4(4-x)}{4(4-4x)} \\ &= \frac{4-x}{4-4x} \\ \text{misalkan } \frac{4-x}{4-4x} &= a \\ f(x \cdot f(x))^{\frac{f(4x)}{f(x)}} &= 256 \\ a^a &= 256 \\ a^a &= 4^4 \\ a &= 4 \\ \frac{4-x}{4-4x} &= 4 \\ 4-x &= 16-16x \\ 15x &= 12 \\ x &= \frac{4}{5} \\ \end{align} </math> </div></div> # Fungsi <math>f(x) = \frac{kx}{2x+1} \text{dengan } x \neq -\frac{1}{2}</math>. Dengan f(f(x)) = x maka tentukan nilai k! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} f(x) &= \frac{kx}{2x+1} \\ f(f(x)) &= x \\ f(\frac{kx}{2x+1}) &= x \\ \frac{k(\frac{kx}{2x+1})}{2(\frac{kx}{2x+1})+1} &= x \\ \frac{\frac{k^2x}{2x+1}}{\frac{2kx+2x+1}{2x+1}} &= x \\ \frac{k^2x}{2kx+2x+1} &= x \\ \frac{k^2}{2kx+2x+1} &= 1 \\ k^2 &= 2kx+2x+1 \\ k^2-2kx &= 2x+1 \\ k^2-2kx+x^2 &= x^2+2x+1 \\ (k-x)^2 &= (x+1)^2 \\ (k-x)^2-(x+1)^2 &= 0 \\ (k-x+x+1)(k-x-(x+1)) &= 0 \\ k=-1 &\text{ atau } k=2x+1 &\text{ (TM) } \\ \end{align} </math> </div></div> # Jika n = 2023²+2024² maka berapa hasil dari <math>\sqrt{2n-1}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n &= 2023^2+2024^2 \\ &= 2023^2+(2023+1)^2 \\ \text{Misalkan 2023 = p} \\ n &= p^2+(p+1)^2 \\ &= p^2+p^2+2p+1 \\ &= 2p^2+2p+1 \\ \sqrt{2n-1} &= \sqrt{2(2p^2+2p+1)-1} \\ &= \sqrt{4p^2+4p+2-1} \\ &= \sqrt{4p^2+4p+1} \\ &= \sqrt{(2p+1)^2} \\ &= 2p+1 \\ &= 2(2023)+1 \\ &= 4046+1 \\ &= 4047 \\ \end{align} </math> </div></div> # tentukan nilai dari (a-c)^b jika <math>\frac{ab}{a+b} = \frac{1}{3}</math>, <math>\frac{bc}{b+c} = \frac{1}{4}</math> dan <math>\frac{ac}{a+c} = \frac{1}{9}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{ab}{a+b} &= \frac{1}{3} \\ \frac{a+b}{ab} &= 3 \text{ (terbalik posisinya)} \\ \frac{1}{b} + \frac{1}{a} &= 3 \\ \frac{bc}{b+c} &= \frac{1}{4} \\ \frac{b+c}{bc} &= 4 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{b} &= 4 \\ \frac{ac}{a+c} &= \frac{1}{9} \\ \frac{a+c}{ac} &= 9 \text{ (terbalik posisinya)} \\ \frac{1}{c} + \frac{1}{a} &= 9 \\ \text{Misalkan 1/a = x, 1/b = y dan 1/c = z} \\ x+y &= 3 \\ y+z &= 4 \\ x+z &= 9 \\ x+y &= 3 \\ y+z &= 4 \\ x-z &= -1 \\ x-z &= -1 \\ x+z &= 9 \\ 2x &= 8 \\ x &= 4 \\ x-z &= -1 \\ 4-z &= -1 \\ z &= 5 \\ x+y &= 3 \\ 4+y &= 3 \\ y &= -1 \\ \frac{1}{a} &= 4 \\ a &= \frac{1}{4} \\ \frac{1}{b} &= -1 \\ b &= -1 \\ \frac{1}{c} &= 5 \\ c &= \frac{1}{5} \\ (a-c)^b &= (\frac{1}{4} - \frac{1}{5})^{-1} \\ &= (\frac{5-4}{20})^{-1} \\ &= (\frac{1}{20})^{-1} \\ &= 20 \\ \end{align} </math> </div></div> # tentukan nilai dari a, b dan c jika <math>\frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5}</math> dan a+2b+3c=28? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan k untuk semua ketiga persamaan tersebut } \\ \frac{a+b}{2}=\frac{a+c}{4}=\frac{b+c}{5} &= k \\ a+b &= 2k \\ a+c &= 4k \\ b+c &= 5k \\ 2a+b+c &= 6k \\ 2a+5k &= 6k \\ k &= 2a \\ a &= \frac{k}{2} \\ b &= \frac{3k}{2} \\ c &= \frac{7k}{2} \\ a+2b+3c &= 28 \\ \frac{k}{2}+2(\frac{3k}{2})+3(\frac{7k}{2}) &= 28 \\ k+6k+21k &= 56 \\ 28k &= 56 \\ k &= 2 \\ a &= \frac{k}{2} \\ &= \frac{2}{2} = 1 \\ b &= \frac{3k}{2} \\ &= \frac{3(2)}{2} = 3 \\ c &= \frac{7k}{2} \\ &= \frac{7(2)}{2} = 7 \\ \end{align} </math> </div></div> # tentukan nilai dari (b+c)^a jika <math>\frac{a+b+c}{2} = \sqrt{a-2}+\sqrt{b-1}+\sqrt{c}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{a+b+c}{2} &= \sqrt{a-2}+\sqrt{b-1}+\sqrt{c} \\ a+b+c &= 2(\sqrt{a-2}+\sqrt{b-1}+\sqrt{c}) \\ a-2\sqrt{a-2}+b-2\sqrt{b-1}+c-2\sqrt{c} &= 0 \\ a-2-2\sqrt{a-2}+1+b-1-2\sqrt{b-1}+1+c-2\sqrt{c}+1 &= 0 \\ (\sqrt{a-2}-1)^2+(\sqrt{b-1}-1)^2+(\sqrt{c}-1)^2 &= 0 \\ (\sqrt{a-2}-1)^2 &= 0 \\ \sqrt{a-2}-1 &= 0 \\ \sqrt{a-2} &= 1 \\ a-2 &= 1 \\ a &= 3 \\ (\sqrt{b-1}-1)^2 &= 0 \\ \sqrt{b-1}-1 &= 0 \\ \sqrt{b-1} &= 1 \\ b-1 &= 1 \\ b &= 1 \\ (\sqrt{c}-1)^2 &= 0 \\ \sqrt{c}-1 &= 0 \\ \sqrt{c} &= 1 \\ c &= 1 \\ (b+c)^a &= (2+1)^3 \\ &= 3^3 \\ &= 27 \\ \end{align} </math> </div></div> # x dan y merupakan bilangan tak nol. Jika xy = <math>\frac{x}{y}</math> = x-y maka berapa nilai x+y? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} xy &= \frac{x}{y} \\ y^2 &= 1 \\ y^2 - 1 &= 0 \\ (y-1)(y+1) &= 0 \\ y = 1 &\text{ atau } y = -1 \\ \frac{x}{y} &= x-y \\ x &= xy-y^2 \\ x-xy &= -y^2 \\ x(1-y) &= -y^2 \\ x &= \frac{-y^2}{1-y} \\ \text{cek y=1 } \\ x &= \frac{-1^2}{1-1} \\ \text{tidak memenuhi syarat } \\ \text{cek y=-1 } \\ x &= \frac{-(-1)^2}{1-(-1)} \\ &= \frac{-1}{2} \\ x+y &= -1-\frac{1}{2} \\ &= -\frac{3}{2} \\ \end{align} </math> </div></div> # berapa nilai x dari <math>(\frac{a}{b})^3+(\frac{b}{a})^3 = 2\sqrt{x}</math> jika <math>\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{1}{a}+\frac{1}{b} &= \frac{1}{a+b} \\ \frac{a+b}{ab} &= \frac{1}{a+b} \\ (a+b)^2 &= ab \\ a^2+2ab+b^2 &= ab \\ a^2+b^2 &= -ab \\ \text{misalkan } \frac{a}{b}+\frac{b}{a} = n \\ \frac{a}{b}+\frac{b}{a} &= n \\ \frac{a^2+b^2}{ab} &= n \\ a^2+b^2 &= nab \\ n &= -1 \\ \frac{a}{b}+\frac{b}{a} &= n \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3(\frac{a}{b}+\frac{b}{a}) &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3+3n &= n^3 \\ (\frac{a}{b})^3+(\frac{b}{a})^3 &= n^3-3n \\ &= (-1)^3-3(-1) \\ &= 2 \\ 2\sqrt{x} &= 2 \\ \sqrt{x} &= 1 \\ x &= 1 \\ \end{align} </math> </div></div> # berapa nilai m dari <math>x^2-mx-1=0</math> jika <math>\sqrt[3]{x_1}+\sqrt[3]{x_2}=1</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \sqrt[3]{x_1} &= a \\ x_1 &= a^3 \\ \sqrt[3]{x_2} &= b \\ x_2 &= b^3 \\ \sqrt[3]{x_1}+\sqrt[3]{x_2} &= 1 \\ a+b &= 1 \\ x^2-mx-1 &= 0 \\ x_1+x_2 &= m \\ x_1 \cdot x_2 &= -1 \\ x_1+x_2 &= m \\ a^3+b^3 &= m \\ x_1 \cdot x_2 &= -1 \\ a^3 \cdot b^3 &= -1 \\ (ab)^2 &= (-1)^3 \\ ab &= -1 \\ (a+b)^3 &= a^3+b^3+3ab(a+b) \\ (1)^3 &= m+3(-1)(1) \\ 1 &= m-3 \\ m &= 4 \\ \end{align} </math> </div></div> # berapa nilai <math>\frac{x_1}{x_2}</math> dari <math>ax^2-18x-b=0</math> jika <math>ab=45</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} ab &= 45 \\ b &= \frac{45}{a} \\ ax^2-18x-b &= 0 \\ ax^2-18x-\frac{45}{a} &= 0 \\ a^2x^2-18ax-45 &= 0 \\ (ax-3)(ax-15) &= 0 \\ ax-3 &= 0 \\ x &= \frac{3}{a} \\ ax-15 &= 0 \\ x &= \frac{15}{a} \\ \frac{x_1}{x_2} &= \frac{\frac{3}{a}}{\frac{15}{a}} \\ &= \frac{3}{15} \\ &= \frac{1}{5} \\ \frac{x_1}{x_2} &= \frac{\frac{15}{a}}{\frac{3}{a}} \\ &= \frac{15}{3} \\ &= 5 \\ \end{align} </math> </div></div> # Jika <math>\frac{u_3}{u_1+u_2} = \frac{7}{8}</math> merupakan barisan aritmetika maka berapa dari <math>\frac{u_2+u_3}{u_1}</math>? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{u_3}{u_1+u_2} &= \frac{7}{8} \\ \frac{a+2b}{a+a+b} &= \frac{7}{8} \\ \frac{a+2b}{2a+b} &= \frac{7}{8} \\ 8(a+2b) &= 7(2a+b) \\ 8a+16b &= 14a+7b \\ 9b &= 6a \\ b &= \frac{2a}{3} \\ \frac{u_2+u_3}{u_1} &= \frac{a+b+a+2b}{a} \\ &= \frac{2a+3b}{a} \\ &= \frac{2a+3(\frac{2a}{3})}{a} \\ &= \frac{2a+2a}{a} \\ &= \frac{4a}{a} \\ &= 4 \\ \end{align} </math> </div></div> # Jika 2p+q, 7p+q, 17p+q membentuk barisan geometri maka berapa rasionya? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \frac{7p+q}{2p+q} &= \frac{17p+q}{7p+q} \\ (7p+q)^2 &= (17p+q)(2p+q) \\ 49p^2+14pq+q^2 &= 34p^2+19pq+q^2 \\ 15p^2 &= 5pq \\ 3p &= q \\ \frac{7p+q}{2p+q} &= \frac{7p+3p}{2p+3p} \\ &= \frac{10p}{5p} \\ &= 2 \\ \end{align} </math> </div></div> # Rataan geometris a dan b adalah kurangnya 24 dari b serta rataan aritmatik a dan b adalah lebihnya 15 dari a maka berapa nilai a+b? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{rataan geometris } \\ \sqrt{a \cdot b} &= b-24 \\ a \cdot b &= (b-24)^2 \\ \text{rataan aritmatik } \\ \frac{a+b}{2} &= a+15 \\ a+b &= 2(a+15) \\ a+b &= 2a+30 \\ a &= b-30 \\ a \cdot b &= (b-24)^2 \\ (b-30)b &= (b-24)^2 \\ b^2-30b &= b^2-48b+576 \\ 18b &= 576 \\ b &= 32 \\ a &= b-30 \\ &= 32-30 \\ &= 2 \\ a+b &= 32+2 \\ &= 34 \\ \end{align} </math> </div></div> # Segitiga lancip ABC dengan <math>\frac{a^4+b^4+c^4+a^2b^2}{c^2(a^2+b^2)}=2</math>. tentukan nilai sudut C? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{syarat segitiga lancip semua sudut masing-masing kurang dari } 90^\circ \\ c^2 &= a^2+b^2-2ab cos C \\ cos C &= \frac{a^2+b^2-c^2}{2ab} \\ a^4+b^4+c^4+a^2b^2 &= 2c^2(a^2+b^2) \\ a^4+b^4+a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-a^2b^2+c^4 &= 2c^2(a^2+b^2) \\ (a^2+b^2)^2-2c^2(a^2+b^2)+(c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= a^2b^2 \\ (a^2+b^2-c^2)^2 &= (ab)^2 \\ a^2+b^2-c^2 &= \pm ab \\ cos C &= \pm \frac{ab}{2ab} \\ &= \pm \frac{1}{2} \\ &= \frac{1}{2} \text{ (karena sudut harus kurang dari } 90^\circ) \\ C &= 60^\circ \\ \end{align} </math> </div></div> # Segitiga siku-siku CAB titik D diantara C dan A dan titik E diantara B dan A. Panjang CD adalah 9 cm, panjang BE 5 cm serta panjang DA = EA. Berapakah panjang BC jika luasnya 45 cm²? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{misalkan panjang DA dan EA } = x \text{ dan panjang AB } = y \\ \text{luas segitiga CAB } &= \frac{CA \cdot AB}{2} \\ 45 &= \frac{(x+9)(x+5)}{2} \\ 90 &= x^2+14x+45 \\ x^2+14x &= 45 \\ y^2 &= (x+9)^2+(x+5)^2 \\ &= x^2+18x+81+x^2+10x+25 \\ &= 2x^2+28x+106 \\ &= 2(x^2+14x)+106 \\ &= 2(45)+106 \\ &= 196 \\ y &= 14 \\ \end{align} </math> jadi panjang BC adalah 14 cm </div></div> # Persegi panjang ABCD memiliki AD 15 cm dan DC 12 cm. E dan F merupakan perpanjangan DC yaitu CE 6 cm serta EF = DC. G merupakan titik potong antara BC dan AE maka berapa luas daerah BFEG? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \text{kita cari ukuran GC yaitu } \\ \frac{GC}{AD} &= \frac{CE}{DE} \\ \frac{GC}{15} &= \frac{6}{18} \\ GC &= 5 \\ \text{luas BEFG = luas segitiga BFC - luas segitiga GEC } \\ &= \frac{1}{2} \cdot BC \cdot CF - \frac{1}{2} \cdot GC \cdot CE \\ &= \frac{1}{2} \cdot 15 \cdot 18 - \frac{1}{2} \cdot 5 \cdot 6 \\ &= 135 - 15 \\ &= 120 \\ \end{align} </math> jadi luas daerah BFEG adalah 120 cm² </div></div> # Dua buah persegi masing-masing yaitu ABCD dan EFGH. persegi ABCD berhimpit dengan EFGH. I terletak antara A dengan F. Sisi persegi ABCD 4 cm dan EFGH 6 cm. Perbandingan AI:AF adalah 1:5 maka berapa luas daerah segitiga IGD? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ AI &= \frac{1}{5} AF \\ &= \frac{1}{5} 10 \\ &= 2 \\ IF &= AF-AI \\ &= 10-2 \\ &= 8 \\ \text{luas trapesium AFGD } &= \frac{(AD+EF) \cdot AF}{2} \\ &= \frac{(4+6)10}{2} \\ &= 50 \\ \text{luas segitiga AID } &= \frac{AI \cdot AF}{2} \\ &= \frac{(2)4}{2} \\ &= 4 \\ \text{luas segitiga IFG } &= \frac{IF \cdot FG}{2} \\ &= \frac{(8)6}{2} \\ &= 24 \\ \text{luas daerah segitiga IGD } &= \text{luas trapesium AFGD-luas segitiga AI—luas segitiga IFG } \\ &= 50-4-24 \\ &= 22 \\ \end{align} </math> jadi luas daerah segitiga IGD adalah 22 cm² </div></div> # Sebuah balok tertutup memiliki alas yang berbentuk persegi dengan tinggi 12 cm. Di dalam balok terdapat kerucut yang alasnya menempel serta titik tinggi tepat di atas baloknya dimana tingginya sama dengan tinggi balok. Volume antara luar kerucut dan dalam balok adalah 100(3-<math>\pi</math>) cm³ maka berapa luas permukaan kerucut tersebut? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} \\ \text{volume balok} \\ V_b &= x^2(12) \\ \text{volume kerucut} \\ V_b &= \frac{1}{3}\pi x^2(12) \\ &= 4\pi x^2 \\ V_{b-k} &= Vb-Vk \\ 100(3-\pi) &= 12x^2-4\pi x^2 \\ 100(3-\pi) &= 4x^2(3-\pi) \\ x^2 &= 25 \\ x &= 5 \\ s &= \sqrt{12^2+5^2} \\ &= \sqrt{144+25} \\ &= \sqrt{169} \\ &= 13 \\ \text{luas permukaan kerucut } &= \pi r(r+s) \\ &= \pi(5)(5+13) \\ &= 90\pi \\ \end{align} </math> jadi luas daerah permukaan kerucut adalah 90<math>\pi</math> cm² </div></div> # Suatu bilangan bulat positif A dan B masing-masing dibagi 3 bersisa 1 dan 2 maka berapa sisa pembagian A(A+1)+3B dibagi 9? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 3a+1 \\ B &= 3b+2 \\ A(A+1)+3B \\ (3a+1)(3a+1+1)+3(3b+2) \\ (3a+1)(3a+2)+9b+6 \\ 9a^2+9a+2+9b+6 \\ 9a^2+9a+9b+8 \\ 9(a^2+a+b)+8 \\ \text{sisa pembagiannya adalah } 8 \\ \end{align} </math> </div></div> # Suatu bilangan bulat positif A dan B masing-masing dibagi 9 bersisa 7 dan 8 maka berapa sisa pembagian A(A-5)+9B dibagi 81? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A &= 9a+7 \\ B &= 9b+8 \\ A(A-5)+9B \\ (9a+7)(9a+7-5)+9(9b+8) \\ (9a+7)(9a+2)+81b+72 \\ 81a^2+81a+14+81b+72 \\ 81a^2+81a+81b+86 \\ 81a^2+81a+81b+81+5 \\ 81(a^2+a+b+1)+5 \\ \text{sisa pembagiannya adalah } 5 \\ \end{align} </math> </div></div> # Jika <math>\begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix}</math> maka berapa hasil dari A²¹+A²⁵+A⁴⁶? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} A^2 &= A \cdot A \\ &= \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \\ A^3 &= A^2 \cdot A \\ &= \begin{bmatrix} 2 & 7 \\ -1 & -3 \\ \end{bmatrix} \cdot \begin{bmatrix} 3 & 7 \\ -1 & -2 \\ \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ &= - \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= -I \\ A^{21}+A^{25}+A^{46} &= A^{21} \cdot (I+A^4+A^{25}) \\ &= A^{21} \cdot (I+A^3 \cdot A +A^{24} \cdot A) \\ &= (A^3)^7 \cdot (I+A^3 \cdot A +(A^3)^8 \cdot A) \\ &= (-I)^7 \cdot (I-I \cdot A +(-I)^8 \cdot A) \\ &= -I \cdot (I-A+A) \\ &= -I \cdot I \\ &= -I \\ &= -\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\ \end{align} </math> </div></div> # Ida menuliskan 8 buah bilangan bulat positif berbeda yang kurang dari 16 sehingga tidak ada jumlah 2 bilangan dari 8 bilangan yang jumlahnya 16. Bilangan berapa yang pasti ditulis Ida? : bilangan yang kurang dari 16 yaitu 1,2,3,4,5,6, … , 15 : ditulis 7 buah bilangan berbeda yang jumlahnya 8 yaitu (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9). : ditulis 8 buah bilangan sama yang jumlahnya 8 yaitu (8,8) : maka Ida menulis bilangan 8. # Berapa banyaknya bilangan lima digit 743ab habis dibagi 5 dan 9? : Perhatikan angka terakhir pasti 0 atau 5 karena dibagi 5 dulu. : untuk 0 yaitu 743a0 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74340 saja. : untuk 5 yaitu 743a5 maka aturannya habis dibagi 9 yaitu semua jumlah angka-angka harus dibagi 9. Jadi hanya berarti 74385 saja. : Jadi banyaknya bilangan mungkin 2. # Buktikan bahwa 8ⁿ dibagi 7 hasil sisa selalu 1 untuk semua n adalah bilangan asli! ;cara 1 # 8¹ = 1 # 8² = 1 (8²=8¹x8¹ sama dengan 1x1) # 8³ = 1 (8³=8¹x8² sama dengan 1x1) # 8⁴ = 1 (8⁴=8¹x8³ sama dengan 1x1 atau 8⁴=(8²)² sama dengan 1^2) # 8⁵ = 1 # 8ⁿ = 1 (semua n untuk bilangan asli) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli ;cara 2 # 8ⁿ = b mod 7 # 8¹ = 1 mod 7 (cari hasil 1 sebagai hasil terendah dimana 8¹ dianggap pangkat terkecil) # (8¹)ⁿ = 1ⁿ mod 7 (pangkat n kedua ruasnya) # 8ⁿ = 1ⁿ mod 7 # 8ⁿ = 1 mod 7 (berapapun pangkatnya dimana 1 hasilnya 1) Terbukti 8ⁿ dibagi 7 pasti bersisa 1 untuk semua n adalah bilangan asli # Berapa hasil sisa dari 17⁹⁹ dibagi 5? ;cara 1 # 1 & 6 = sisa 1, 2 & 7 = sisa 2, 3 & 8 = sisa 3, 4 & 9 = sisa 4 serta 5 = sisa 0 # 7¹ = 7 (sisa 1) # 7² = 49 (sisa 2) # 7³ = 343 (sisa 3) # 7⁴ = 2,401 (sisa 0) # 7⁵ = 16,807 # 7⁶ = 117,649 nah 99 : 4 hasilnya 24 sisa 3 jadi 3 itu 343 lalu 343 dibagi 5 bersisa 3 ;cara 2 :17¹ = 2 :17² = 4 :17³ = 3 :17⁴ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 4 x 24 + 3 :17⁹⁹ = (17⁴)²⁴ x 17³ Untuk 17⁴ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 3. Jadi 17⁹⁹ dibagi 7 bersisa 3 ;cara 3 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 5, yaitu 17⁴ ::17⁴ = 1 mod 5 ::(17⁴)²⁴ = 1²⁴ mod 5 ::17⁹⁶ = 1²⁴ mod 5 ::17⁹⁶ = 1 mod 5 ::17⁹⁶ x 17³ = 1 x 17³ mod 5 ::17⁹⁹ = 17³ mod 5 ::17⁹⁹ = 17 x 17 x 17 mod 5 ::17⁹⁹ = 2 x 2 x 2 mod 5 ::17⁹⁹ = 8 mod 5 ::17⁹⁹ = 3 mod 5 Jadi 17⁹⁹ dibagi 5 bersisa 3 # Berapa hasil sisa dari 17⁹⁹ dibagi 7? ;cara 1 :17¹ = 3 :17² = 2 :17³ = 6 :17⁴ = 4 :17⁵ = 5 :17⁶ = 1 (sampai disini karena pangkat selanjutnya yang menghasilkan angka berulang dari semula diatas) Bahwa 99 = 6 x 16 + 3 :17⁹⁹ = (17⁶)¹⁶ x 17³ Untuk 17⁶ hasilnya 1 jadi berapapun pangkat bilangan asli pasti tetap 1. sisa 17⁹⁹ dibagi 7 sama dengan sisa 17³ dibagi 7 yaitu 6. Jadi 17⁹⁹ dibagi 7 bersisa 6 ;cara 2 :Mulailah dari bilangan terkecil diatas yang bersisa 1 yang dibagi 7, yaitu 17⁶ ::17⁶ = 1 mod 7 ::(17⁶)¹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1¹⁶ mod 7 ::17⁹⁶ = 1 mod 7 ::17⁹⁶ x 17³ = 1 x 17³ mod 7 ::17⁹⁹ = 17³ mod 7 ::17⁹⁹ = 17 x 17 x 17 mod 7 ::17⁹⁹ = 3 x 3 x 3 mod 7 ::17⁹⁹ = 27 mod 7 ::17⁹⁹ = 6 mod 7 Jadi 17⁹⁹ dibagi 7 bersisa 6 # Berapa hasil sisa dari 41²⁰²⁴ dibagi 33? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 41^{2024} &= 41^{2024} \text{ mod } 33 \\ &= (33 \times 3 + 2)^{2024} \text{ mod } 33 \\ &= 2^{2024} \text{ mod } 33 \\ &= 2^{2020} 2^4 \text{ mod } 33 \\ &= (2^5)^{404} 2^4 \text{ mod } 33 \\ &= (33 - 1)^{404} 2^4 \text{ mod } 33 \\ &= (-1)^{404} 2^4 \text{ mod } 33 \\ &= 2^4 \text{ mod } 33 \\ &= 16 \text{ mod } 33 \\ \text{Jadi hasil sisa adalah } 16 \\ \end{align} </math> </div></div> # Berapa nilai bilangan n terbesar sehingga 243ⁿ membagi 99⁹⁹? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 99^{99} &= (3^2 \times 11)^{99} \\ &= 3^{198} \times 11^{99} \\ 243^n &= (3^5)^n \\ &= 3^{5n} \\ \text{agar bisa membagi, maka} \\ 5n &= 198 \\ n &= 39.6 \\ \text{jadi bilangan n terbesar adalah } 39 \\ \end{align} </math> </div></div> # Berapa nilai bilangan n terbesar sehingga 512ⁿ membagi 88⁸⁸? <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} 88^{88} &= (8 \times 11)^{88} \\ &= 8^{88} \times 11^{88} \\ &= 8^{87} \times 8 \times 11^{88} \\ &= (8^3)^{29} \times 8 \times 11^{88} \\ &= 512^{29} \times 8 \times 11^{88} \\ 512^n &= 512^{29} \\ \text{jadi bilangan n terbesar adalah } 29 \\ \end{align} </math> </div></div> # Tentukan bilangan bulat positif terkecil jika dibagi 3 bersisa 1, jika dibagi 5 bersisa 2 dan jika dibagi dengan 7 bersisa 6! ; Cara 1 : KPK dari 3,5 dan 7 adalah 105. Misalkan N adalah bilangan bulat positif jadi N < 105. : N dibagi 3 sisa 1 : N dibagi 5 sisa 2 : N dibagi 7 sisa 6 FPB dari 3,5 dan 7 adalah 1 maka cari bilangan KPK dari b dan c bersisa 1 dibagi a : KPK 5 dan 7 (35,70,105,dst) dibagi 3 sisa 1 yaitu 70 : KPK 3 dan 7 (21,42,63,dst) dibagi 5 sisa 1 yaitu 21 : KPK 3 dan 5 (15,30,45,dst) dibagi 7 sisa 1 yaitu 15 Jadi N = 1 x 70 + 2 x 21 + 6 x 15 = 202 tetapi diminta bilangan bulat terkecil jadi 202-105=97 ; Cara 2 : Carilah 2 bilangan pembagi terbesar yaitu 5 dan 7 kemudian KPK dari 5 dan 7 adalah 35 : kemudian ditambahkan sisa masing-masing sesuai dengan KPK. : KPK 3 bersisa 1: 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, <b>97</b> : KPK 5 bersisa 2: 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, <b>97</b> : KPK 7 bersisa 6: 41, 48, 55, 62, 69, 76, 83, 90, <b>97</b> Jadi bilangan bulat positif adalah 97 :: NB: kalau ditanyakan bilangan bulat tiga digit maka menjawabnya 202 # Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 1 liter untuk satu ember? ; Cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |- | 4 || 0 || Semua isi ember B dibuang |- | 1 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. ; Cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |} nah ada ember A berisi 1 liter. # Ada dua ember berisi 5 liter dan 3 liter. Tanpa menggunakan alat-alat lain bagaimana mengisi 4 liter untuk satu ember? ; Cara 1 {| class="wikitable" |+ |- ! Ember A (5 l) !! Ember B (3 l) !! Keterangan |- | 5 || 0 || Isikan 5 l ke ember A |- | 2 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A tersisa 2 |- | 2 || 0 || Semua isi ember B dibuang |- | 0 || 2 || Tuangkan sisa ember A ke B |- | 5 || 2 || Isikan 5 l ke ember A |- | 4 || 3 || Tuangkan 1 l dari ember A ke B sehingga ember A tersisa 4 |} nah ada ember A berisi 4 liter. ; Cara 2 {| class="wikitable" |+ |- ! Ember A (3 l) !! Ember B (5 l) !! Keterangan |- | 3 || 0 || Isikan 3 l ke ember A |- | 0 || 3 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |- | 3 || 3 || Isikan 3 l ke ember A |- | 1 || 5 || Tuangkan 2 l dari ember A ke B sehingga ember A tersisa 1 |- | 1 || 0 || Semua isi ember B dibuang |- | 0 || 1 || Tuangkan 1 l dari ember A ke B sehingga ember A kosong |- | 3 || 1 || Isikan 3 l ke ember A |- | 0 || 4 || Tuangkan 3 l dari ember A ke B sehingga ember A kosong |} nah ada ember B berisi 4 liter. [[Kategori:Soal-Soal Matematika]] 378lm0mabglotszrpd38gmwb0km8ruz 0 25843 115033 112439 2026-04-27T06:21:05Z ~2026-25546-23 43043 115033 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{\frac{x_1 \dot x_2 \dot x_3 + \cdots \cdot x_n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean Harmonik (HM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{log x_i}{n}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] jvr40nyedrh92fuofm3denhw9hst003 115034 115033 2026-04-27T06:22:55Z ~2026-25546-23 43043 /* Jenis mean */ 115034 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{\frac{x_1 \dot x_2 \dot x_3 + \cdots \cdot x_n} = \sum\limits_{i=0}^{n}\frac{log x_i}{n}</math> ; Mean Harmonik (HM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{log x_i}{n}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] 0iqhqy6jibi305l93p9b3cgs706f9q3 115035 115034 2026-04-27T06:23:52Z ~2026-25546-23 43043 /* Jenis mean */ 115035 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{\frac{x_1 \cdot x_2 \cdot x_3 + \cdots \cdot x_n} = \sum\limits_{i=0}^{n}\frac{log x_i}{n}</math> ; Mean Harmonik (HM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{log x_i}{n}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] jxk8mly64vn61zbp47kglsjz2gfkekw 115036 115035 2026-04-27T06:25:56Z ~2026-25546-23 43043 /* Jenis mean */ 115036 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{x_1 \cdot x_2 \cdot x_3 + \cdots \cdot x_n} = \sum\limits_{i=0}^{n}\frac{log \, x_i}{n}</math> ; Mean Harmonik (HM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{log x_i}{n}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] btt78dybm7pfde79g3slysvf09p2883 115037 115036 2026-04-27T06:26:28Z ~2026-25546-23 43043 /* Jenis mean */ 115037 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{x_1 \cdot x_2 \cdot x_3 \cdot \cdots \cdot x_n} = \sum\limits_{i=0}^{n}\frac{log \, x_i}{n}</math> ; Mean Harmonik (HM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{log x_i}{n}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] qqxi6wjnfcoe3i8t0mqg6niljte4h18 115038 115037 2026-04-27T06:34:06Z ~2026-25546-23 43043 /* Jenis mean */ 115038 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{x_1 \cdot x_2 \cdot x_3 \cdot \cdots \cdot x_n} = \Pi\limits_{i=0}^{n} (x_i)^{\frac{1}{n}}</math> : <math display="block">log \, GM = \sum\limits_{i=0}^{n}\frac{log \, x_i}{n}</math> ; Mean Harmonik (HM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{log x_i}{n}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] 32p8xqm32yggno8mp9an0jo64kbibt8 115039 115038 2026-04-27T06:35:42Z ~2026-25546-23 43043 /* Jenis mean */ 115039 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{x_1 \cdot x_2 \cdot x_3 \cdot \cdots \cdot x_n} = \sum\limits_{i=0}^{n} (x_i)^{\frac{1}{n}}</math> : <math display="block">log \, GM = \sum\limits_{i=0}^{n}\frac{log \, x_i}{n}</math> ; Mean Harmonik (HM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{log x_i}{n}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] kiqcgpaycs2xx2icg2if8fr5ax21ixn 115040 115039 2026-04-27T06:36:56Z ~2026-25546-23 43043 /* Jenis mean */ 115040 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{x_1 \cdot x_2 \cdot x_3 \cdot \cdots \cdot x_n} = \Phi\limits_{i=0}^{n} (x_i)^{\frac{1}{n}}</math> : <math display="block">log \, GM = \sum\limits_{i=0}^{n}\frac{log \, x_i}{n}</math> ; Mean Harmonik (HM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{log x_i}{n}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] n6tivhrrfvrctb21j89z87x0vqzm0cu 115041 115040 2026-04-27T06:37:10Z ~2026-25546-23 43043 /* Jenis mean */ 115041 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{x_1 \cdot x_2 \cdot x_3 \cdot \cdots \cdot x_n} = \Pi\limits_{i=0}^{n} (x_i)^{\frac{1}{n}}</math> : <math display="block">log \, GM = \sum\limits_{i=0}^{n}\frac{log \, x_i}{n}</math> ; Mean Harmonik (HM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{log x_i}{n}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] 32p8xqm32yggno8mp9an0jo64kbibt8 115042 115041 2026-04-27T06:37:40Z ~2026-25546-23 43043 /* Jenis mean */ 115042 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{x_1 \cdot x_2 \cdot x_3 \cdot \cdots \cdot x_n} = \pi\limits_{i=0}^{n} (x_i)^{\frac{1}{n}}</math> : <math display="block">log \, GM = \sum\limits_{i=0}^{n}\frac{log \, x_i}{n}</math> ; Mean Harmonik (HM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{log x_i}{n}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] a9rnmnjwd97dommt3bt6zqumweinkor 115043 115042 2026-04-27T06:38:39Z ~2026-25546-23 43043 /* Jenis mean */ 115043 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{x_1 \cdot x_2 \cdot x_3 \cdot \cdots \cdot x_n} = \Phi\limits_{i=0}^{n} (x_i)^{\frac{1}{n}}</math> : <math display="block">log \, GM = \sum\limits_{i=0}^{n}\frac{log \, x_i}{n}</math> <math>\Pi</math> ; Mean Harmonik (HM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{log x_i}{n}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] mdlmyqkv8yn023tdwnzgl5wvbnmihh5 115044 115043 2026-04-27T06:39:19Z ~2026-25546-23 43043 /* Jenis mean */ 115044 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{x_1 \cdot x_2 \cdot x_3 \cdot \cdots \cdot x_n} = \Pi\limits_{i=0}^{n} (x_i)^{\frac{1}{n}}</math> : <math display="block">log \, GM = \sum\limits_{i=0}^{n}\frac{log \, x_i}{n}</math> ; Mean Harmonik (HM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{log x_i}{n}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] 32p8xqm32yggno8mp9an0jo64kbibt8 115045 115044 2026-04-27T06:43:15Z ~2026-25546-23 43043 /* Jenis mean */ 115045 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{x_1 \cdot x_2 \cdot x_3 \cdot \cdots \cdot x_n} = \Pi\limits_{i=0}^{n} (x_i)^{\frac{1}{n}}</math> : <math display="block">log \, GM = \sum\limits_{i=0}^{n}\frac{log \, x_i}{n}</math> ; Mean Harmonik (HM) : <math display="block">\bar{x} = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} + \cdots + \frac{1}{x_n}} = \frac{n}{\sum\limits_{i=0}^{n}\frac{1}{x_i}}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] lyrz0odo1ixns03akmvl366496zgju4 115046 115045 2026-04-27T06:44:10Z ~2026-25546-23 43043 /* Jenis mean */ 115046 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{x_1 \cdot x_2 \cdot x_3 \cdot \cdots \cdot x_n} = \Pi\limits_{i=0}^{n} (x_i)^{\frac{1}{n}}</math> : <math display="block">log \, GM = \sum\limits_{i=0}^{n}\frac{log \, x_i}{n}</math> ; Mean Harmonik (HM) : <math display="block">HM = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} + \cdots + \frac{1}{x_n}} = \frac{n}{\sum\limits_{i=0}^{n}\frac{1}{x_i}}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] owcmkgjaj33rx6qemaulges1xfqge1z 115047 115046 2026-04-27T06:45:05Z ~2026-25546-23 43043 /* Jenis mean */ 115047 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{x_1 \cdot x_2 \cdot x_3 \cdot \cdots \cdot x_n} = \Pi\limits_{i=0}^{n} x_i^{\frac{1}{n}}</math> : <math display="block">log \, GM = \sum\limits_{i=0}^{n}\frac{log \, x_i}{n}</math> ; Mean Harmonik (HM) : <math display="block">HM = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} + \cdots + \frac{1}{x_n}} = \frac{n}{\sum\limits_{i=0}^{n}\frac{1}{x_i}}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] 6jonk7dxsby1w6gcu54tdyi0s3o2o1z 115048 115047 2026-04-27T08:55:49Z ~2026-25546-23 43043 /* Jenis mean */ 115048 wikitext text/x-wiki == Distribusi == Ada tujuh jenis distribusi yaitu: # Distribusi seragam ## Fungsi: f(x,k) = <math>\frac{1}{k}</math> untuk x=x₁,x₂,x₃, ……,x_k # Distribusi Bernoulli ## Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1 ## Nilai harapan: E(x)=p # Distribusi binomial ## Fungsi: f(x;n,p) = (<math>^n_x</math>) p^xq^n-x untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np contoh soal: # Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{2}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \frac{1}{16} \\ &= \frac{4 \cdot 3 \cdot 2!}{2! \cdot 2 \cdot 1} \frac{1}{16} \\ &= 6 \frac{1}{16} \\ &= \frac{3}{8} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah koin muncul: 2⁴ = 16 : muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6 : P(x=2G) = <math>\frac{n(A)}{n(S)}=\frac{6}{16}=\frac{3}{8}</math> # Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang: ## mata dadu 1 muncul satu kali? ## kelipatan 3 muncul dua kali? ## mata dadu 3 muncul paling sedikit tiga kali? ## mata dadu 4 atau 6 muncul empat kali? ## ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} * n = 4, x = 1, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(1,4,\frac{1}{6}) &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^{4-1} \\ &= (^4_1) (\frac{1}{6})^1 \cdot (\frac{5}{6})^3 \\ &= \frac{4!}{1!(4-1)!} \frac{1}{6} \cdot \frac{125}{216} \\ &= \frac{4!}{1!3!} \cdot \frac{125}{1296} \\ &= 4 \cdot \frac{125}{1296} \\ &= \frac{125}{324} \\ * n = 4, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{4-2} \\ &= (^4_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{9} \cdot \frac{4}{9} \\ &= \frac{4!}{2!2!} \cdot \frac{4}{81} \\ &= 6 \cdot \frac{4}{81} \\ &= \frac{24}{81} \\ * \text{muncul 3 kali } \\ n = 4, x = 3, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{1}{6}) &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^{4-3} \\ &= (^4_3) (\frac{1}{6})^3 \cdot (\frac{5}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{1}{216} \cdot \frac{5}{6} \\ &= \frac{4!}{3!1!} \cdot \frac{5}{1296} \\ &= 4 \cdot \frac{5}{1296} \\ &= \frac{20}{2196} \\ \text{muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{6}, q = \frac{5}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{6}) &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4} \\ &= (^4_4) (\frac{1}{6})^4 \cdot (\frac{5}{6})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{1296} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{1296} \\ &= \frac{1}{1296} \\ \text{total } = \frac{20}{1296}+\frac{1}{1296} = \frac{21}{1296} \\ * n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ * \text{ganjil muncul 2 kali } \\ n = 4, x = 2, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,4,\frac{1}{3}) &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^{4-2} \\ &= (^4_2) (\frac{1}{2})^2 \cdot (\frac{1}{2})^2 \\ &= \frac{4!}{2!(4-2)!} \frac{1}{4} \cdot \frac{1}{4} \\ &= \frac{4!}{2!2!} \cdot \frac{1}{16} \\ &= 6 \cdot \frac{1}{16} \\ &= \frac{3}{8} \\ \text{mata dadu 2 atau 5 muncul 4 kali } \\ n = 4, x = 4, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,4,\frac{1}{3}) &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^{4-4} \\ &= (^4_4) (\frac{1}{3})^4 \cdot (\frac{2}{3})^0 \\ &= \frac{4!}{4!(4-4)!} \frac{1}{81} \\ &= \frac{4!}{4!0!} \cdot \frac{1}{81} \\ &= \frac{1}{81} \\ \text{total } = \frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648} \\ \end{align} </math> </div></div> * cara 2 : mata dadu 1 muncul 1 kali : p=1/6 dan q=5/6 : beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324 : P(x=1) = <math>\frac{n(A)}{n(S)}=\frac{125}{324}</math> : kelipatan 3 muncul 2 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81 : P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{24}{81}</math> : mata dadu 3 muncul paling sedikit 3 kali : muncul paling sedikit 3 kali berarti 3 kali atau 4 kali : p=1/6 dan q=5/6 : untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324 : P(x=3) = <math>\frac{n(A)}{n(S)}=\frac{20}{1296}</math> : untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{1296}</math> : jadi P(x>=3) = P(x=3)+P(x=4) = <math>\frac{20}{1296}+\frac{1}{1296}=\frac{21}{1296}</math> : mata dadu 4 atau 6 muncul 4 kali : p=1/3 dan q=2/3 : beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 : P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> : ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali :: ganjil muncul 2 kali :: p=1/2 dan q=1/2 :: beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8 :: P(x=2) = <math>\frac{n(A)}{n(S)}=\frac{3}{8}</math> :: mata dadu 2 atau 5 muncul 4 kali :: p=1/3 dan q=2/3 :: beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81 :: P(x=4) = <math>\frac{n(A)}{n(S)}=\frac{1}{81}</math> :: irisan untuk pernyataan independen: P(x=2) . P(x=4) = <math>\frac{3}{8} \cdot \frac{1}{81} = \frac{3}{648}</math> # Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 2, x = 2, p = \frac{3}{6}, q = \frac{3}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,2,\frac{3}{6}) &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^{2-2} \\ &= (^2_2) (\frac{3}{6})^2 \cdot (\frac{3}{6})^0 \\ &= \frac{2!}{2!(2-2)!} \frac{3^2}{6^2} \\ &= \frac{2!}{2!0!} \frac{9}{36} \\ &= \frac{9}{36} \\ &= \frac{1}{4} \\ \end{align} </math> </div></div> * cara 2 : jumlah seluruh sebuah dadu muncul: 6² = 36 : muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9 : P(x=2P) = <math>\frac{n(A)}{n(S)}=\frac{9}{36}=\frac{1}{4}</math> # Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 5, x = 3, p = \frac{1}{2}, q = \frac{1}{2} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,5,\frac{1}{2}) &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^{5-3} \\ &= (^5_3) (\frac{1}{2})^3 \cdot (\frac{1}{2})^2 \\ &= \frac{5!}{3!(5-3)!} \frac{1}{2^5} \\ &= \frac{5!}{3!2!} \frac{1}{32} \\ &= \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} \frac{1}{32} \\ &= 10 \frac{1}{32} \\ &= \frac{5}{16} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul lima anak: 2⁵ = 32 : muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10 : P(x=3L) = <math>\frac{n(A)}{n(S)}=\frac{10}{32}=\frac{5}{16}</math> # Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 4, x = 3, p = \frac{4}{6}, q = \frac{2}{6} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(3,4,\frac{4}{6}) &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^{4-3} \\ &= (^4_3) (\frac{4}{6})^3 \cdot (\frac{2}{6})^1 \\ &= \frac{4!}{3!(4-3)!} \frac{64}{216} \cdot \frac{2}{6} \\ &= \frac{4!}{3!1!} \frac{128}{1296} \\ &= \frac{4 \cdot 3!}{3! \cdot 1} \frac{128}{1296} \\ &= 4 \frac{128}{1296} \\ &= \frac{512}{1296} \\ \end{align} </math> </div></div> * cara 2 : jumlah muncul terambil empat buah: <math>C^6_4 = \frac{6!}{4!2!} = \frac{6 \cdot 5 \cdot 4!}{4! \cdot 2 \cdot 1} = 15</math> : muncul terambil 3 buah baik: <math>C^2_1 \cdot C^4_3 = \frac{2!}{1!1!} \cdot \frac{4!}{3!1!} = 2 \cdot 4 = 8</math> atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8 : P(x=3BB) = <math>\frac{n(A)}{n(S)}=\frac{8}{15}</math> # Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{4}{5}, q = \frac{1}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{4}{5}) &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{3}{5})^{3-2} \\ &= (^3_2) (\frac{4}{5})^2 \cdot (\frac{1}{5})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{16}{25} \cdot \frac{1}{5} \\ &= \frac{3 \cdot 2!}{2!} \frac{16}{125} \\ &= 3 \frac{16}{125} \\ &= \frac{48}{125} \\ \end{align} </math> </div></div> * cara 2 : p=4/5 dan q=1/5 : tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125 # Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran? * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 3, x = 2, p = \frac{1}{3}, q = \frac{2}{3} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(2,3,\frac{1}{3}) &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^{3-2} \\ &= (^3_2) (\frac{1}{3})^2 \cdot (\frac{2}{3})^1 \\ &= \frac{3!}{2!(3-2)!} \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{3 \cdot 2!}{2!} \frac{1}{9} \cdot \frac{2}{3} \\ &= 3 \frac{1}{9} \cdot \frac{2}{3} \\ &= \frac{2}{9} \\ \end{align} </math> </div></div> * cara 2 : p=1/3 dan q=2/3 : tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal) : jadi P(x=2) = P(MMG) + P(MGM) + P(GMM) : = 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9 # Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu! * cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} n = 12, x = 4, p = \frac{2}{5}, q = \frac{3}{5} \\ f(x;n,p) &= (^n_x) p^x \cdot q^{n-x} \\ f(4,12,\frac{2}{5}) &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^{12-4} \\ &= (^{12}_4) (\frac{2}{5})^4 \cdot (\frac{3}{5})^8 \\ &= \frac{12!}{4!(12-4)!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12!}{4!8!} \frac{2^4}{5^4} \cdot \frac{3^8}{5^8} \\ &= \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{4 \cdot 3 \cdot 2 \cdot 1 \cdot 8!} \frac{2^4 \cdot 3^8}{5^{12}} \\ &= 495 \frac{2^4 \cdot 3^8}{5^{12}} \\ \end{align} </math> </div></div> # Distribusi normal ## Fungsi: f(x;<math>\sigma,\mu</math>) = <math>\frac{1}{\sqrt{2\pi\sigma}}(\frac{x-\mu}{\sigma})^2</math> untuk <math>\infty<x<\infty</math> ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi Possion ## Fungsi: f(x;<math>\mu</math>) = <math>\frac{e^{-\mu} \mu^{x}}{x!}</math> untuk x=0,1,2,3,4, …. ## Nilai harapan: E(x)=<math>\mu</math> # Distribusi geometrik ## Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, …. ## Nilai harapan: E(x)=<math>\frac{1}{p}</math> # Distribusi hipergeometrik ## Fungsi: f(x;N,n,k) = <math>\frac{(^k_x)(^{N-k}_{n-x})}{^N_n}</math> untuk x=0,1,2,3,4, ….,n ## Nilai harapan: E(x)=np dimana p=<math>\frac{k}{N}</math> == Jenis mean == ; Mean aritmatik (AM) : <math display="block">\bar{x} = \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} = \sum\limits_{i=0}^{n}\frac{x_i}{n}</math> ; Mean geometrik (GM) : <math display="block">GM = \sqrt[n]{x_1 \cdot x_2 \cdot x_3 \cdot \cdots \cdot x_n} = \sum\limits_{i=0}^{n} x_i^{\frac{1}{n}}</math> : <math display="block">log \, GM = \sum\limits_{i=0}^{n}\frac{log \, x_i}{n}</math> ; Mean Harmonik (HM) : <math display="block">HM = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} + \cdots + \frac{1}{x_n}} = \frac{n}{\sum\limits_{i=0}^{n}\frac{1}{x_i}}</math> == Regresi linear == [[Kategori:Soal-Soal Matematika]] 84j2qia6r1w5srre62thy99vmbxzoeb 0 27087 115012 114710 2026-04-26T14:29:20Z FILLA1A 42749 Integritas nasional 115012 wikitext text/x-wiki '''Pasir Kemilu''' adalah sebuah desa yang terdapat di Kecamatan Rengat, Kabupaten Indragiri Hulu, Provinsi Riau. Nama kejadian: Banjir di Desa Pasir Kemilu Kapan terjadi:19 Januari 2024 Kronologi kejadian: Pada 19 Januari 2024, terjadi banjir di Desa Pasir Kemilu, Kecamatan Rengat, Kabupaten Indragiri Hulu, Riau. Banjir terjadi karena curah hujan yang tinggi sehingga Sungai Indragiri meluap dan air masuk ke permukiman warga. Beberapa rumah dan jalan desa terendam air sehingga aktivitas masyarakat terganggu. Pemerintah dan beberapa pihak kemudian memberikan bantuan kepada masyarakat yang terdampak banjir.  Keadaan masyarakat saat itu: Masyarakat merasa khawatir karena rumah dan lingkungan mereka terendam air. Namun warga tetap saling membantu dan bekerja sama untuk mengatasi dampak banjir. Apakah generasi sekarang mengetahui peristiwa tersebut: Ya, sebagian besar generasi sekarang mengetahui kejadian tersebut karena terjadi pada tahun 2024 dan banyak warga yang menyaksikan atau mengalaminya secara langsung. [[Kategori:Pendidikan Kewarganegaraan]] [[Kategori:Kebudayaan Melayu Indragiri]] Di desa ini sering juga terdengar istilah, "Numpang Laman" yang disebut Perilaku meminta izin menggunakan halaman atau tanah kosong milik orang lain untuk menjemur hasil panen (padi, kopi, atau pinang).Dan pemilik tanah memberikan izin secara cuma-cuma (pakai je lah) sebagai bentuk dukungan sesama petani. Imbalannya biasanya hanya sekadar pemberian sedikit hasil panen sebagai tanda terima kasih. Tanggapan pelaksanaan demokrasi di Desa pasir kemilu Pelaksanaan demokrasi di Desa Pasir Kemilu dapat dilihat dari kegiatan pemilihan kepala desa (Pilkades) dan juga pemilu seperti pemilihan presiden maupun legislatif. Salah satu kasus yang sering terjadi di desa ini adalah saat berlangsungnya Pilkades, di mana masyarakat berpartisipasi langsung dalam memilih pemimpin desa. Pada hari pemungutan suara, warga datang ke TPS yang telah disediakan. Suasana biasanya cukup ramai dan tertib. Panitia desa bersama aparat setempat mengawasi jalannya pemilihan agar berjalan aman dan lancar. Antusiasme masyarakat cukup tinggi karena mereka ingin menentukan pemimpin yang dianggap mampu membawa perubahan bagi desa. Namun, dalam pelaksanaannya terdapat fenomena yang nyata terjadi di masyarakat, yaitu adanya pengaruh hubungan kekeluargaan dan kedekatan sosial. Sebagian warga memilih calon karena masih memiliki hubungan saudara atau karena sudah saling mengenal dekat, bukan sepenuhnya karena program kerja yang ditawarkan. Selain itu, terdapat juga kasus di mana sebagian masyarakat kurang memahami visi dan misi calon kepala desa karena minimnya keikutsertaan dalam kegiatan sosialisasi atau kampanye. Hal ini menyebabkan pilihan yang diambil kurang berdasarkan pertimbangan yang matang. Meski demikian, setelah hasil pemilihan diumumkan, masyarakat Desa Pasir Kemilu umumnya dapat menerima hasil dengan baik. Tidak terjadi konflik besar antar pendukung calon. Warga tetap menjaga kerukunan dan kembali menjalankan aktivitas sehari-hari seperti biasa. Di masyarakat pedesaan, ketika seseorang ingin menanam tanaman seperti cabai, pisang, atau bunga, biasanya mereka meminta bibit kepada tetangga atau keluarga yang sudah memiliki tanaman tersebut. Permintaan dilakukan dengan sopan, dan biasanya bibit diberikan secara gratis sebagai bentuk kepedulian. Setelah tanaman tumbuh dan berbuah, orang yang menerima bibit sering membalas dengan memberikan sebagian hasil panen atau membantu saat dibutuhkan. Tradisi ini mencerminkan nilai kebersamaan, saling tolong-menolong, serta menjaga keberlanjutan kegiatan agribisnis secara sederhana dan alami. oewqs0wgnmkrr04k2ent6k5ctjn1est 115013 115012 2026-04-26T14:33:30Z FILLA1A 42749 Konstitusi di Indonesia 115013 wikitext text/x-wiki '''Pasir Kemilu''' adalah sebuah desa yang terdapat di Kecamatan Rengat, Kabupaten Indragiri Hulu, Provinsi Riau. Nama kejadian: Banjir di Desa Pasir Kemilu Kapan terjadi:19 Januari 2024 Kronologi kejadian: Pada 19 Januari 2024, terjadi banjir di Desa Pasir Kemilu, Kecamatan Rengat, Kabupaten Indragiri Hulu, Riau. Banjir terjadi karena curah hujan yang tinggi sehingga Sungai Indragiri meluap dan air masuk ke permukiman warga. Beberapa rumah dan jalan desa terendam air sehingga aktivitas masyarakat terganggu. Pemerintah dan beberapa pihak kemudian memberikan bantuan kepada masyarakat yang terdampak banjir.  Keadaan masyarakat saat itu: Masyarakat merasa khawatir karena rumah dan lingkungan mereka terendam air. Namun warga tetap saling membantu dan bekerja sama untuk mengatasi dampak banjir. Apakah generasi sekarang mengetahui peristiwa tersebut: Ya, sebagian besar generasi sekarang mengetahui kejadian tersebut karena terjadi pada tahun 2024 dan banyak warga yang menyaksikan atau mengalaminya secara langsung. [[Kategori:Pendidikan Kewarganegaraan]] [[Kategori:Kebudayaan Melayu Indragiri]] Di desa ini sering juga terdengar istilah, "Numpang Laman" yang disebut Perilaku meminta izin menggunakan halaman atau tanah kosong milik orang lain untuk menjemur hasil panen (padi, kopi, atau pinang).Dan pemilik tanah memberikan izin secara cuma-cuma (pakai je lah) sebagai bentuk dukungan sesama petani. Imbalannya biasanya hanya sekadar pemberian sedikit hasil panen sebagai tanda terima kasih. Tanggapan pelaksanaan demokrasi di Desa pasir kemilu Pelaksanaan demokrasi di Desa Pasir Kemilu dapat dilihat dari kegiatan pemilihan kepala desa (Pilkades) dan juga pemilu seperti pemilihan presiden maupun legislatif. Salah satu kasus yang sering terjadi di desa ini adalah saat berlangsungnya Pilkades, di mana masyarakat berpartisipasi langsung dalam memilih pemimpin desa. Pada hari pemungutan suara, warga datang ke TPS yang telah disediakan. Suasana biasanya cukup ramai dan tertib. Panitia desa bersama aparat setempat mengawasi jalannya pemilihan agar berjalan aman dan lancar. Antusiasme masyarakat cukup tinggi karena mereka ingin menentukan pemimpin yang dianggap mampu membawa perubahan bagi desa. Namun, dalam pelaksanaannya terdapat fenomena yang nyata terjadi di masyarakat, yaitu adanya pengaruh hubungan kekeluargaan dan kedekatan sosial. Sebagian warga memilih calon karena masih memiliki hubungan saudara atau karena sudah saling mengenal dekat, bukan sepenuhnya karena program kerja yang ditawarkan. Selain itu, terdapat juga kasus di mana sebagian masyarakat kurang memahami visi dan misi calon kepala desa karena minimnya keikutsertaan dalam kegiatan sosialisasi atau kampanye. Hal ini menyebabkan pilihan yang diambil kurang berdasarkan pertimbangan yang matang. Meski demikian, setelah hasil pemilihan diumumkan, masyarakat Desa Pasir Kemilu umumnya dapat menerima hasil dengan baik. Tidak terjadi konflik besar antar pendukung calon. Warga tetap menjaga kerukunan dan kembali menjalankan aktivitas sehari-hari seperti biasa. Di masyarakat pedesaan, ketika seseorang ingin menanam tanaman seperti cabai, pisang, atau bunga, biasanya mereka meminta bibit kepada tetangga atau keluarga yang sudah memiliki tanaman tersebut. Permintaan dilakukan dengan sopan, dan biasanya bibit diberikan secara gratis sebagai bentuk kepedulian. Setelah tanaman tumbuh dan berbuah, orang yang menerima bibit sering membalas dengan memberikan sebagian hasil panen atau membantu saat dibutuhkan. Tradisi ini mencerminkan nilai kebersamaan, saling tolong-menolong, serta menjaga keberlanjutan kegiatan agribisnis secara sederhana dan alami. Salah satu aturan tidak tertulis di Desa adanya larangan mengambil hasil kebun orang lain tanpa izin pemiliknya. Masyarakat sangat menjunjung tinggi kejujuran, sehingga siapa pun yang melanggar akan mendapat teguran sosial dan dianggap tidak menjaga kepercayaan bersama. Aturan ini bertujuan menjaga ketertiban, keamanan hasil pertanian, serta memperkuat nilai integritas dalam kehidupan masyarakat desa. shaeh3w8zgrn68gmb37w0nhinj2i7an 115014 115013 2026-04-26T14:38:47Z FILLA1A 42749 Kewajiban WNI 115014 wikitext text/x-wiki '''Pasir Kemilu''' adalah sebuah desa yang terdapat di Kecamatan Rengat, Kabupaten Indragiri Hulu, Provinsi Riau. Nama kejadian: Banjir di Desa Pasir Kemilu Kapan terjadi:19 Januari 2024 Kronologi kejadian: Pada 19 Januari 2024, terjadi banjir di Desa Pasir Kemilu, Kecamatan Rengat, Kabupaten Indragiri Hulu, Riau. Banjir terjadi karena curah hujan yang tinggi sehingga Sungai Indragiri meluap dan air masuk ke permukiman warga. Beberapa rumah dan jalan desa terendam air sehingga aktivitas masyarakat terganggu. Pemerintah dan beberapa pihak kemudian memberikan bantuan kepada masyarakat yang terdampak banjir.  Keadaan masyarakat saat itu: Masyarakat merasa khawatir karena rumah dan lingkungan mereka terendam air. Namun warga tetap saling membantu dan bekerja sama untuk mengatasi dampak banjir. Apakah generasi sekarang mengetahui peristiwa tersebut: Ya, sebagian besar generasi sekarang mengetahui kejadian tersebut karena terjadi pada tahun 2024 dan banyak warga yang menyaksikan atau mengalaminya secara langsung. [[Kategori:Pendidikan Kewarganegaraan]] [[Kategori:Kebudayaan Melayu Indragiri]] Di desa ini sering juga terdengar istilah, "Numpang Laman" yang disebut Perilaku meminta izin menggunakan halaman atau tanah kosong milik orang lain untuk menjemur hasil panen (padi, kopi, atau pinang).Dan pemilik tanah memberikan izin secara cuma-cuma (pakai je lah) sebagai bentuk dukungan sesama petani. Imbalannya biasanya hanya sekadar pemberian sedikit hasil panen sebagai tanda terima kasih. Tanggapan pelaksanaan demokrasi di Desa pasir kemilu Pelaksanaan demokrasi di Desa Pasir Kemilu dapat dilihat dari kegiatan pemilihan kepala desa (Pilkades) dan juga pemilu seperti pemilihan presiden maupun legislatif. Salah satu kasus yang sering terjadi di desa ini adalah saat berlangsungnya Pilkades, di mana masyarakat berpartisipasi langsung dalam memilih pemimpin desa. Pada hari pemungutan suara, warga datang ke TPS yang telah disediakan. Suasana biasanya cukup ramai dan tertib. Panitia desa bersama aparat setempat mengawasi jalannya pemilihan agar berjalan aman dan lancar. Antusiasme masyarakat cukup tinggi karena mereka ingin menentukan pemimpin yang dianggap mampu membawa perubahan bagi desa. Namun, dalam pelaksanaannya terdapat fenomena yang nyata terjadi di masyarakat, yaitu adanya pengaruh hubungan kekeluargaan dan kedekatan sosial. Sebagian warga memilih calon karena masih memiliki hubungan saudara atau karena sudah saling mengenal dekat, bukan sepenuhnya karena program kerja yang ditawarkan. Selain itu, terdapat juga kasus di mana sebagian masyarakat kurang memahami visi dan misi calon kepala desa karena minimnya keikutsertaan dalam kegiatan sosialisasi atau kampanye. Hal ini menyebabkan pilihan yang diambil kurang berdasarkan pertimbangan yang matang. Meski demikian, setelah hasil pemilihan diumumkan, masyarakat Desa Pasir Kemilu umumnya dapat menerima hasil dengan baik. Tidak terjadi konflik besar antar pendukung calon. Warga tetap menjaga kerukunan dan kembali menjalankan aktivitas sehari-hari seperti biasa. Di masyarakat pedesaan, ketika seseorang ingin menanam tanaman seperti cabai, pisang, atau bunga, biasanya mereka meminta bibit kepada tetangga atau keluarga yang sudah memiliki tanaman tersebut. Permintaan dilakukan dengan sopan, dan biasanya bibit diberikan secara gratis sebagai bentuk kepedulian. Setelah tanaman tumbuh dan berbuah, orang yang menerima bibit sering membalas dengan memberikan sebagian hasil panen atau membantu saat dibutuhkan. Tradisi ini mencerminkan nilai kebersamaan, saling tolong-menolong, serta menjaga keberlanjutan kegiatan agribisnis secara sederhana dan alami. Salah satu aturan tidak tertulis di Desa adanya larangan mengambil hasil kebun orang lain tanpa izin pemiliknya. Masyarakat sangat menjunjung tinggi kejujuran, sehingga siapa pun yang melanggar akan mendapat teguran sosial dan dianggap tidak menjaga kepercayaan bersama. Aturan ini bertujuan menjaga ketertiban, keamanan hasil pertanian, serta memperkuat nilai integritas dalam kehidupan masyarakat desa. Salah satu hal menarik di Desa ini terkait kehidupan sosial adalah tradisi rewang saat acara pernikahan, di mana warga secara sukarela membantu tuan rumah dalam menyiapkan makanan, perlengkapan, dan kebutuhan acara tanpa meminta imbalan. Tradisi ini mencerminkan nilai kebersamaan, gotong royong, serta tanggung jawab sosial masyarakat, sekaligus menunjukkan bagaimana warga menjalankan kewajiban untuk saling membantu dalam kehidupan bermasyarakat. ljaj495hu6n8w8m87rguh2bhjshxxjh