Wikibuku idwikibooks https://id.wikibooks.org/wiki/Halaman_Utama MediaWiki 1.45.0-wmf.8 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 0 5785 108055 108016 2025-07-03T02:42:05Z Alfarq 799 /* Daftar Obat (2) */ 108055 wikitext text/x-wiki {{Wikipedia|Farmasi}} == Istilah Latin == === Uc === * ℞ (racipe): Ambillah. * Supp (suppositorium): Supositoria, gentel. * No (nomero): Jumlah, sejumlah. * S (signa): Tanda, tandai, tandailah. * Uc (usus cognitus): Pemakaian diketahui. * Pro: Untuk. ℞ Miconazole 2% cream tube No. I S uc ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. C (20 tahun) ℞ Anusol supp No. V S uc ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. A (21 tahun) === Imm === * Imm (in manus medicine): Serahkan ke tangan paramedis (dokter, perawat, bidan, mantri). * Inf (infus): Infus. * Flab (flabot): Flabot. ℞ Natrium Chlorida 0,9% inf flab No. III Abbocath no. 22 No. I Cum infus set No. I S imm ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. B (22 tahun) === Prn === * Tab (tabletta, tabella): Tablet. * Mg (miligram): Miligram. * Prn (pro renatera), sns (si necesse sit): Jika perlu. Singkatan ini digunakan untuk obat-obat simtomatis (penghilang gejala). * Dd (de die): Tiap hari, setiap hari. * Aggr febr (agrediante febre): saat demam. ℞ Paracetamol tab mg 500 No. X S prn 1–3 dd tab I aggr febr ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. C (23 tahun) '''Perhatian:''' * Penulisan setelah Signa tanpa tanda kurung. Berikut adalah penulisan yang salah: ℞ Paracetamol tab mg 500 No. X S prn (1–3) dd tab I aggr febr ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. C (23 tahun) === Up === * Up (usus propius): Pemakaian sendiri (untuk dokter). Tidak perlu memakai Pro. ℞ Amoxycillin tab mg 500 No. X S up ––––––––––––––––––––––––––––––––––––––––––––––– ₰ === Dcf === * Dcf, dc form (da cun formula): Sesuai obat yang tertulis. * Sol, solut (solutio): Larutan. ℞ Alkohol 70% sol fl No. I S imm dcf ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. E (25 tahun) === Sine confectiones === * Cap (capsule): Kapsul. * Sine confectiones: Tanpa kemasan. ℞ Amoxsan cap mg 500 No. X (sine confectiones) S 3 dd cap I ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. F (26 tahun) === Adde === * Adde: Tambahkan. * Dry: Kering. * Syr (syrupus): Sirup. * Aq coct (aqua cocta): Air masak. * Ad: Sampai, hingga. * Cc (cubic centimetre): sentimeter kubik. * C (cochlea): sendok makan (15 cc). * Cth (cochlea tea): sendok teh (5 cc). ℞ Ampicillin dry syr fl No. I Adde aq coct ad cc 60 S 3 dd cth I ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. G (27 tahun) === Ad lib === * Ad lib, ad libit (ad libitum): Sesuka hati. ℞ Oralit sachet No. XX S ad lib ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. H (28 tahun) === Ac === * Ac (antec cibos, ante cibum, ante coenam): Sebelum makan. ℞ Antasida DOEN tab No. X S 3 dd tab I ac ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. I (29 tahun) === Dc === * Dc (durante coenam): Saat makan, ketika makan, selagi makan. ℞ Enzyplex tab No. XII S 3 dd tab I dc ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. J (30 tahun) === Pc === * Pc (post coenam): Setelah makan, sesudah makan, sehabis makan. ℞ Rifampicin tab mg 450 No. VII S 1 dd tab I pc ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. K (31 tahun) === Hora spatio === * Octa: Delapan. * Hora spatio: Selang sekian jam. * Octa hora spatio: Selang delapan jam. ℞ Chloramphenicol cap mg 500 No. X S 3 dd cap I octa hora spatio ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. L (32 tahun) === Agita ante sumendum === Agita ante sumendum: kocok sebelum digunakan R/ Mylanta Syr fl no I ʃ 3 dd CI ac agitatio ante sumundum ---- Pro: Tn J (29th) === Hora ieiune === Saat perut kosong === Sive simile === * Boleh diganti * Obat paten boleh diganti dengan obat paten lain dengan kandungan, bentuk, sediaan, dan dosis yang sama === Mds === === use know === === Da in caps === dimasukkan ke dalam kapsul === Cum === === Cito, Urgent, PIM (Periculum In Mora) === Segera, Penting, Berbahaya bila ditunda. Resep yang terdapat tanda seperti diatas, meskipun masuk belakangan harus di dahulukan pengerjaannya. (Ilmu Meracik Obat Teori dan Praktik, Moh. Anief, Gadjah Mada University Presentasi, 1987, Halaman 7) === Iter === Diulang === Ne iter === tidak boleh diulang === Qs === quantum satis (secukupnya) === Ad gram 20 === Hingga 20 g === Ante defecatio === Sebelum defekasi === Omni noctem per vaginal === == Daftar Obat (1) == # Cefixime Trihydrate kapsul 100 mg 2 x 1 # Loperamide HCl tablet 2 mg 3 x 1 (Stop jika sudah tidak diare) # Omeprazole (OMZ) 20 mg kapsul pelepasan lambat 2 x 1 # Paracetamol tablet 500 mg 3 x 1 # Zinc dispersible tablet 20 mg 1 x 1 == Daftar Obat (2) == {| {{Prettytable}} ! Merk ! Kandungan |- | Acifar | * Acyclovir |- | Alleron | * Chlorpheniramina maleat 4 mg |- | Alofar 100 | * Allopurinol 100 mg |- | Alofar 300 | * Allopurinol 300 mg |- | Aspilet | * Asam asetilsalisilat (Aspirin) |- | Akita | * Attapulgite 600 mg * Pektin 50 mg |- | Beneuron | * Vitamin B1 (Thiamine Mononitrate) 100 mg * Vitamin B6 (Pyridoxine Hydrochloride) 200 mg * Vitamin B12 (Cyanocobalamin) 200 mcg |- | Arkavit-C | * Vitamin B1 (Tiamin) 50 mg * Vitamin B2 (Riboflavin) 25 mg * Vitamin B3 (Niasin) 50 mg * Vitamin B5 (Asam pantotenat) 20 mg * Vitamin B6 (Piridoksin) 10 mg * Vitamin B12 (Sianokonalamin) 5 mcg * Vitamin C (Asam askorbat) 500 mg |- | Bufacaryl | * Dexamethasone 0,5 mg * Dexchlorpheniramine maleate 2 mg |- | Cavicur | * Ekstrak Curcuma xanthorriza (temulawak) 20 mg (mengandung curcuminoid ~15%) * Vitamin A palmitat * Vitamin B1, B2, B6, B12 * Vitamin D * Calsium pantothenate * Calsium glycerophosphate * Cod liver oil |- | Cavicur Syr | Tiap 5 ml mengandung: * Ekstrak Curcuma xanthorrhiza 10 mg * Vitamin A palmitat 850 IU * Vitamin B1 (tiamin HCl) 3 mg * Vitamin B2 (riboflavin) 1,5 mg * Vitamin B6 (piridoksin HCl) 0,5 mg * Vitamin B12 5 mcg * Vitamin D 100 IU * Calsium pantothenate 5 mg * Calsium glycerophosphate 100 mg * Cod liver oil (ekstrak minyak ikan kod) 2,5 mg |- | Caviplex | * Vitamin ** A 4.000 IU ** D3 400 IU ** B1 3 mg, B2 3-4 mg, B6 4 mg, B12 12 mcg ** C 75 mg ** E 10 mg ** B3 (Nikotinamid) 20 mg ** B5 (Ca pantotenat) 5 mg ** Biotin 0,1 mg ** Asam folat 1 mg * Mineral & lainnya ** Zat besi (Fe) ~30-90 mg ** Kalsium (Ca) ~70-100 mg ** Magnesium ~25-87,5 mg ** Zinc 15 mg ** Tembaga 0,5 mg ** Mangan 0,5 mg ** Fluor 0,5 mg ** Iodium 0,15 mg ** Asam glutamat 50 mg |- | Cardipin 5 | * Nifedipine 5 mg |- | Cardipin 10 | * Nifedipine 10 mg |- | Carmeson 4 | * Ondansetron 4 mg |- | Carmeson 8 | * Ondansetron 8 mg |- | Cendo Cenfresh | * Carboxymethylcellulose sodium 5 mg/ml |- | Cendo Eyefresh | * Hypromellose (HPMC) 3 mg/ml * Dextran 70 1 mg/ml |- | Cendo Genta | * Gentamisin sulfat 0,3% (3 mg/ml) |- | Cendo Lytrees | * Sodium chloride (Natrium klorida) 4,4 mg * Potassium chloride (Kalium klorida) 0,8 mg |- | Cendo Timol | * Timolol Maleat setara Timolol 0,25% atau 0,5% (2,5 mg atau 5 mg per ml) |- | Cendo Xitrol | * Dexamethasone sodium phosphate (kortikosteroid anti-inflamasi) 0,1% (setara ±1 mg) * Neomycin sulfate (antibiotik aminoglikosida) setara 3,5 mg neomycin base * Polymyxin B sulfate (antibiotik) antara 6.000 ‒ 10.000 IU |- | Dionicol | * Kloramfenikol (Chloramphenicol) 5 mg |- | Dionicol Syr | * Kloramfenikol (Chloramphenicol) 125 mg per 5 ml |} # Demacolin # Dexaharsen 0,5 # Dexyl Syr # Etaflusin # Etambion # Farsifen 400 # Farsifen 100 ml/5 ml Ifar # Farsifen Plus # Fasidol # Fasidol 100 mg/5 ml Ifar # Fasidol Drop # Fasidol Forte # Fasiprim Forte # Flutop-C Syr # Grafazol # Gencetron 8 # Gencobal # Oxicobal # Histigo # Grantusif # Hufamag Plus # Hufagrip Forte # Intunal-F # Infalgin # Inamid 2 # Lerzin 10 # Lecozink Syr # Laxana 5 # Lodecon # Lokev # Loctacef 125 mg/5 ml # Nexitra # Novakal # Omedom # Relaxon # Suprabiotik # Spasminal # Teosal # Tifestan Forte # Samcodin # Vesperum Syr # Vosea 5 # Vosea Syr # Yekaprim Syr # Yusimox Syr # Zengesic # Zelona 50 == Daftar Pustaka == # {{id}} Ikatan Apoteker Indonesia. Informasi Spesialite Obat Indonesia. Volume 45 – 2010 s/d 2011. Jakarta: PT. ISFI Penerbitan; 2010. ISSN 0854-4492. {{Catatan Dokter Muda-Stase}} [[Kategori:Stase|{{SUBPAGENAME}}]] 008cqdpsnzer5dcfv3okuesanj3m0uc 108056 108055 2025-07-03T04:28:16Z Alfarq 799 /* Daftar Obat (2) */ 108056 wikitext text/x-wiki {{Wikipedia|Farmasi}} == Istilah Latin == === Uc === * ℞ (racipe): Ambillah. * Supp (suppositorium): Supositoria, gentel. * No (nomero): Jumlah, sejumlah. * S (signa): Tanda, tandai, tandailah. * Uc (usus cognitus): Pemakaian diketahui. * Pro: Untuk. ℞ Miconazole 2% cream tube No. I S uc ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. C (20 tahun) ℞ Anusol supp No. V S uc ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. A (21 tahun) === Imm === * Imm (in manus medicine): Serahkan ke tangan paramedis (dokter, perawat, bidan, mantri). * Inf (infus): Infus. * Flab (flabot): Flabot. ℞ Natrium Chlorida 0,9% inf flab No. III Abbocath no. 22 No. I Cum infus set No. I S imm ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. B (22 tahun) === Prn === * Tab (tabletta, tabella): Tablet. * Mg (miligram): Miligram. * Prn (pro renatera), sns (si necesse sit): Jika perlu. Singkatan ini digunakan untuk obat-obat simtomatis (penghilang gejala). * Dd (de die): Tiap hari, setiap hari. * Aggr febr (agrediante febre): saat demam. ℞ Paracetamol tab mg 500 No. X S prn 1–3 dd tab I aggr febr ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. C (23 tahun) '''Perhatian:''' * Penulisan setelah Signa tanpa tanda kurung. Berikut adalah penulisan yang salah: ℞ Paracetamol tab mg 500 No. X S prn (1–3) dd tab I aggr febr ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. C (23 tahun) === Up === * Up (usus propius): Pemakaian sendiri (untuk dokter). Tidak perlu memakai Pro. ℞ Amoxycillin tab mg 500 No. X S up ––––––––––––––––––––––––––––––––––––––––––––––– ₰ === Dcf === * Dcf, dc form (da cun formula): Sesuai obat yang tertulis. * Sol, solut (solutio): Larutan. ℞ Alkohol 70% sol fl No. I S imm dcf ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. E (25 tahun) === Sine confectiones === * Cap (capsule): Kapsul. * Sine confectiones: Tanpa kemasan. ℞ Amoxsan cap mg 500 No. X (sine confectiones) S 3 dd cap I ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. F (26 tahun) === Adde === * Adde: Tambahkan. * Dry: Kering. * Syr (syrupus): Sirup. * Aq coct (aqua cocta): Air masak. * Ad: Sampai, hingga. * Cc (cubic centimetre): sentimeter kubik. * C (cochlea): sendok makan (15 cc). * Cth (cochlea tea): sendok teh (5 cc). ℞ Ampicillin dry syr fl No. I Adde aq coct ad cc 60 S 3 dd cth I ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. G (27 tahun) === Ad lib === * Ad lib, ad libit (ad libitum): Sesuka hati. ℞ Oralit sachet No. XX S ad lib ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. H (28 tahun) === Ac === * Ac (antec cibos, ante cibum, ante coenam): Sebelum makan. ℞ Antasida DOEN tab No. X S 3 dd tab I ac ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. I (29 tahun) === Dc === * Dc (durante coenam): Saat makan, ketika makan, selagi makan. ℞ Enzyplex tab No. XII S 3 dd tab I dc ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. J (30 tahun) === Pc === * Pc (post coenam): Setelah makan, sesudah makan, sehabis makan. ℞ Rifampicin tab mg 450 No. VII S 1 dd tab I pc ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. K (31 tahun) === Hora spatio === * Octa: Delapan. * Hora spatio: Selang sekian jam. * Octa hora spatio: Selang delapan jam. ℞ Chloramphenicol cap mg 500 No. X S 3 dd cap I octa hora spatio ––––––––––––––––––––––––––––––––––––––––––––––– ₰ Pro: Tn. L (32 tahun) === Agita ante sumendum === Agita ante sumendum: kocok sebelum digunakan R/ Mylanta Syr fl no I ʃ 3 dd CI ac agitatio ante sumundum ---- Pro: Tn J (29th) === Hora ieiune === Saat perut kosong === Sive simile === * Boleh diganti * Obat paten boleh diganti dengan obat paten lain dengan kandungan, bentuk, sediaan, dan dosis yang sama === Mds === === use know === === Da in caps === dimasukkan ke dalam kapsul === Cum === === Cito, Urgent, PIM (Periculum In Mora) === Segera, Penting, Berbahaya bila ditunda. Resep yang terdapat tanda seperti diatas, meskipun masuk belakangan harus di dahulukan pengerjaannya. (Ilmu Meracik Obat Teori dan Praktik, Moh. Anief, Gadjah Mada University Presentasi, 1987, Halaman 7) === Iter === Diulang === Ne iter === tidak boleh diulang === Qs === quantum satis (secukupnya) === Ad gram 20 === Hingga 20 g === Ante defecatio === Sebelum defekasi === Omni noctem per vaginal === == Daftar Obat (1) == # Cefixime Trihydrate kapsul 100 mg 2 x 1 # Loperamide HCl tablet 2 mg 3 x 1 (Stop jika sudah tidak diare) # Omeprazole (OMZ) 20 mg kapsul pelepasan lambat 2 x 1 # Paracetamol tablet 500 mg 3 x 1 # Zinc dispersible tablet 20 mg 1 x 1 == Daftar Obat (2) == {| {{Prettytable}} ! Merk ! Kandungan |- | Acifar | * Acyclovir |- | Alleron | * Chlorpheniramina maleat 4 mg |- | Alofar 100 | * Allopurinol 100 mg |- | Alofar 300 | * Allopurinol 300 mg |- | Aspilet | * Asam asetilsalisilat (Aspirin) |- | Akita | * Attapulgite 600 mg * Pektin 50 mg |- | Beneuron | * Vitamin B1 (Thiamine Mononitrate) 100 mg * Vitamin B6 (Pyridoxine Hydrochloride) 200 mg * Vitamin B12 (Cyanocobalamin) 200 mcg |- | Arkavit-C | * Vitamin B1 (Tiamin) 50 mg * Vitamin B2 (Riboflavin) 25 mg * Vitamin B3 (Niasin) 50 mg * Vitamin B5 (Asam pantotenat) 20 mg * Vitamin B6 (Piridoksin) 10 mg * Vitamin B12 (Sianokonalamin) 5 mcg * Vitamin C (Asam askorbat) 500 mg |- | Bufacaryl | * Dexamethasone 0,5 mg * Dexchlorpheniramine maleate 2 mg |- | Cavicur | * Ekstrak Curcuma xanthorriza (temulawak) 20 mg (mengandung curcuminoid ~15%) * Vitamin A palmitat * Vitamin B1, B2, B6, B12 * Vitamin D * Calsium pantothenate * Calsium glycerophosphate * Cod liver oil |- | Cavicur Syr | Tiap 5 ml mengandung: * Ekstrak Curcuma xanthorrhiza 10 mg * Vitamin A palmitat 850 IU * Vitamin B1 (tiamin HCl) 3 mg * Vitamin B2 (riboflavin) 1,5 mg * Vitamin B6 (piridoksin HCl) 0,5 mg * Vitamin B12 5 mcg * Vitamin D 100 IU * Calsium pantothenate 5 mg * Calsium glycerophosphate 100 mg * Cod liver oil (ekstrak minyak ikan kod) 2,5 mg |- | Caviplex | * Vitamin ** A 4.000 IU ** D3 400 IU ** B1 3 mg, B2 3-4 mg, B6 4 mg, B12 12 mcg ** C 75 mg ** E 10 mg ** B3 (Nikotinamid) 20 mg ** B5 (Ca pantotenat) 5 mg ** Biotin 0,1 mg ** Asam folat 1 mg * Mineral & lainnya ** Zat besi (Fe) ~30-90 mg ** Kalsium (Ca) ~70-100 mg ** Magnesium ~25-87,5 mg ** Zinc 15 mg ** Tembaga 0,5 mg ** Mangan 0,5 mg ** Fluor 0,5 mg ** Iodium 0,15 mg ** Asam glutamat 50 mg |- | Cardipin 5 | * Nifedipine 5 mg |- | Cardipin 10 | * Nifedipine 10 mg |- | Carmeson 4 | * Ondansetron 4 mg |- | Carmeson 8 | * Ondansetron 8 mg |- | Cendo Cenfresh | * Carboxymethylcellulose sodium 5 mg/ml |- | Cendo Eyefresh | * Hypromellose (HPMC) 3 mg/ml * Dextran 70 1 mg/ml |- | Cendo Genta | * Gentamisin sulfat 0,3% (3 mg/ml) |- | Cendo Lytrees | * Sodium chloride (Natrium klorida) 4,4 mg * Potassium chloride (Kalium klorida) 0,8 mg |- | Cendo Timol | * Timolol Maleat setara Timolol 0,25% atau 0,5% (2,5 mg atau 5 mg per ml) |- | Cendo Xitrol | * Dexamethasone sodium phosphate (kortikosteroid anti-inflamasi) 0,1% (setara ±1 mg) * Neomycin sulfate (antibiotik aminoglikosida) setara 3,5 mg neomycin base * Polymyxin B sulfate (antibiotik) antara 6.000 ‒ 10.000 IU |- | Dionicol | * Kloramfenikol (Chloramphenicol) 5 mg |- | Dionicol Syr | * Kloramfenikol (Chloramphenicol) 125 mg per 5 ml |- | Demacolin | * Paracetamol 500 mg * Phenylpropanolamine HCl 12,5 mg * Chlorpheniramine maleate (CTM) 2 mg |} # Dexaharsen 0,5 # Dexyl Syr # Etaflusin # Etambion # Farsifen 400 # Farsifen 100 ml/5 ml Ifar # Farsifen Plus # Fasidol # Fasidol 100 mg/5 ml Ifar # Fasidol Drop # Fasidol Forte # Fasiprim Forte # Flutop-C Syr # Grafazol # Gencetron 8 # Gencobal # Oxicobal # Histigo # Grantusif # Hufamag Plus # Hufagrip Forte # Intunal-F # Infalgin # Inamid 2 # Lerzin 10 # Lecozink Syr # Laxana 5 # Lodecon # Lokev # Loctacef 125 mg/5 ml # Nexitra # Novakal # Omedom # Relaxon # Suprabiotik # Spasminal # Teosal # Tifestan Forte # Samcodin # Vesperum Syr # Vosea 5 # Vosea Syr # Yekaprim Syr # Yusimox Syr # Zengesic # Zelona 50 == Daftar Pustaka == # {{id}} Ikatan Apoteker Indonesia. Informasi Spesialite Obat Indonesia. Volume 45 – 2010 s/d 2011. Jakarta: PT. ISFI Penerbitan; 2010. ISSN 0854-4492. {{Catatan Dokter Muda-Stase}} [[Kategori:Stase|{{SUBPAGENAME}}]] 3bpuy4blvi3aqejr9tzg19kbuldmp7v 0 20603 108052 101637 2025-07-02T14:26:03Z CommonsDelinker 656 Removing [[:c:File:Dushanbe_Presidential_Palace_01.jpg|Dushanbe_Presidential_Palace_01.jpg]], it has been deleted from Commons by [[:c:User:Wdwd|Wdwd]] because: per [[:c:Commons:Deletion requests/File:Dushanbe Presidential Palace 01.jpg|]]. 108052 wikitext text/x-wiki {{DISPLAYTITLE:<span style="display:block;text-align:center;font-size:200%;color:red;font-style:bold;line-height:1em;">Wikijunior:Asia</span>}} {{status|25%}} {{/Pengantar/}} <center> <gallery> Buddha_statue,_Nha_Trang.jpg|Patung Buddha di Nha Trang, [[Wikijunior:Asia/Vietnam|Vietnam]]. Negombo Beach, Sri Lanka.jpg|Pantai Negombo di Negombo, [[Wikijunior:Asia/Sri Lanka|Sri Lanka]]. Young orang utan.JPG|Seekor orang utan di Pusat Orangutan Bohorok, [[Wikijunior:Asia/Indonesia|Indonesia]]. Manora Beach 1100641.JPG|Pesisir pantai di [[Wikijunior:Asia/Pakistan|Pakistan]]. Banaue_Philippines_Ifugao-Tribesman-01.jpg|Suku Ifugao di [[Wikijunior:Asia/Filipina|Filipina]]. Chinese Garden Bridge.jpg|Taman Chinese Garden di [[Wikijunior:Asia/Singapura|Singapura]]. Little AraratDSC 3125.jpg|Gunung Little Ararat di [[Wikijunior:Asia/Turki|Turki]]. Beauty of Phewa lake, Pokhara.jpg|Pemandangan Danau Phewa di [[Wikijunior:Asia/Nepal|Nepal]]. The_Great_Wall_of_China_at_Jinshanling-edit.jpg|Tembok Besar [[Wikijunior:Asia/Tiongkok|Tiongkok]], di kota Jinshanling. Insidemasjedolharam4.JPG|Masjid Al-Haram di Mekkah, [[Wikijunior:Asia/Arab Saudi|Arab Saudi]]. Saigon_Trade_Center_21112013.JPG|Saigon Trade Center, di Kota Hồ Chí Minh, [[Wikijunior:Asia/Vietnam|Vietnam]]. Sultan Omar Ali Saifuddien Mosque. Brunei.. (14406928611).jpg|Masjid Sultan Omar Ali Saifuddin, [[Wikijunior:Asia/Brunei|Brunei]] A Famosa, remnant of the Portuguese fortress (12738512135).jpg|A Famosa, sisa benteng Portugis di Melaka, [[Wikijunior:Asia/Malaysia|Malaysia]]. Wat Arun BangkokImg 3941.jpg|Kuil Wat Arun di [[Wikijunior:Asia/Thailand|Thailand]]. </gallery> </center> == Daftar negara == {{Wikijunior:Asia}} {| style="width:100%; margin:auto;" | style="vertical-align:top; width:25%;" | {{color box|Lime|Asia Barat}} *[[/Armenia|Armenia]] {{stage|0}} *[[/Azerbaijan|Azerbaijan]] {{stage|0}} *[[/Bahrain|Bahrain]] {{stage|0}} *[[/Siprus|Siprus]] {{stage|0}} *[[/Georgia|Georgia]] {{stage|0}} *[[/Iran|Iran]] {{stage|0}} *[[/Irak|Irak]] {{stage|0}} *[[/Israel|Israel]] {{stage|0}} *[[/Yordania|Yordania]] {{stage|0}} *[[/Kuwait|Kuwait]] {{stage|0}} *[[/Libanon|Libanon]] {{stage|0}} *[[/Oman|Oman]] {{stage|0}} *[[/Palestina|Palestina]] {{stage|0}} *[[/Qatar|Qatar]] {{stage|0}} *[[/Arab Saudi|Arab Saudi]] {{stage|0}} *[[/Suriah|Suriah]] {{stage|0}} *[[/Turki|Turki]] {{stage|0}} *[[/Uni Emirat Arab|Uni Emirat Arab]] {{stage|0}} *[[/Yaman|Yaman]] {{stage|0}} | style="vertical-align:top; width:25%;" | {{color box|Blue|Asia Utara}} *[[/Rusia|Rusia]] {{stage|0}} {{color box|Magenta|Asia Tengah|White}} *[[/Kazakstan|Kazakstan]] {{stage|0}} *[[/Kirgizstan|Kirgizstan]] {{stage|0}} *[[/Tajikistan|Tajikistan]] {{stage|0}} *[[/Turkmenistan|Turkmenistan]] {{stage|0}} *[[/Uzbekistan|Uzbekistan]] {{stage|0}} {{color box|Yellow|Asia Timur}} *[[/Tiongkok|Tiongkok]] {{stage|0}} *[[/Jepang|Jepang]] {{stage|0}} *[[/Makau|Makau]] {{stage|0}} *[[/Mongolia|Mongolia]] {{stage|0}} *[[/Korea Utara|Korea Utara]] {{stage|0}} *[[/Korea Selatan|Korea Selatan]] {{stage|0}} *[[/Taiwan|Taiwan]] {{stage|0}} | style="vertical-align:top; width:25%;" | {{color box|Red|Asia Selatan|White}} *[[/Afganistan|Afganistan]] {{stage|0}} *[[/Bhutan|Bhutan]] {{stage|0}} *[[/Bangladesh|Bangladesh]] {{stage|0}} *[[/India|India]] {{stage|0}} *[[/Maladewa|Maladewa]] {{stage|0}} *[[/Nepal|Nepal]] {{stage|0}} *[[/Pakistan|Pakistan]] {{stage|0}} *[[/Sri Lanka|Sri Lanka]] {{stage|0}} | style="vertical-align:top; width:25%;" | {{color box|Orange|Asia Tenggara}} *[[/Brunei|Brunei]] {{stage|0}} *[[/Kamboja|Kamboja]] {{stage|0}} *[[/Indonesia|Indonesia]] {{stage|100}} *[[/Laos|Laos]] {{stage|0}} *[[/Malaysia|Malaysia]] {{stage|0}} *[[/Myanmar|Myanmar]] {{stage|0}} *[[/Filipina|Filipina]] {{stage|}} *[[/Singapura|Singapura]] {{stage|100}} *[[/Thailand|Thailand]] {{stage|0}} *[[/Timor Leste|Timor Leste]] {{Stage|0}} *[[/Vietnam|Vietnam]] {{stage|0}} |} ===Pertanyaan yang wajib ada di tiap halaman=== #Bagaimana sejarah negara ini? #Bagaimana keadaan alamnya? #Seperti apa penduduknya? #Apa tempat yang populer? ===Pertanyaan bagus lain=== #Seperti apa sekolah di sana? #Seperti apa makanan mereka? #Apa musik yang populer? #Apa olahraga yang populer? == Lihat pula == [[Berkas:Asia (orthographic projection).svg|frameless|right]] *[[Wikijunior:Asia/Kuis]] *{{tlx|User book|Wikijunior:Asia}} *[[w:Asia|Asia di Wikipedia]] *[http://www.worldatlas.com/webimage/countrys/as.htm Asia] di WorldAtlas *[[Wikijunior:Afrika]] *[[Wikijunior:Amerika Selatan]] *[[Wikijunior:Amerika Utara]] *[[Wikijunior:Eropa]] *[[Wikijunior:Daftar negara A-Z]] {{WikijuniorDunia}} {{shelves|Wikijunior Geografi}} ff62y8z2snr7a9pnr2xt6fa26rktenw 0 23139 108053 107962 2025-07-03T02:28:33Z Akuindo 8654 /* Rumus sederhana */ 108053 wikitext text/x-wiki == Kaidah umum == :<math>\left({cf}\right)' = cf'</math> :<math>\left({f + g}\right)' = f' + g'</math> :<math>\left({f - g}\right)' = f' - g'</math> ;[[Kaidah darab]] :<math>\left({fg}\right)' = f'g + fg'</math> ;[[Kaidah timbalbalik]] :<math>\left(\frac{1}{f}\right)' = \frac{-f'}{f^2}, \qquad f \ne 0</math> ;[[Kaidah hasil-bagi]] :<math>\left({f \over g}\right)' = {f'g - fg' \over g^2}, \qquad g \ne 0</math> ;[[Kaidah rantai]] :<math>(f \circ g)' = (f' \circ g)g'</math> ;Turunan [[fungsi invers]] :<math>(f^{-1})' =\frac{1}{f' \circ f^{-1}}</math> untuk setiap fungsi terdiferensialkan ''f'' dengan argumen riil dan dengan nilai riil, bila komposisi dan invers ada ;Kaidah pangkat umum :<math>(f^g)'=f^g \left( g'\ln f + \frac{g}{f} f' \right)</math> == Rumus sederhana == : <math>c' = 0 \, </math> : <math>x' = 1 \, </math> : <math>(cx)' = c \, </math> : <math>|x|' = {x \over |x|} = \sgn x, \qquad x \ne 0</math> <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| \\ y^2 &= x^2 \\ 2y y' &= 2x \\ y' &= \frac{x}{y} \\ &= \frac{x}{|x|} \\ \end{align} </math> </div></div> : <math>(x^c)' = cx^{c-1} \qquad \mbox{baik } x^c \mbox{ maupun } cx^{c-1} \mbox { terdefinisi}</math> : <math>\left({1 \over x}\right)' = \left(x^{-1}\right)' = -x^{-2} = -{1 \over x^2}</math> : <math>\left({1 \over x^c}\right)' = \left(x^{-c}\right)' = -cx^{-(c+1)} = -{c \over x^{c+1}}</math> : <math>\left(\sqrt{x}\right)' = \left(x^{1\over 2}\right)' = {1 \over 2} x^{-{1\over 2}} = {1 \over 2 \sqrt{x}}, \qquad x > 0</math> : <math>\left(x^n\right)' = n \cdot x^{n-1}</math> : <math>\left(u^n\right)' = n \cdot u' \cdot u^{n-1}</math> ; Eksponen dan logaritma :<math> \left(c^x\right)' = c^x \ln c,\qquad c > 0</math> :<math> \left(e^x\right)' = e^x</math> :<math> \left(^c\log x\right)' = \frac{1}{x \ln c}, \qquad c > 0</math> :<math> \left(\ln x\right)' = \frac{1}{x}</math> ; Trigonometri {| style="width:100%; background:transparent; margin-left:2em;" |width=50%|<math> (\sin x)' = \cos x \,</math> |width=50%|<math> (\arcsin x)' = { 1 \over \sqrt{1 - x^2}} \,</math> |- |<math> (\cos x)' = -\sin x \,</math> |<math> (\arccos x)' = {-1 \over \sqrt{1 - x^2}} \,</math> |- |<math> (\tan x)' = \sec^2 x = { 1 \over \cos^2 x} = 1 + \tan^2 x \,</math> |<math> (\arctan x)' = { 1 \over 1 + x^2} \,</math> |- |<math> (\sec x)' = \sec x \tan x \,</math> |<math> (\arcsec x)' = { 1 \over |x|\sqrt{x^2 - 1}} \,</math> |- |<math> (\csc x)' = -\csc x \cot x \,</math> |<math> (\arccsc x)' = {-1 \over |x|\sqrt{x^2 - 1}} \,</math> |- |<math> (\cot x)' = -\csc^2 x = { -1 \over \sin^2 x} = -(1 + \cot^2 x) \,</math> |<math> (\arccot x)' = {-1 \over 1 + x^2} \,</math> |} * Tambahkan: ** <math> (\sin^n x)' = n \sin^{n-1} x cos x \,</math> ** <math> (\sin u)' = u' \cos u \,</math> ** <math> (\sin^n u)' = n u' \sin^{n-1} u cos u \,</math> ; Hiperbolik Perhatikan sebagai berikut : <math>sinh x = \frac{e^x - e^{-x}}{2}</math> : <math>cosh x = \frac{e^x + e^{-x}}{2}</math> {| style="width:100%; background:transparent; margin-left:2em;" |width=50%|<math>(\sinh x )'= \cosh x</math> |width=50%|<math>(\operatorname{arcsinh}\,x)' = { 1 \over \sqrt{x^2 + 1}}</math> |- |<math>(\cosh x )'= \sinh x</math> |<math>(\operatorname{arccosh}\,x)' = { 1 \over \sqrt{x^2 - 1}}, x<1</math> |- |<math>(\tanh x )'= \operatorname{sech}^2\,x</math> |<math>(\operatorname{arctanh}\,x)' = { 1 \over 1 - x^2}, |x|<1</math> |- |<math>(\operatorname{sech}\,x)' = - \tanh x\,\operatorname{sech}\,x</math> |<math>(\operatorname{arcsech}\,x)' = {-1 \over x\sqrt{1 - x^2}}, 0<x<1</math> |- |<math>(\operatorname{csch}\,x)' = -\,\operatorname{coth}\,x\,\operatorname{csch}\,x</math> |<math>(\operatorname{arccsch}\,x)' = {-1 \over |x|\sqrt{1 + x^2}}, x \neq 0</math> |- |<math>(\operatorname{coth}\,x )' = -\,\operatorname{csch}^2\,x</math> |<math>(\operatorname{arccoth}\,x)' = { 1 \over 1-x^2}, x>1</math> |} :: catatan: jika x diganti u maka merumuskan seperti trigonometri. ; implisit *cara 1 : <math>ax + by = 1</math> : <math>a + b y' = 0</math> : <math>y' = -\frac{a}{b}</math> *cara 2 persamaan F(x,y) dibuat hasil nol kemudian diubah menjadi <math>\frac{d_y}{d_x} = - \frac{F_x(x,y)}{F_y(x,y)}</math> Fungsi implisit dibagi 2 jenis yaitu: # eksplisit artinya fungsi implisit dapat diubah menjadi fungsi eksplisit. Contoh: x²y+7=8xy+5x, y²-8x=5-6y # in-eksplisit artinya fungsi implisit tidak dapat diubah menjadi fungsi eksplisit. Contoh: xy+8x=y³-11 == Laju perubahan (rata-rata) == Rumus laju perubahan f(x) pada interval x1 dan x2 adalah V(rata-rata) = <math>\frac{\Delta y}{\Delta x} = \frac{f'(x_2)-f'(x_1)}{x_2-x_1}</math> == Nilai ekstrem, interval serta titik stasioner/diam == : Persamaan kuadrat :: Nilai minimum persamaan kuadrat adalah titik terendah (titik stasioner/diam) serta intervalnya turun-naik. :: Nilai maksimum persamaan kuadrat adalah titik tertinggi (titik stasioner/diam) serta intervalnya naik-turun. : Persamaan kubik Ada 4 kemungkinan yang berdasarkan interval sebagai berikut: :: Nilai maksimum-minimum dan intervalnya naik-turun-naik. :: Nilai maksimum-minimum dan intervalnya turun secara monoton. :: Nilai minimum-maksimum dan intervalnya turun-naik-turun. :: Nilai maksimum-minimum dan intervalnya naik secara monoton. Untuk mengetahui nilainya harus turunan kedua (jika kuadrat berarti hasilnya konstanta sedangkan berpangkat lebih dari 2 berarti masukkan x dari hasil turunan pertama untuk memperoleh hasilnya). Jika konstanta > 0 maka itu berarti nilai minimum, konstanta < 0 maka itu berarti nilai maksimum serta konstanta = 0 itu berarti titik belok. Posisi gradien adalah turunan pertama bernilai nol untuk mencari nilai x tersebut. Titik belok berarti turunan kedua bernilai nol tapi turunan ketiga tidak boleh bernilai nol. == Persamaan garis singgung kurva == Gradien (m) pada Persamaan garis singgung kurva y = f(x) pada titik A (a, f(a)) adalah m = f’(a) = <math>\lim\limits_{\Delta x \to 0} \frac{f(a+\Delta x)-f(a)}{\Delta x}</math> contoh: # Berapa laju perubahaan f(x)=x²-4x+3 pada: ## x=5! ## interval 2<x<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(x) &= x^2-4x+3 \\ f'(x) &= 2x-4 \\ v &= \frac{f'(x)}{x} \\ &= \frac{f'(5)}{5} \\ &= \frac{2(5)-4}{5} \\ &= \frac{6}{5} \\ &= 1.2 \\ v &= \frac{f'(x_2)-f'(x_1)}{x_2-x_1} \\ &= \frac{f'(3)-f'(2)}{3-2} \\ &= \frac{2(3)-4-(2(2)-5)}{3-2} \\ &= \frac{3}{1} \\ &= 3 \\ \end{align} </math> </div></div> # Berapa laju perubahaan f(x)=sin x-cos x pada: ## x=<math>\frac{\pi}{6}</math>! ## interval <math>0<x<\frac{\pi}{2}</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) &= sin x-cos x \\ f'(x) &= cos x+sin x \\ v &= \frac{f'(x)}{x} \\ &= \frac{f'(\frac{\pi}{6})}{\frac{\pi}{6}} \\ &= \frac{cos \frac{\pi}{6}+sin \frac{\pi}{6}}{\frac{\pi}{6}} \\ &= \frac{\frac{\pi \sqrt{3}}{2}+\frac{\pi}{2}}{\frac{\pi}{6}} \\ &= \frac{\frac{\pi}{2} (\sqrt{3}+1)}{\frac{\pi}{6}} \\ &= 3 (\sqrt{3}+1) \\ v &= \frac{f'(x_2)-f'(x_1)}{x_2-x_1} \\ &= \frac{f'(\frac{\pi}{2})-f'(0)}{\frac{\pi}{2}-0} \\ &= \frac{cos \frac{\pi}{2}+sin \frac{\pi}{2}-(cos 0+sin 0)}{\frac{\pi}{2}-0} \\ &= \frac{(0+1)-(1+0)}{\frac{\pi}{2}} \\ &= \frac{0}{\frac{\pi}{2}} \\ &= 0 \\ \end{align} </math> </div></div> # Tentukan nilai ekstrem, titik stasioner, titik belok serta interval dari <math>f(x)=\frac{x^3}{3}-3x^2-16x+36</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{align} f(x) &= \frac{x^3}{3}-3x^2-16x+36 \\ f'(x) &= x^2-6x-16 \\ \text{untuk menentukan nilai x ketika f'(x)=0 } \\ x^2-6x-16 &= 0 \\ (x+2)(x-8) &= 0 \\ x=-2 &\text{ atau } x=8 \\ \text{ menentukan nilai ekstrem } \\ f''(x) &= 2x-6 \\ f''(-2) &= 2(2)-6 = -2 < 0 \\ f''(8) &= 2(8)-6 = 10 > 0 \\ \text{f''(-2) negatif maka nilai maksimum sedangkan f''(8) positif maka nilai minimum } \\ \text{untuk menentukan titik stasionernya adalah } \\ f(-2) &= \frac{(-2)^3}{3}-3(-2)^2-16(-2)+36 \\ &= \frac{160}{3} \\ f(8) &= \frac{8^3}{3}-3(8)^2-16(8)+36 \\ &= -\frac{1.386}{3} \\ \text{jadi titik stasionernya adalah } (-2,\frac{160}{3}) \text{ dan } (8,-\frac{1.386}{3}) \\ \text{untuk menentukan titik beloknya ketika f''(x)=0 } \\ f''(x) &= 2x-6 \\ 2x-6 &= 0 \\ x &= 3 \\ \text{untuk menentukan titik beloknya adalah } \\ f(3) &= \frac{3^3}{3}-3(3)^2-16(3)+36 \\ f(3) &= -30 \\ \text{jadi titik beloknya adalah } (3, -30) \\ \text{untuk menentukan intervalnya, buatlah irisan pada titik stasionernya dengan bantuan x = (-4; 0; 9) maka naik jika } x<-2 \text{ atau } x>8 \text{ dan turun jika } -2<x<8 \\ \end{align} </math> </div></div> # Tentukan nilai ekstrem, titik stasioner, titik belok serta interval dari <math>f(x)=1-2cos 2x</math> pada batas-batas interval <math>0<x<\frac{\pi}{2}</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{align} f(x) &= 1-2cos 2x \\ f'(x) &= 4sin 2x \\ \text{untuk menentukan nilai x ketika f'(x)=0 } \\ 4sin 2x &= 0 \\ 8sin x \cdot cos x &= 0 \\ (sin x)(cos x) &= 0 \\ x=0 &\text{ atau } x=\frac{\pi}{2} \\ \text{ menentukan nilai ekstrem } \\ f''(x) &= 8cos 2x \\ f''(0) &= 8cos 2(0) = 8 > 0 \\ f''(\frac{\pi}{2}) &= 8cos 2(\frac{\pi}{2}) = -8 < 0 \\ \text{f''(0) positif maka nilai minimum sedangkan } f''(\frac{\pi}{2}) \text{ negatif maka nilai maksimum } \\ \text{untuk menentukan titik stasionernya adalah } \\ f(0) &= 1-2cos 2(0) \\ &= -1 \\ f(\frac{\pi}{2}) &= 1-2cos 2(\frac{\pi}{2}) \\ &= 3 \\ \text{jadi titik stasionernya adalah } (0,-1) \text{ dan } (\frac{\pi}{2},3) \\ \text{untuk menentukan titik beloknya ketika f''(x)=0 } \\ f''(x) &= 8cos 2x \\ 8cos 2x &= 0 \\ cos 2x &= 0 \\ cos 2x &= cos \frac{\pi}{2} \\ 2x &= \frac{\pi}{2} \\ x &= \frac{\pi}{4} \\ \text{untuk menentukan titik beloknya adalah } \\ f(\frac{\pi}{4}) &= 1-2cos 2(\frac{\pi}{4}) \\ f(\frac{\pi}{4}) &= 1 \\ \text{jadi titik beloknya adalah } (\frac{\pi}{4}, 1) \\ \text{kondisi intervalnya pada batas-batas tersebut adalah naik } \\ \end{align} </math> </div></div> # Tentukan persamaan garis singgung kurva <math>f(x)=\frac{x^3}{3}-x^2-8x</math> pada: ## di titik (-3,-3)! ## berabsis 2! :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{align} f(x) &= \frac{x^3}{3}-x^2-8x \\ f'(x) &= x^2-2x-8 \\ f'(x) &= (-3)^2-2(3)-8 \\ f'(x) &= -5 \\ (y-y_1) &= m (x-x_1) \\ y-(-3) &= -5 (x-(-3)) \\ y+3 &= -5x-15 \\ y &= -5x-18 \\ \end{align} </math> </div></div> ## <div class="toccolours mw-collapsible mw-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{untuk mengetahui y dari berabsis (x) 2 yaitu } \frac{2^3}{3}-2^2-8(2) = -\frac{52}{3} \\ f(x) &= \frac{x^3}{3}-x^2-8x \\ f'(x) &= x^2-2x-8 \\ f'(x) &= 2^2-2(2)-8 \\ f'(x) &= -8 \\ (y-y_1) &= m (x-x_1) \\ y-(-\frac{52}{3}) &= -8 (x-2) \\ y+\frac{52}{3} &= -8x+16 \\ y &= -8x-\frac{4}{3} \\ \end{align} </math> </div></div> # Tentukan persamaan garis singgung kurva <math>f(x)=3x^2-11x+10</math> dan: ## sejajar dengan 7x-y=21! ## tegak lurus dengan 5y-x=10! :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{align} 7x-y &= 21 \\ y &= 7x-21 \\ m_1 &= 7 \\ \text{karena bergradien sejajar maka } m_2 = m_1 \\ m_2 &= m_1 \\ m_2 &= 7 \\ f(x) &= 3x^2-11x+10 \\ f'(x) &= 6x-11 \\ 7 &= 6x-11 \\ 6x &= 18 \\ x &= 3 \\ \text{untuk mencari nilai y dari 3 } \\ f(x) &= 3x^2-11x+10 \\ &= 3(3)^2-11(3)+10 \\ &= 4 \\ (y-y_1) &= m (x-x_1) \\ y-4 &= 7 (x-3) \\ y-4 &= 7x-21 \\ y &= 7x-17 \\ \end{align} </math> </div></div> ## <div class="toccolours mw-collapsible mw-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} 5y-x &= 10 \\ 5y &= x+10 \\ y &= \frac{x}{5}+2 \\ m_1 &= \frac{1}{5} \\ \text{karena bergradien tegak lurus maka } m_2 = -\frac{1}{m_1} \\ m_2 &= -\frac{1}{\frac{1}{5}} \\ m_2 &= -5 \\ f(x) &= 3x^2-11x+10 \\ f'(x) &= 6x-11 \\ -5 &= 6x-11 \\ 6x &= 6 \\ x &= 1 \\ \text{untuk mencari nilai y dari 1 } \\ f(x) &= 3x^2-11x+10 \\ &= 3(1)^2-11(1)+10 \\ &= 2 \\ (y-y_1) &= m (x-x_1) \\ y-2 &= -5 (x-1) \\ y-2 &= -5x+5 \\ y &= -5x+7 \\ \end{align} </math> </div></div> # Sehelai karton berbentuk persegipanjang dengan ukuran 45 x 24 cm. Karton ini akan dibuat kotak tanpa tutup dengan cara memotong keempat pojoknya berupa persegi dan melipatnya. Tentukan ukuran kotak agar volume maksimum! <div class="toccolours mw-collapsible mw-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{Misal tinggi adalah x } \\ t = x, p &= 45-2x, l = 24-2x \\ v &= (45-2x) \cdot (24-2x) \cdot x \\ &= 4x^3-138x^2-1080x \\ \text{agar volume maksimum adalah v'(x) = 0} \\ v(x) &= 4x^3-138x^2-1080x \\ v'(x) &= 12x^2-376x-1080 \\ v'(x) &= 0 \\ 12x^2-376x-1080 &= 0 \\ 12(x^2-23x-90) &= 0 \\ x^2-23x-90 &= 0 \\ (x-18)(x-5) &= 0 \\ x = 18 &\text{ atau } x = 5 \\ \end{align} </math> jadi yang memenuhi x adalah 5 maka ukurannya adalah 35x14x5 cm </div></div> # Sebuah kawat berbentuk persegipanjang dengan ukuran p x l cm. Lebar kawat ini akan dibuat kawat baru dengan cara memasangnya dengan cara setengah ukuran lebar masing-masing dari ujung ke ujung sehingga kelilingnya adalah 130 cm. Tentukan ukuran kawat agar luas maksimum! <div class="toccolours mw-collapsible mw-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{Misal panjang adalah p dan lebar adalah l } \\ k &= 3p + 4l \\ 130 &= 3p + 4l \\ 4l &= 130 - 3p \\ l &= 32,5-0,75p \\ L(x) &= pl \\ &= p(32,5-0,75p) \\ &= 32,5p-0,75p^2 \\ \text{agar luas maksimum adalah L'(x) = 0} \\ L(x) &= 32,5p-0,75p^2 \\ L'(x) &= 32,5-1,5p \\ L'(x) &= 0 \\ 32,5-1,5p &= 0 \\ p(32,5-1,5p) &= 0 \\ p = 0 &\text{ atau } p = \frac{65}{3} \\ l &= 32,5-0,75(\frac{65}{3}) \\ &= 32,5-16,25 \\ &= 16,25 \\ \end{align} </math> jadi ukurannya adalah <math>\frac{65}{3}x\frac{65}{4}</math> cm </div></div> # Jumlah kedua bilangan adalah 40. Tentukan hasil kali agar nilainya maksimum! <div class="toccolours mw-collapsible mw-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{Misal kedua bilangan masing-masing adalah x dan y } \\ x + y &= 40 \\ y &= 40 - x \\ \text{agar nilai hasil kali maksimum adalah h'(x) = 0 } \\ h(x) &= x \cdot y \\ h(x) &= x \cdot (40 - x) \\ h(x) &= 40x - x^2 \\ h'(x) &= 40 - 2x \\ h'(x) &= 0 \\ 40 - 2x &= 0 \\ 2x &= 40 \\ x &= 20 \\ h(x) &= 20 \cdot (40 - 20) \\ &= 20 \cdot 20 \\ &= 400 \\ \end{align} </math> jadi hasil kalinya adalah 400 </div></div> * Tentukan hasil turunan pertama dari x^x! <div class="toccolours mw-collapsible mw-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^x \\ ln y &= ln x^x \\ ln y &= x ln x \\ \frac{1}{y} y' &= ln x + x \frac{1}{x} \\ y' &= y (ln x + 1) \\ &= x^x (ln x + 1) \\ \end{align} </math> </div></div> * Tentukan hasil turunan pertama dari x³-y²=15! ;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} x^3-y^2 &= 15 \\ y^2 &= x^3-15 \\ y &= \sqrt{x^3-15} \\ y' &= \frac{3x^2}{2 \sqrt{x^3-15}} \\ &= \frac{3x^2 \sqrt{x^3-15}}{2(x^3-15)} \\ &= \frac{3x^2 \sqrt{x^3-15}}{2x^3-30} \\ \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} \text{untuk mencari nilai y maka } \\ x^3-y^2 &= 15 \\ y^2 &= x^3-15 \\ y &= \sqrt{x^3-15} \\ x^3-y^2 &= 15 \\ 3x^2-2yy' &= 0 \\ 2yy' &= 3x^2 \\ y' &= \frac{3x^2}{2y} \\ &= \frac{3x^2}{2 \sqrt{x^3-15}} \\ &= \frac{3x^2 \sqrt{x^3-15}}{2(x^3-15)} \\ &= \frac{3x^2 \sqrt{x^3-15}}{2x^3-30} \\ \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} x^3-y^2 &= 15 \\ x^3-y^2-15 &= 0 \\ y'(x) &= 3x^2 \\ y'(y) &= -2y \\ y' = \frac{dy}{dx} &= -\frac{y'(x)}{y'(y)} \\ &= -\frac{3x^2}{-2y} \\ &= \frac{3x^2}{2y} \\ \text{untuk mencari nilai y maka } \\ x^3-y^2 &= 15 \\ y^2 &= x^3-15 \\ y &= \sqrt{x^3-15} \\ y' &= \frac{3x^2}{2y} \\ &= \frac{3x^2}{2 \sqrt{x^3-15}} \\ &= \frac{3x^2 \sqrt{x^3-15}}{2(x^3-15)} \\ &= \frac{3x^2 \sqrt{x^3-15}}{2x^3-30} \\ \end{align} </math> </div></div> * Tentukan hasil turunan pertama dari 3x²y+7y=x³-2! ;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} 3x^2y+7y &= x^3-2 \\ y(3x^2+7) &= x^3-2 \\ y &= \frac{x^3-2}{3x^2+7} \\ y' &= \frac{3x^2(3x^2+7)-(x^3-2)6x}{(3x^2+7)^2} \\ &= \frac{9x^4+21x^2-6x^4+12x}{81x^8+756x^6+2.646x^4+4.116x^2+2.401} \\ &= \frac{3x^4+21x^2+12x}{81x^8+756x^6+2.646x^4+4.116x^2+2.401} \\ \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} \text{untuk mencari nilai y maka } \\ 3x^2y+7y &= x^3-2 \\ (3x^2+7)y &= x^3-2 \\ y &= \frac{x^3-2}{3x^2+7} \\ 3x^2y+7y &= x^3-2 \\ 6xy+3x^2y'+7y' &= 3x^2 \\ 3x^2y'+7y' &= 3x^2-6xy \\ (3x^2+7)y' &= 3x^2-6xy \\ y' &= \frac{3x^2-6xy}{3x^2+7} \\ &= \frac{3x^2-6x(\frac{x^3-2}{3x^2+7})}{3x^2+7} \\ &= \frac{3x^2(3x^2+7)-6x(x^3-2)}{(3x^2+7)^2} \\ &= \frac{9x^4+21x^2-6x^4+12x}{81x^8+756x^6+2.646x^4+4.116x^2+2.401} \\ &= \frac{3x^4+21x^2+12x}{81x^8+756x^6+2.646x^4+4.116x^2+2.401} \\ \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} 3x^2y+7y &= x^3-2 \\ 3x^2y+7y-x^3+2 &= 0 \\ y'(x) &= 6xy-3x^2 \\ y'(y) &= 3x^2+7 \\ y' = \frac{dy}{dx} &= -\frac{y'(x)}{y'(y)} \\ &= -\frac{6xy-3x^2}{3x^2+7} \\ &= \frac{3x^2-6xy}{3x^2+7} \\ \text{untuk mencari nilai y maka } \\ 3x^2y+7y &= x^3-2 \\ (3x^2+7)y &= x^3-2 \\ y &= \frac{x^3-2}{3x^2+7} \\ y' &= \frac{3x^2-6xy}{3x^2+7} \\ &= \frac{3x^2-6x(\frac{x^3-2}{3x^2+7})}{3x^2+7} \\ &= \frac{3x^2(3x^2+7)-6x(x^3-2)}{(3x^2+7)^2} \\ &= \frac{9x^4+21x^2-6x^4+12x}{81x^8+756x^6+2.646x^4+4.116x^2+2.401} \\ &= \frac{3x^4+21x^2+12x}{81x^8+756x^6+2.646x^4+4.116x^2+2.401} \\ \end{align} </math> </div></div> * Tentukan hasil turunan pertama dari x²y+xy²=x+2y! ;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} x^2y+xy^2 &= x+2y \\ 2xy+x^2y'+y^2+x2yy' &= 1+2y' \\ x^2y'+2xyy'-2y' &= 1-2xy-y^2 \\ (x^2+2xy-2)y' &= -y^2-2xy+1 \\ y' &= \frac{-y^2-2xy+1}{x^2+2xy-2} \\ \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} x^2y+xy^2 &= x+2y \\ x^2y+xy^2-x-2y &= 0 \\ y'(x) &= 2xy+y^2-1 \\ y'(y) &= x^2+x2y-2 \\ y' = \frac{dy}{dx} &= -\frac{y'(x)}{y'(y)} \\ &= -\frac{2xy+y^2-1}{x^2+x2y-2} \\ &= \frac{-y^2-2xy+1}{x^2+2xy-2} \\ \end{align} </math> [[Kategori:Soal-Soal Matematika]] 60i5edkyqtnj6onpb7mhgdm9uav2ah3 108054 108053 2025-07-03T02:31:51Z Akuindo 8654 /* Rumus sederhana */ 108054 wikitext text/x-wiki == Kaidah umum == :<math>\left({cf}\right)' = cf'</math> :<math>\left({f + g}\right)' = f' + g'</math> :<math>\left({f - g}\right)' = f' - g'</math> ;[[Kaidah darab]] :<math>\left({fg}\right)' = f'g + fg'</math> ;[[Kaidah timbalbalik]] :<math>\left(\frac{1}{f}\right)' = \frac{-f'}{f^2}, \qquad f \ne 0</math> ;[[Kaidah hasil-bagi]] :<math>\left({f \over g}\right)' = {f'g - fg' \over g^2}, \qquad g \ne 0</math> ;[[Kaidah rantai]] :<math>(f \circ g)' = (f' \circ g)g'</math> ;Turunan [[fungsi invers]] :<math>(f^{-1})' =\frac{1}{f' \circ f^{-1}}</math> untuk setiap fungsi terdiferensialkan ''f'' dengan argumen riil dan dengan nilai riil, bila komposisi dan invers ada ;Kaidah pangkat umum :<math>(f^g)'=f^g \left( g'\ln f + \frac{g}{f} f' \right)</math> == Rumus sederhana == : <math>c' = 0 \, </math> : <math>x' = 1 \, </math> : <math>(cx)' = c \, </math> : <math>|x|' = {x \over |x|} = \sgn x, \qquad x \ne 0</math> <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| \\ y^2 &= x^2 \\ 2y y' &= 2x \\ y' &= \frac{x}{y} \\ &= \frac{x}{|x|} \\ \end{align} </math> </div></div> : <math>(x^c)' = cx^{c-1} \qquad \mbox{baik } x^c \mbox{ maupun } cx^{c-1} \mbox { terdefinisi}</math> : <math>\left({1 \over x}\right)' = \left(x^{-1}\right)' = -x^{-2} = -{1 \over x^2}</math> : <math>\left({1 \over x^c}\right)' = \left(x^{-c}\right)' = -cx^{-(c+1)} = -{c \over x^{c+1}}</math> : <math>\left(\sqrt{x}\right)' = \left(x^{1\over 2}\right)' = {1 \over 2} x^{-{1\over 2}} = {1 \over 2 \sqrt{x}}, \qquad x > 0</math> : <math>\left(x^n\right)' = n \cdot x^{n-1}</math> : <math>\left(u^n\right)' = n \cdot u' \cdot u^{n-1}</math> ; Eksponen dan logaritma :<math> \left(c^x\right)' = c^x \ln c,\qquad c > 0</math> :<math> \left(e^x\right)' = e^x</math> :<math> \left(^c\log x\right)' = \frac{1}{x \ln c}, \qquad c > 0</math> :<math> \left(\ln x\right)' = \frac{1}{x}</math> ; Trigonometri {| style="width:100%; background:transparent; margin-left:2em;" |width=50%|<math> (\sin x)' = \cos x \,</math> |width=50%|<math> (\arcsin x)' = { 1 \over \sqrt{1 - x^2}}, x \neq \pm 1 \,</math> |- |<math> (\cos x)' = -\sin x \,</math> |<math> (\arccos x)' = {-1 \over \sqrt{1 - x^2}}, x \neq \pm 1 \,</math> |- |<math> (\tan x)' = \sec^2 x = { 1 \over \cos^2 x} = 1 + \tan^2 x \,</math> |<math> (\arctan x)' = { 1 \over 1 + x^2} \,</math> |- |<math> (\sec x)' = \sec x \tan x \,</math> |<math> (\arcsec x)' = { 1 \over |x|\sqrt{x^2 - 1}}, x \neq {\pm 1, 0} \,</math> |- |<math> (\csc x)' = -\csc x \cot x \,</math> |<math> (\arccsc x)' = {-1 \over |x|\sqrt{x^2 - 1}}, x \neq {\pm 1, 0} \,</math> |- |<math> (\cot x)' = -\csc^2 x = { -1 \over \sin^2 x} = -(1 + \cot^2 x) \,</math> |<math> (\arccot x)' = {-1 \over 1 + x^2} \,</math> |} * Tambahkan: ** <math> (\sin^n x)' = n \sin^{n-1} x cos x \,</math> ** <math> (\sin u)' = u' \cos u \,</math> ** <math> (\sin^n u)' = n u' \sin^{n-1} u cos u \,</math> ; Hiperbolik Perhatikan sebagai berikut : <math>sinh x = \frac{e^x - e^{-x}}{2}</math> : <math>cosh x = \frac{e^x + e^{-x}}{2}</math> {| style="width:100%; background:transparent; margin-left:2em;" |width=50%|<math>(\sinh x )'= \cosh x</math> |width=50%|<math>(\operatorname{arcsinh}\,x)' = { 1 \over \sqrt{x^2 + 1}}</math> |- |<math>(\cosh x )'= \sinh x</math> |<math>(\operatorname{arccosh}\,x)' = { 1 \over \sqrt{x^2 - 1}}, x>1</math> |- |<math>(\tanh x )'= \operatorname{sech}^2\,x</math> |<math>(\operatorname{arctanh}\,x)' = { 1 \over 1 - x^2}, |x|<1</math> |- |<math>(\operatorname{sech}\,x)' = - \tanh x\,\operatorname{sech}\,x</math> |<math>(\operatorname{arcsech}\,x)' = {-1 \over x\sqrt{1 - x^2}}, 0<x<1</math> |- |<math>(\operatorname{csch}\,x)' = -\,\operatorname{coth}\,x\,\operatorname{csch}\,x</math> |<math>(\operatorname{arccsch}\,x)' = {-1 \over |x|\sqrt{1 + x^2}}, x \neq 0</math> |- |<math>(\operatorname{coth}\,x )' = -\,\operatorname{csch}^2\,x</math> |<math>(\operatorname{arccoth}\,x)' = { 1 \over 1-x^2}, |x|>1</math> |} :: catatan: jika x diganti u maka merumuskan seperti trigonometri. ; implisit *cara 1 : <math>ax + by = 1</math> : <math>a + b y' = 0</math> : <math>y' = -\frac{a}{b}</math> *cara 2 persamaan F(x,y) dibuat hasil nol kemudian diubah menjadi <math>\frac{d_y}{d_x} = - \frac{F_x(x,y)}{F_y(x,y)}</math> Fungsi implisit dibagi 2 jenis yaitu: # eksplisit artinya fungsi implisit dapat diubah menjadi fungsi eksplisit. Contoh: x²y+7=8xy+5x, y²-8x=5-6y # in-eksplisit artinya fungsi implisit tidak dapat diubah menjadi fungsi eksplisit. Contoh: xy+8x=y³-11 == Laju perubahan (rata-rata) == Rumus laju perubahan f(x) pada interval x1 dan x2 adalah V(rata-rata) = <math>\frac{\Delta y}{\Delta x} = \frac{f'(x_2)-f'(x_1)}{x_2-x_1}</math> == Nilai ekstrem, interval serta titik stasioner/diam == : Persamaan kuadrat :: Nilai minimum persamaan kuadrat adalah titik terendah (titik stasioner/diam) serta intervalnya turun-naik. :: Nilai maksimum persamaan kuadrat adalah titik tertinggi (titik stasioner/diam) serta intervalnya naik-turun. : Persamaan kubik Ada 4 kemungkinan yang berdasarkan interval sebagai berikut: :: Nilai maksimum-minimum dan intervalnya naik-turun-naik. :: Nilai maksimum-minimum dan intervalnya turun secara monoton. :: Nilai minimum-maksimum dan intervalnya turun-naik-turun. :: Nilai maksimum-minimum dan intervalnya naik secara monoton. Untuk mengetahui nilainya harus turunan kedua (jika kuadrat berarti hasilnya konstanta sedangkan berpangkat lebih dari 2 berarti masukkan x dari hasil turunan pertama untuk memperoleh hasilnya). Jika konstanta > 0 maka itu berarti nilai minimum, konstanta < 0 maka itu berarti nilai maksimum serta konstanta = 0 itu berarti titik belok. Posisi gradien adalah turunan pertama bernilai nol untuk mencari nilai x tersebut. Titik belok berarti turunan kedua bernilai nol tapi turunan ketiga tidak boleh bernilai nol. == Persamaan garis singgung kurva == Gradien (m) pada Persamaan garis singgung kurva y = f(x) pada titik A (a, f(a)) adalah m = f’(a) = <math>\lim\limits_{\Delta x \to 0} \frac{f(a+\Delta x)-f(a)}{\Delta x}</math> contoh: # Berapa laju perubahaan f(x)=x²-4x+3 pada: ## x=5! ## interval 2<x<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(x) &= x^2-4x+3 \\ f'(x) &= 2x-4 \\ v &= \frac{f'(x)}{x} \\ &= \frac{f'(5)}{5} \\ &= \frac{2(5)-4}{5} \\ &= \frac{6}{5} \\ &= 1.2 \\ v &= \frac{f'(x_2)-f'(x_1)}{x_2-x_1} \\ &= \frac{f'(3)-f'(2)}{3-2} \\ &= \frac{2(3)-4-(2(2)-5)}{3-2} \\ &= \frac{3}{1} \\ &= 3 \\ \end{align} </math> </div></div> # Berapa laju perubahaan f(x)=sin x-cos x pada: ## x=<math>\frac{\pi}{6}</math>! ## interval <math>0<x<\frac{\pi}{2}</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) &= sin x-cos x \\ f'(x) &= cos x+sin x \\ v &= \frac{f'(x)}{x} \\ &= \frac{f'(\frac{\pi}{6})}{\frac{\pi}{6}} \\ &= \frac{cos \frac{\pi}{6}+sin \frac{\pi}{6}}{\frac{\pi}{6}} \\ &= \frac{\frac{\pi \sqrt{3}}{2}+\frac{\pi}{2}}{\frac{\pi}{6}} \\ &= \frac{\frac{\pi}{2} (\sqrt{3}+1)}{\frac{\pi}{6}} \\ &= 3 (\sqrt{3}+1) \\ v &= \frac{f'(x_2)-f'(x_1)}{x_2-x_1} \\ &= \frac{f'(\frac{\pi}{2})-f'(0)}{\frac{\pi}{2}-0} \\ &= \frac{cos \frac{\pi}{2}+sin \frac{\pi}{2}-(cos 0+sin 0)}{\frac{\pi}{2}-0} \\ &= \frac{(0+1)-(1+0)}{\frac{\pi}{2}} \\ &= \frac{0}{\frac{\pi}{2}} \\ &= 0 \\ \end{align} </math> </div></div> # Tentukan nilai ekstrem, titik stasioner, titik belok serta interval dari <math>f(x)=\frac{x^3}{3}-3x^2-16x+36</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{align} f(x) &= \frac{x^3}{3}-3x^2-16x+36 \\ f'(x) &= x^2-6x-16 \\ \text{untuk menentukan nilai x ketika f'(x)=0 } \\ x^2-6x-16 &= 0 \\ (x+2)(x-8) &= 0 \\ x=-2 &\text{ atau } x=8 \\ \text{ menentukan nilai ekstrem } \\ f''(x) &= 2x-6 \\ f''(-2) &= 2(2)-6 = -2 < 0 \\ f''(8) &= 2(8)-6 = 10 > 0 \\ \text{f''(-2) negatif maka nilai maksimum sedangkan f''(8) positif maka nilai minimum } \\ \text{untuk menentukan titik stasionernya adalah } \\ f(-2) &= \frac{(-2)^3}{3}-3(-2)^2-16(-2)+36 \\ &= \frac{160}{3} \\ f(8) &= \frac{8^3}{3}-3(8)^2-16(8)+36 \\ &= -\frac{1.386}{3} \\ \text{jadi titik stasionernya adalah } (-2,\frac{160}{3}) \text{ dan } (8,-\frac{1.386}{3}) \\ \text{untuk menentukan titik beloknya ketika f''(x)=0 } \\ f''(x) &= 2x-6 \\ 2x-6 &= 0 \\ x &= 3 \\ \text{untuk menentukan titik beloknya adalah } \\ f(3) &= \frac{3^3}{3}-3(3)^2-16(3)+36 \\ f(3) &= -30 \\ \text{jadi titik beloknya adalah } (3, -30) \\ \text{untuk menentukan intervalnya, buatlah irisan pada titik stasionernya dengan bantuan x = (-4; 0; 9) maka naik jika } x<-2 \text{ atau } x>8 \text{ dan turun jika } -2<x<8 \\ \end{align} </math> </div></div> # Tentukan nilai ekstrem, titik stasioner, titik belok serta interval dari <math>f(x)=1-2cos 2x</math> pada batas-batas interval <math>0<x<\frac{\pi}{2}</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{align} f(x) &= 1-2cos 2x \\ f'(x) &= 4sin 2x \\ \text{untuk menentukan nilai x ketika f'(x)=0 } \\ 4sin 2x &= 0 \\ 8sin x \cdot cos x &= 0 \\ (sin x)(cos x) &= 0 \\ x=0 &\text{ atau } x=\frac{\pi}{2} \\ \text{ menentukan nilai ekstrem } \\ f''(x) &= 8cos 2x \\ f''(0) &= 8cos 2(0) = 8 > 0 \\ f''(\frac{\pi}{2}) &= 8cos 2(\frac{\pi}{2}) = -8 < 0 \\ \text{f''(0) positif maka nilai minimum sedangkan } f''(\frac{\pi}{2}) \text{ negatif maka nilai maksimum } \\ \text{untuk menentukan titik stasionernya adalah } \\ f(0) &= 1-2cos 2(0) \\ &= -1 \\ f(\frac{\pi}{2}) &= 1-2cos 2(\frac{\pi}{2}) \\ &= 3 \\ \text{jadi titik stasionernya adalah } (0,-1) \text{ dan } (\frac{\pi}{2},3) \\ \text{untuk menentukan titik beloknya ketika f''(x)=0 } \\ f''(x) &= 8cos 2x \\ 8cos 2x &= 0 \\ cos 2x &= 0 \\ cos 2x &= cos \frac{\pi}{2} \\ 2x &= \frac{\pi}{2} \\ x &= \frac{\pi}{4} \\ \text{untuk menentukan titik beloknya adalah } \\ f(\frac{\pi}{4}) &= 1-2cos 2(\frac{\pi}{4}) \\ f(\frac{\pi}{4}) &= 1 \\ \text{jadi titik beloknya adalah } (\frac{\pi}{4}, 1) \\ \text{kondisi intervalnya pada batas-batas tersebut adalah naik } \\ \end{align} </math> </div></div> # Tentukan persamaan garis singgung kurva <math>f(x)=\frac{x^3}{3}-x^2-8x</math> pada: ## di titik (-3,-3)! ## berabsis 2! :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{align} f(x) &= \frac{x^3}{3}-x^2-8x \\ f'(x) &= x^2-2x-8 \\ f'(x) &= (-3)^2-2(3)-8 \\ f'(x) &= -5 \\ (y-y_1) &= m (x-x_1) \\ y-(-3) &= -5 (x-(-3)) \\ y+3 &= -5x-15 \\ y &= -5x-18 \\ \end{align} </math> </div></div> ## <div class="toccolours mw-collapsible mw-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{untuk mengetahui y dari berabsis (x) 2 yaitu } \frac{2^3}{3}-2^2-8(2) = -\frac{52}{3} \\ f(x) &= \frac{x^3}{3}-x^2-8x \\ f'(x) &= x^2-2x-8 \\ f'(x) &= 2^2-2(2)-8 \\ f'(x) &= -8 \\ (y-y_1) &= m (x-x_1) \\ y-(-\frac{52}{3}) &= -8 (x-2) \\ y+\frac{52}{3} &= -8x+16 \\ y &= -8x-\frac{4}{3} \\ \end{align} </math> </div></div> # Tentukan persamaan garis singgung kurva <math>f(x)=3x^2-11x+10</math> dan: ## sejajar dengan 7x-y=21! ## tegak lurus dengan 5y-x=10! :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{align} 7x-y &= 21 \\ y &= 7x-21 \\ m_1 &= 7 \\ \text{karena bergradien sejajar maka } m_2 = m_1 \\ m_2 &= m_1 \\ m_2 &= 7 \\ f(x) &= 3x^2-11x+10 \\ f'(x) &= 6x-11 \\ 7 &= 6x-11 \\ 6x &= 18 \\ x &= 3 \\ \text{untuk mencari nilai y dari 3 } \\ f(x) &= 3x^2-11x+10 \\ &= 3(3)^2-11(3)+10 \\ &= 4 \\ (y-y_1) &= m (x-x_1) \\ y-4 &= 7 (x-3) \\ y-4 &= 7x-21 \\ y &= 7x-17 \\ \end{align} </math> </div></div> ## <div class="toccolours mw-collapsible mw-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} 5y-x &= 10 \\ 5y &= x+10 \\ y &= \frac{x}{5}+2 \\ m_1 &= \frac{1}{5} \\ \text{karena bergradien tegak lurus maka } m_2 = -\frac{1}{m_1} \\ m_2 &= -\frac{1}{\frac{1}{5}} \\ m_2 &= -5 \\ f(x) &= 3x^2-11x+10 \\ f'(x) &= 6x-11 \\ -5 &= 6x-11 \\ 6x &= 6 \\ x &= 1 \\ \text{untuk mencari nilai y dari 1 } \\ f(x) &= 3x^2-11x+10 \\ &= 3(1)^2-11(1)+10 \\ &= 2 \\ (y-y_1) &= m (x-x_1) \\ y-2 &= -5 (x-1) \\ y-2 &= -5x+5 \\ y &= -5x+7 \\ \end{align} </math> </div></div> # Sehelai karton berbentuk persegipanjang dengan ukuran 45 x 24 cm. Karton ini akan dibuat kotak tanpa tutup dengan cara memotong keempat pojoknya berupa persegi dan melipatnya. Tentukan ukuran kotak agar volume maksimum! <div class="toccolours mw-collapsible mw-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{Misal tinggi adalah x } \\ t = x, p &= 45-2x, l = 24-2x \\ v &= (45-2x) \cdot (24-2x) \cdot x \\ &= 4x^3-138x^2-1080x \\ \text{agar volume maksimum adalah v'(x) = 0} \\ v(x) &= 4x^3-138x^2-1080x \\ v'(x) &= 12x^2-376x-1080 \\ v'(x) &= 0 \\ 12x^2-376x-1080 &= 0 \\ 12(x^2-23x-90) &= 0 \\ x^2-23x-90 &= 0 \\ (x-18)(x-5) &= 0 \\ x = 18 &\text{ atau } x = 5 \\ \end{align} </math> jadi yang memenuhi x adalah 5 maka ukurannya adalah 35x14x5 cm </div></div> # Sebuah kawat berbentuk persegipanjang dengan ukuran p x l cm. Lebar kawat ini akan dibuat kawat baru dengan cara memasangnya dengan cara setengah ukuran lebar masing-masing dari ujung ke ujung sehingga kelilingnya adalah 130 cm. Tentukan ukuran kawat agar luas maksimum! <div class="toccolours mw-collapsible mw-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{Misal panjang adalah p dan lebar adalah l } \\ k &= 3p + 4l \\ 130 &= 3p + 4l \\ 4l &= 130 - 3p \\ l &= 32,5-0,75p \\ L(x) &= pl \\ &= p(32,5-0,75p) \\ &= 32,5p-0,75p^2 \\ \text{agar luas maksimum adalah L'(x) = 0} \\ L(x) &= 32,5p-0,75p^2 \\ L'(x) &= 32,5-1,5p \\ L'(x) &= 0 \\ 32,5-1,5p &= 0 \\ p(32,5-1,5p) &= 0 \\ p = 0 &\text{ atau } p = \frac{65}{3} \\ l &= 32,5-0,75(\frac{65}{3}) \\ &= 32,5-16,25 \\ &= 16,25 \\ \end{align} </math> jadi ukurannya adalah <math>\frac{65}{3}x\frac{65}{4}</math> cm </div></div> # Jumlah kedua bilangan adalah 40. Tentukan hasil kali agar nilainya maksimum! <div class="toccolours mw-collapsible mw-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{Misal kedua bilangan masing-masing adalah x dan y } \\ x + y &= 40 \\ y &= 40 - x \\ \text{agar nilai hasil kali maksimum adalah h'(x) = 0 } \\ h(x) &= x \cdot y \\ h(x) &= x \cdot (40 - x) \\ h(x) &= 40x - x^2 \\ h'(x) &= 40 - 2x \\ h'(x) &= 0 \\ 40 - 2x &= 0 \\ 2x &= 40 \\ x &= 20 \\ h(x) &= 20 \cdot (40 - 20) \\ &= 20 \cdot 20 \\ &= 400 \\ \end{align} </math> jadi hasil kalinya adalah 400 </div></div> * Tentukan hasil turunan pertama dari x^x! <div class="toccolours mw-collapsible mw-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^x \\ ln y &= ln x^x \\ ln y &= x ln x \\ \frac{1}{y} y' &= ln x + x \frac{1}{x} \\ y' &= y (ln x + 1) \\ &= x^x (ln x + 1) \\ \end{align} </math> </div></div> * Tentukan hasil turunan pertama dari x³-y²=15! ;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} x^3-y^2 &= 15 \\ y^2 &= x^3-15 \\ y &= \sqrt{x^3-15} \\ y' &= \frac{3x^2}{2 \sqrt{x^3-15}} \\ &= \frac{3x^2 \sqrt{x^3-15}}{2(x^3-15)} \\ &= \frac{3x^2 \sqrt{x^3-15}}{2x^3-30} \\ \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} \text{untuk mencari nilai y maka } \\ x^3-y^2 &= 15 \\ y^2 &= x^3-15 \\ y &= \sqrt{x^3-15} \\ x^3-y^2 &= 15 \\ 3x^2-2yy' &= 0 \\ 2yy' &= 3x^2 \\ y' &= \frac{3x^2}{2y} \\ &= \frac{3x^2}{2 \sqrt{x^3-15}} \\ &= \frac{3x^2 \sqrt{x^3-15}}{2(x^3-15)} \\ &= \frac{3x^2 \sqrt{x^3-15}}{2x^3-30} \\ \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} x^3-y^2 &= 15 \\ x^3-y^2-15 &= 0 \\ y'(x) &= 3x^2 \\ y'(y) &= -2y \\ y' = \frac{dy}{dx} &= -\frac{y'(x)}{y'(y)} \\ &= -\frac{3x^2}{-2y} \\ &= \frac{3x^2}{2y} \\ \text{untuk mencari nilai y maka } \\ x^3-y^2 &= 15 \\ y^2 &= x^3-15 \\ y &= \sqrt{x^3-15} \\ y' &= \frac{3x^2}{2y} \\ &= \frac{3x^2}{2 \sqrt{x^3-15}} \\ &= \frac{3x^2 \sqrt{x^3-15}}{2(x^3-15)} \\ &= \frac{3x^2 \sqrt{x^3-15}}{2x^3-30} \\ \end{align} </math> </div></div> * Tentukan hasil turunan pertama dari 3x²y+7y=x³-2! ;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} 3x^2y+7y &= x^3-2 \\ y(3x^2+7) &= x^3-2 \\ y &= \frac{x^3-2}{3x^2+7} \\ y' &= \frac{3x^2(3x^2+7)-(x^3-2)6x}{(3x^2+7)^2} \\ &= \frac{9x^4+21x^2-6x^4+12x}{81x^8+756x^6+2.646x^4+4.116x^2+2.401} \\ &= \frac{3x^4+21x^2+12x}{81x^8+756x^6+2.646x^4+4.116x^2+2.401} \\ \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} \text{untuk mencari nilai y maka } \\ 3x^2y+7y &= x^3-2 \\ (3x^2+7)y &= x^3-2 \\ y &= \frac{x^3-2}{3x^2+7} \\ 3x^2y+7y &= x^3-2 \\ 6xy+3x^2y'+7y' &= 3x^2 \\ 3x^2y'+7y' &= 3x^2-6xy \\ (3x^2+7)y' &= 3x^2-6xy \\ y' &= \frac{3x^2-6xy}{3x^2+7} \\ &= \frac{3x^2-6x(\frac{x^3-2}{3x^2+7})}{3x^2+7} \\ &= \frac{3x^2(3x^2+7)-6x(x^3-2)}{(3x^2+7)^2} \\ &= \frac{9x^4+21x^2-6x^4+12x}{81x^8+756x^6+2.646x^4+4.116x^2+2.401} \\ &= \frac{3x^4+21x^2+12x}{81x^8+756x^6+2.646x^4+4.116x^2+2.401} \\ \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} 3x^2y+7y &= x^3-2 \\ 3x^2y+7y-x^3+2 &= 0 \\ y'(x) &= 6xy-3x^2 \\ y'(y) &= 3x^2+7 \\ y' = \frac{dy}{dx} &= -\frac{y'(x)}{y'(y)} \\ &= -\frac{6xy-3x^2}{3x^2+7} \\ &= \frac{3x^2-6xy}{3x^2+7} \\ \text{untuk mencari nilai y maka } \\ 3x^2y+7y &= x^3-2 \\ (3x^2+7)y &= x^3-2 \\ y &= \frac{x^3-2}{3x^2+7} \\ y' &= \frac{3x^2-6xy}{3x^2+7} \\ &= \frac{3x^2-6x(\frac{x^3-2}{3x^2+7})}{3x^2+7} \\ &= \frac{3x^2(3x^2+7)-6x(x^3-2)}{(3x^2+7)^2} \\ &= \frac{9x^4+21x^2-6x^4+12x}{81x^8+756x^6+2.646x^4+4.116x^2+2.401} \\ &= \frac{3x^4+21x^2+12x}{81x^8+756x^6+2.646x^4+4.116x^2+2.401} \\ \end{align} </math> </div></div> * Tentukan hasil turunan pertama dari x²y+xy²=x+2y! ;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} x^2y+xy^2 &= x+2y \\ 2xy+x^2y'+y^2+x2yy' &= 1+2y' \\ x^2y'+2xyy'-2y' &= 1-2xy-y^2 \\ (x^2+2xy-2)y' &= -y^2-2xy+1 \\ y' &= \frac{-y^2-2xy+1}{x^2+2xy-2} \\ \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} x^2y+xy^2 &= x+2y \\ x^2y+xy^2-x-2y &= 0 \\ y'(x) &= 2xy+y^2-1 \\ y'(y) &= x^2+x2y-2 \\ y' = \frac{dy}{dx} &= -\frac{y'(x)}{y'(y)} \\ &= -\frac{2xy+y^2-1}{x^2+x2y-2} \\ &= \frac{-y^2-2xy+1}{x^2+2xy-2} \\ \end{align} </math> [[Kategori:Soal-Soal Matematika]] 0khotk3vmw16jygukmvumkzigr6oxxp 0 23140 108057 108051 2025-07-03T10:51:32Z Akuindo 8654 /* Integral lipat */ 108057 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_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. == 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-x^2}} = {1 \over a}\arcsin {x \over a} + C</math> :<math>\int {dx \over {a^2+x^2}} = {1 \over a}\arctan {x \over a} + C</math> <math>\int {dx \over x\sqrt{x^2-a^2}} = {1 \over a}\arcsec {x \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 \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| \sec {x} \right|} + C</math> :<math>\int \cot{x} \, dx = -\ln{\left| \csc{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| \cosh x | + C</math> :<math>\int \coth x \, dx = \ln| \sinh x | + C</math> :<math>\int \mbox{sech}\,x \, dx = \arctan (\sinh x) + C</math> :<math>\int \mbox{csch}\,x \, dx = \ln\left| \tanh {x \over2}\right| + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : <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> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : <math>\int \frac{\ln(x)}{x}\,dx</math> : <math>t = \ln(x),\, dt = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, : <math> \begin{aligned} &\; \int \frac{\ln(x)}{x}\,dx \\ =&\; \int t\,dt \\ =&\; \frac{1}{2} t^2 + C \\ =&\; \frac{1}{2} \ln^2 x + C \end{aligned} </math> === 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. : <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, : <math> \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> ;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. : <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, : <math> \begin{aligned} &\; \int x \sin(x)\,dx \\ =&\; (x)(-\cos(x)) - (1)(-\sin(x)) + C \\ =&\; -x \cos(x) + \sin(x) + C \end{aligned} </math> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{dx}{x^2 - 4}</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{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 !! 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}}{4 x} + C \end{aligned} </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> == 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{25-x^2}} 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> # 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]] j0wa9svljclnnq8lr36vnxg38mkc3ym 108058 108057 2025-07-03T10:59:02Z Akuindo 8654 /* Integral lipat */ 108058 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_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. == 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-x^2}} = {1 \over a}\arcsin {x \over a} + C</math> :<math>\int {dx \over {a^2+x^2}} = {1 \over a}\arctan {x \over a} + C</math> <math>\int {dx \over x\sqrt{x^2-a^2}} = {1 \over a}\arcsec {x \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 \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| \sec {x} \right|} + C</math> :<math>\int \cot{x} \, dx = -\ln{\left| \csc{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| \cosh x | + C</math> :<math>\int \coth x \, dx = \ln| \sinh x | + C</math> :<math>\int \mbox{sech}\,x \, dx = \arctan (\sinh x) + C</math> :<math>\int \mbox{csch}\,x \, dx = \ln\left| \tanh {x \over2}\right| + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : <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> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : <math>\int \frac{\ln(x)}{x}\,dx</math> : <math>t = \ln(x),\, dt = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, : <math> \begin{aligned} &\; \int \frac{\ln(x)}{x}\,dx \\ =&\; \int t\,dt \\ =&\; \frac{1}{2} t^2 + C \\ =&\; \frac{1}{2} \ln^2 x + C \end{aligned} </math> === 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. : <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, : <math> \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> ;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. : <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, : <math> \begin{aligned} &\; \int x \sin(x)\,dx \\ =&\; (x)(-\cos(x)) - (1)(-\sin(x)) + C \\ =&\; -x \cos(x) + \sin(x) + C \end{aligned} </math> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{dx}{x^2 - 4}</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{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 !! 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}}{4 x} + C \end{aligned} </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> == 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{25-x^2}} 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{25-x^2}} dx \\ x &= 5 sin A \\ dx &= 5 cos A dA \\ \int \frac{1}{\sqrt{25-x^2}} dx &= \int \frac{5 cos A}{\sqrt{25-(5 sin A)^2}} dA \\ &= \int \frac{5 cos A}{\sqrt{25-(5 sin A)^2}} dA \\ &= \int \frac{5 cos A}{\sqrt{25-25 sin^2 A}} dA \\ &= \int \frac{5 cos A}{\sqrt{25(1-sin^2 A)}} dA \\ &= \int \frac{5 cos A}{\sqrt{25cos^2 A}} dA \\ &= \int \frac{5 cos A}{5 cos A} dA \\ &= \int dA \\ \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]] nlysgu96lyrokgipnifvvl0nods3dpb 108059 108058 2025-07-03T11:02:09Z Akuindo 8654 /* Integral lipat */ 108059 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_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. == 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-x^2}} = {1 \over a}\arcsin {x \over a} + C</math> :<math>\int {dx \over {a^2+x^2}} = {1 \over a}\arctan {x \over a} + C</math> <math>\int {dx \over x\sqrt{x^2-a^2}} = {1 \over a}\arcsec {x \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 \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| \sec {x} \right|} + C</math> :<math>\int \cot{x} \, dx = -\ln{\left| \csc{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| \cosh x | + C</math> :<math>\int \coth x \, dx = \ln| \sinh x | + C</math> :<math>\int \mbox{sech}\,x \, dx = \arctan (\sinh x) + C</math> :<math>\int \mbox{csch}\,x \, dx = \ln\left| \tanh {x \over2}\right| + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : <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> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : <math>\int \frac{\ln(x)}{x}\,dx</math> : <math>t = \ln(x),\, dt = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, : <math> \begin{aligned} &\; \int \frac{\ln(x)}{x}\,dx \\ =&\; \int t\,dt \\ =&\; \frac{1}{2} t^2 + C \\ =&\; \frac{1}{2} \ln^2 x + C \end{aligned} </math> === 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. : <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, : <math> \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> ;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. : <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, : <math> \begin{aligned} &\; \int x \sin(x)\,dx \\ =&\; (x)(-\cos(x)) - (1)(-\sin(x)) + C \\ =&\; -x \cos(x) + \sin(x) + C \end{aligned} </math> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{dx}{x^2 - 4}</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{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 !! 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}}{4 x} + C \end{aligned} </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> == 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{25-x^2}} 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{25-x^2}} dx \\ x &= 5 sin A \\ dx &= 5 cos A dA \\ \int \frac{1}{\sqrt{25-x^2}} dx &= \int \frac{5 cos A}{\sqrt{25-(5 sin A)^2}} dA \\ &= \int \frac{5 cos A}{\sqrt{25-(5 sin A)^2}} dA \\ &= \int \frac{5 cos A}{\sqrt{25-25 sin^2 A}} dA \\ &= \int \frac{5 cos A}{\sqrt{25(1-sin^2 A)}} dA \\ &= \int \frac{5 cos A}{\sqrt{25cos^2 A}} dA \\ &= \int \frac{5 cos A}{5 cos A} dA \\ &= \int dA \\ &= A+C \\ A &= arc sin \frac{x}{5} \\ &= arc sin \frac{x}{5}+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]] dezeyghrr9uj0zflp7ho1qo1fekjuq9 108060 108059 2025-07-03T11:05:57Z Akuindo 8654 /* rumus sederhana */ 108060 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_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. == 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-x^2}} = \arcsin {x \over a} + C</math> :<math>\int {dx \over {a^2+x^2}} = {1 \over a}\arctan {x \over a} + C</math> <math>\int {dx \over x\sqrt{x^2-a^2}} = {1 \over a}\arcsec {x \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 \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| \sec {x} \right|} + C</math> :<math>\int \cot{x} \, dx = -\ln{\left| \csc{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| \cosh x | + C</math> :<math>\int \coth x \, dx = \ln| \sinh x | + C</math> :<math>\int \mbox{sech}\,x \, dx = \arctan (\sinh x) + C</math> :<math>\int \mbox{csch}\,x \, dx = \ln\left| \tanh {x \over2}\right| + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : <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> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : <math>\int \frac{\ln(x)}{x}\,dx</math> : <math>t = \ln(x),\, dt = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, : <math> \begin{aligned} &\; \int \frac{\ln(x)}{x}\,dx \\ =&\; \int t\,dt \\ =&\; \frac{1}{2} t^2 + C \\ =&\; \frac{1}{2} \ln^2 x + C \end{aligned} </math> === 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. : <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, : <math> \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> ;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. : <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, : <math> \begin{aligned} &\; \int x \sin(x)\,dx \\ =&\; (x)(-\cos(x)) - (1)(-\sin(x)) + C \\ =&\; -x \cos(x) + \sin(x) + C \end{aligned} </math> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{dx}{x^2 - 4}</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{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 !! 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}}{4 x} + C \end{aligned} </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> == 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{25-x^2}} 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{25-x^2}} dx \\ x &= 5 sin A \\ dx &= 5 cos A dA \\ \int \frac{1}{\sqrt{25-x^2}} dx &= \int \frac{5 cos A}{\sqrt{25-(5 sin A)^2}} dA \\ &= \int \frac{5 cos A}{\sqrt{25-(5 sin A)^2}} dA \\ &= \int \frac{5 cos A}{\sqrt{25-25 sin^2 A}} dA \\ &= \int \frac{5 cos A}{\sqrt{25(1-sin^2 A)}} dA \\ &= \int \frac{5 cos A}{\sqrt{25cos^2 A}} dA \\ &= \int \frac{5 cos A}{5 cos A} dA \\ &= \int dA \\ &= A+C \\ A &= arc sin \frac{x}{5} \\ &= arc sin \frac{x}{5}+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]] mby7p3t5rztvbimep30na3ma7coyz9w 108061 108060 2025-07-03T11:12:08Z Akuindo 8654 /* Integral lipat */ 108061 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_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. == 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-x^2}} = \arcsin {x \over a} + C</math> :<math>\int {dx \over {a^2+x^2}} = {1 \over a}\arctan {x \over a} + C</math> <math>\int {dx \over x\sqrt{x^2-a^2}} = {1 \over a}\arcsec {x \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 \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| \sec {x} \right|} + C</math> :<math>\int \cot{x} \, dx = -\ln{\left| \csc{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| \cosh x | + C</math> :<math>\int \coth x \, dx = \ln| \sinh x | + C</math> :<math>\int \mbox{sech}\,x \, dx = \arctan (\sinh x) + C</math> :<math>\int \mbox{csch}\,x \, dx = \ln\left| \tanh {x \over2}\right| + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : <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> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : <math>\int \frac{\ln(x)}{x}\,dx</math> : <math>t = \ln(x),\, dt = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, : <math> \begin{aligned} &\; \int \frac{\ln(x)}{x}\,dx \\ =&\; \int t\,dt \\ =&\; \frac{1}{2} t^2 + C \\ =&\; \frac{1}{2} \ln^2 x + C \end{aligned} </math> === 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. : <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, : <math> \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> ;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. : <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, : <math> \begin{aligned} &\; \int x \sin(x)\,dx \\ =&\; (x)(-\cos(x)) - (1)(-\sin(x)) + C \\ =&\; -x \cos(x) + \sin(x) + C \end{aligned} </math> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{dx}{x^2 - 4}</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{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 !! 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}}{4 x} + C \end{aligned} </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> == 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{25-x^2}} dx</math> * <math>\int \frac{1}{\sqrt{x^2-49}} 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{25-x^2}} dx \\ x &= 5 sin A \\ dx &= 5 cos A dA \\ \int \frac{1}{\sqrt{25-x^2}} dx &= \int \frac{5 cos A}{\sqrt{25-(5 sin A)^2}} dA \\ &= \int \frac{5 cos A}{\sqrt{25-(5 sin A)^2}} dA \\ &= \int \frac{5 cos A}{\sqrt{25-25 sin^2 A}} dA \\ &= \int \frac{5 cos A}{\sqrt{25(1-sin^2 A)}} dA \\ &= \int \frac{5 cos A}{\sqrt{25cos^2 A}} dA \\ &= \int \frac{5 cos A}{5 cos A} dA \\ &= \int dA \\ &= A+C \\ A &= arc sin \frac{x}{5} \\ &= arc sin \frac{x}{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}{\sqrt{x^2-49}} dx \\ x &= 7 tan A \\ dx &= 7 tan A sec A dA \\ \int \frac{1}{\sqrt{x^2-49}} dx &= \int \frac{7 tan A sec A}{\sqrt{(7 tan A)^2-49} dA \\ &= \int \frac{7 tan A sec A}{\sqrt{49 sin^2 A)-49}} dA \\ &= \int \frac{5 cos A}{\sqrt{25-25 sin^2 A}} dA \\ &= \int \frac{5 cos A}{\sqrt{25(1-sin^2 A)}} dA \\ &= \int \frac{5 cos A}{\sqrt{25cos^2 A}} dA \\ &= \int \frac{5 cos A}{5 cos A} dA \\ &= \int dA \\ &= A+C \\ A &= arc sin \frac{x}{5} \\ &= arc sin \frac{x}{5}+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]] 1v35a39amyo56crqgwayyca423yx9uz 108062 108061 2025-07-03T11:13:47Z Akuindo 8654 /* Integral lipat */ 108062 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_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. == 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-x^2}} = \arcsin {x \over a} + C</math> :<math>\int {dx \over {a^2+x^2}} = {1 \over a}\arctan {x \over a} + C</math> <math>\int {dx \over x\sqrt{x^2-a^2}} = {1 \over a}\arcsec {x \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 \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| \sec {x} \right|} + C</math> :<math>\int \cot{x} \, dx = -\ln{\left| \csc{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| \cosh x | + C</math> :<math>\int \coth x \, dx = \ln| \sinh x | + C</math> :<math>\int \mbox{sech}\,x \, dx = \arctan (\sinh x) + C</math> :<math>\int \mbox{csch}\,x \, dx = \ln\left| \tanh {x \over2}\right| + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : <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> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : <math>\int \frac{\ln(x)}{x}\,dx</math> : <math>t = \ln(x),\, dt = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, : <math> \begin{aligned} &\; \int \frac{\ln(x)}{x}\,dx \\ =&\; \int t\,dt \\ =&\; \frac{1}{2} t^2 + C \\ =&\; \frac{1}{2} \ln^2 x + C \end{aligned} </math> === 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. : <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, : <math> \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> ;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. : <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, : <math> \begin{aligned} &\; \int x \sin(x)\,dx \\ =&\; (x)(-\cos(x)) - (1)(-\sin(x)) + C \\ =&\; -x \cos(x) + \sin(x) + C \end{aligned} </math> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{dx}{x^2 - 4}</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{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 !! 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}}{4 x} + C \end{aligned} </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> == 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{25-x^2}} dx</math> * <math>\int \frac{1}{\sqrt{x^2-49}} 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{25-x^2}} dx \\ x &= 5 sin A \\ dx &= 5 cos A dA \\ \int \frac{1}{\sqrt{25-x^2}} dx &= \int \frac{5 cos A}{\sqrt{25-(5 sin A)^2}} dA \\ &= \int \frac{5 cos A}{\sqrt{25-25 sin^2 A}} dA \\ &= \int \frac{5 cos A}{\sqrt{25(1-sin^2 A)}} dA \\ &= \int \frac{5 cos A}{\sqrt{25cos^2 A}} dA \\ &= \int \frac{5 cos A}{5 cos A} dA \\ &= \int dA \\ &= A+C \\ A &= arc sin \frac{x}{5} \\ &= arc sin \frac{x}{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}{\sqrt{x^2-49}} dx \\ x &= 7 tan A \\ dx &= 7 tan A sec A dA \\ \int \frac{1}{\sqrt{x^2-49}} dx &= \int \frac{7 tan A sec A}{\sqrt{(7 tan A)^2-49}} dA \\ &= \int \frac{7 tan A sec A}{\sqrt{49 tan^2 A-49}} dA \\ &= \int \frac{5 cos A}{\sqrt{25(1-sin^2 A)}} dA \\ &= \int \frac{5 cos A}{\sqrt{25cos^2 A}} dA \\ &= \int \frac{5 cos A}{5 cos A} dA \\ &= \int dA \\ &= A+C \\ A &= arc sin \frac{x}{5} \\ &= arc sin \frac{x}{5}+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]] feh29sryktsf5hxw6socmolamnjcck6 108063 108062 2025-07-03T11:21:38Z Akuindo 8654 /* Integral lipat */ 108063 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_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. == 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-x^2}} = \arcsin {x \over a} + C</math> :<math>\int {dx \over {a^2+x^2}} = {1 \over a}\arctan {x \over a} + C</math> <math>\int {dx \over x\sqrt{x^2-a^2}} = {1 \over a}\arcsec {x \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 \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| \sec {x} \right|} + C</math> :<math>\int \cot{x} \, dx = -\ln{\left| \csc{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| \cosh x | + C</math> :<math>\int \coth x \, dx = \ln| \sinh x | + C</math> :<math>\int \mbox{sech}\,x \, dx = \arctan (\sinh x) + C</math> :<math>\int \mbox{csch}\,x \, dx = \ln\left| \tanh {x \over2}\right| + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : <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> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : <math>\int \frac{\ln(x)}{x}\,dx</math> : <math>t = \ln(x),\, dt = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, : <math> \begin{aligned} &\; \int \frac{\ln(x)}{x}\,dx \\ =&\; \int t\,dt \\ =&\; \frac{1}{2} t^2 + C \\ =&\; \frac{1}{2} \ln^2 x + C \end{aligned} </math> === 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. : <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, : <math> \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> ;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. : <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, : <math> \begin{aligned} &\; \int x \sin(x)\,dx \\ =&\; (x)(-\cos(x)) - (1)(-\sin(x)) + C \\ =&\; -x \cos(x) + \sin(x) + C \end{aligned} </math> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{dx}{x^2 - 4}</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{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 !! 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}}{4 x} + C \end{aligned} </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> == 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{25-x^2}} dx</math> * <math>\int \frac{1}{x\sqrt{x^2-49}} 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{25-x^2}} dx \\ x &= 5 sin A \\ dx &= 5 cos A dA \\ \int \frac{1}{\sqrt{25-x^2}} dx &= \int \frac{5 cos A}{\sqrt{25-(5 sin A)^2}} dA \\ &= \int \frac{5 cos A}{\sqrt{25-25 sin^2 A}} dA \\ &= \int \frac{5 cos A}{\sqrt{25(1-sin^2 A)}} dA \\ &= \int \frac{5 cos A}{\sqrt{25cos^2 A}} dA \\ &= \int \frac{5 cos A}{5 cos A} dA \\ &= \int dA \\ &= A+C \\ A &= arc sin \frac{x}{5} \\ &= arc sin \frac{x}{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}{x\sqrt{x^2-49}} dx \\ x &= 7 sec A \\ dx &= 7 sec A tan A dA \\ \int \frac{1}{x\sqrt{x^2-49}} dx &= \int \frac{7 sec A tan A}{x\sqrt{(7 sec A)^2-49}} dA \\ &= \int \frac{7 sec A tan A}{x\sqrt{49 sec^2 A-49}} dA \\ &= \int \frac{7 sec A tan A}{\sqrt{49(sec^2 A-1)}} dA \\ &= \int \frac{7 sec A tan A}{\sqrt{49tan^2 A}} dA \\ &= \int \frac{7 sec A tan A}{7 tan A} dA \\ &= \int sec A dA \\ &= A+C \\ A &= arc sin \frac{x}{5} \\ &= arc sin \frac{x}{5}+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]] 8nnozvujk0xicw9akc1zt1b0vgds74w 108064 108063 2025-07-03T11:26:10Z Akuindo 8654 /* Integral lipat */ 108064 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_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. == 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-x^2}} = \arcsin {x \over a} + C</math> :<math>\int {dx \over {a^2+x^2}} = {1 \over a}\arctan {x \over a} + C</math> <math>\int {dx \over x\sqrt{x^2-a^2}} = {1 \over a}\arcsec {x \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 \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| \sec {x} \right|} + C</math> :<math>\int \cot{x} \, dx = -\ln{\left| \csc{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| \cosh x | + C</math> :<math>\int \coth x \, dx = \ln| \sinh x | + C</math> :<math>\int \mbox{sech}\,x \, dx = \arctan (\sinh x) + C</math> :<math>\int \mbox{csch}\,x \, dx = \ln\left| \tanh {x \over2}\right| + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : <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> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : <math>\int \frac{\ln(x)}{x}\,dx</math> : <math>t = \ln(x),\, dt = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, : <math> \begin{aligned} &\; \int \frac{\ln(x)}{x}\,dx \\ =&\; \int t\,dt \\ =&\; \frac{1}{2} t^2 + C \\ =&\; \frac{1}{2} \ln^2 x + C \end{aligned} </math> === 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. : <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, : <math> \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> ;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. : <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, : <math> \begin{aligned} &\; \int x \sin(x)\,dx \\ =&\; (x)(-\cos(x)) - (1)(-\sin(x)) + C \\ =&\; -x \cos(x) + \sin(x) + C \end{aligned} </math> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{dx}{x^2 - 4}</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{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 !! 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}}{4 x} + C \end{aligned} </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> == 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{25-x^2}} dx</math> * <math>\int \frac{1}{x\sqrt{x^2-49}} 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{25-x^2}} dx \\ x &= 5 sin A \\ dx &= 5 cos A dA \\ \int \frac{1}{\sqrt{25-x^2}} dx &= \int \frac{5 cos A}{\sqrt{25-(5 sin A)^2}} dA \\ &= \int \frac{5 cos A}{\sqrt{25-25 sin^2 A}} dA \\ &= \int \frac{5 cos A}{\sqrt{25(1-sin^2 A)}} dA \\ &= \int \frac{5 cos A}{\sqrt{25cos^2 A}} dA \\ &= \int \frac{5 cos A}{5 cos A} dA \\ &= \int dA \\ &= A+C \\ A &= arc sin \frac{x}{5} \\ &= arc sin \frac{x}{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}{x\sqrt{x^2-49}} dx \\ x &= 7 sec A \\ dx &= 7 sec A tan A dA \\ \int \frac{1}{x\sqrt{x^2-49}} dx &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{(7 sec A)^2-49}} dA \\ &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{49 sec^2 A-49}} dA \\ &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{49(sec^2 A-1)}} dA \\ &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{49tan^2 A}} dA \\ &= \int \frac{7 sec A tan A}{49 sec A tan A} dA \\ &= \int \frac{1}{7} dA \\ &= \frac{A}{7}+C \\ A &= arc sec \frac{x}{7}} \\ &= \frac{arc sec \frac{x}{7}}{7}+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]] d69gr25dz3f519yqdzdwln9ksv2hb5d 108065 108064 2025-07-03T11:26:44Z Akuindo 8654 /* Integral lipat */ 108065 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_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. == 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-x^2}} = \arcsin {x \over a} + C</math> :<math>\int {dx \over {a^2+x^2}} = {1 \over a}\arctan {x \over a} + C</math> <math>\int {dx \over x\sqrt{x^2-a^2}} = {1 \over a}\arcsec {x \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 \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| \sec {x} \right|} + C</math> :<math>\int \cot{x} \, dx = -\ln{\left| \csc{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| \cosh x | + C</math> :<math>\int \coth x \, dx = \ln| \sinh x | + C</math> :<math>\int \mbox{sech}\,x \, dx = \arctan (\sinh x) + C</math> :<math>\int \mbox{csch}\,x \, dx = \ln\left| \tanh {x \over2}\right| + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : <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> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : <math>\int \frac{\ln(x)}{x}\,dx</math> : <math>t = \ln(x),\, dt = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, : <math> \begin{aligned} &\; \int \frac{\ln(x)}{x}\,dx \\ =&\; \int t\,dt \\ =&\; \frac{1}{2} t^2 + C \\ =&\; \frac{1}{2} \ln^2 x + C \end{aligned} </math> === 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. : <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, : <math> \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> ;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. : <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, : <math> \begin{aligned} &\; \int x \sin(x)\,dx \\ =&\; (x)(-\cos(x)) - (1)(-\sin(x)) + C \\ =&\; -x \cos(x) + \sin(x) + C \end{aligned} </math> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{dx}{x^2 - 4}</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{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 !! 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}}{4 x} + C \end{aligned} </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> == 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{25-x^2}} dx</math> * <math>\int \frac{1}{x\sqrt{x^2-49}} 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{25-x^2}} dx \\ x &= 5 sin A \\ dx &= 5 cos A dA \\ \int \frac{1}{\sqrt{25-x^2}} dx &= \int \frac{5 cos A}{\sqrt{25-(5 sin A)^2}} dA \\ &= \int \frac{5 cos A}{\sqrt{25-25 sin^2 A}} dA \\ &= \int \frac{5 cos A}{\sqrt{25(1-sin^2 A)}} dA \\ &= \int \frac{5 cos A}{\sqrt{25cos^2 A}} dA \\ &= \int \frac{5 cos A}{5 cos A} dA \\ &= \int dA \\ &= A+C \\ A &= arc sin \frac{x}{5} \\ &= arc sin \frac{x}{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}{x\sqrt{x^2-49}} dx \\ x &= 7 sec A \\ dx &= 7 sec A tan A dA \\ \int \frac{1}{x\sqrt{x^2-49}} dx &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{(7 sec A)^2-49}} dA \\ &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{49 sec^2 A-49}} dA \\ &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{49(sec^2 A-1)}} dA \\ &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{49tan^2 A}} dA \\ &= \int \frac{7 sec A tan A}{49 sec A tan A} dA \\ &= \int \frac{1}{7} dA \\ &= \frac{A}{7}+C \\ A &= arc sec \frac{x}{7} \\ &= \frac{arc sec \frac{x}{7}}{7}+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]] gbnumt82x61vqoxrsa8jzb899knmxu7 108066 108065 2025-07-03T11:35:29Z Akuindo 8654 /* Integral lipat */ 108066 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_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. == 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-x^2}} = \arcsin {x \over a} + C</math> :<math>\int {dx \over {a^2+x^2}} = {1 \over a}\arctan {x \over a} + C</math> <math>\int {dx \over x\sqrt{x^2-a^2}} = {1 \over a}\arcsec {x \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 \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| \sec {x} \right|} + C</math> :<math>\int \cot{x} \, dx = -\ln{\left| \csc{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| \cosh x | + C</math> :<math>\int \coth x \, dx = \ln| \sinh x | + C</math> :<math>\int \mbox{sech}\,x \, dx = \arctan (\sinh x) + C</math> :<math>\int \mbox{csch}\,x \, dx = \ln\left| \tanh {x \over2}\right| + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : <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> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : <math>\int \frac{\ln(x)}{x}\,dx</math> : <math>t = \ln(x),\, dt = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, : <math> \begin{aligned} &\; \int \frac{\ln(x)}{x}\,dx \\ =&\; \int t\,dt \\ =&\; \frac{1}{2} t^2 + C \\ =&\; \frac{1}{2} \ln^2 x + C \end{aligned} </math> === 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. : <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, : <math> \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> ;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. : <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, : <math> \begin{aligned} &\; \int x \sin(x)\,dx \\ =&\; (x)(-\cos(x)) - (1)(-\sin(x)) + C \\ =&\; -x \cos(x) + \sin(x) + C \end{aligned} </math> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{dx}{x^2 - 4}</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{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 !! 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}}{4 x} + C \end{aligned} </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> == 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{25-x^2}} dx</math> * <math>\int \frac{1}{{x^2+81}} dx</math> * <math>\int \frac{1}{x\sqrt{x^2-49}} 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{25-x^2}} dx \\ x &= 5 sin A \\ dx &= 5 cos A dA \\ \int \frac{1}{\sqrt{25-x^2}} dx &= \int \frac{5 cos A}{\sqrt{25-(5 sin A)^2}} dA \\ &= \int \frac{5 cos A}{\sqrt{25-25 sin^2 A}} dA \\ &= \int \frac{5 cos A}{\sqrt{25(1-sin^2 A)}} dA \\ &= \int \frac{5 cos A}{\sqrt{25cos^2 A}} dA \\ &= \int \frac{5 cos A}{5 cos A} dA \\ &= \int dA \\ &= A+C \\ A &= arc sin \frac{x}{5} \\ &= arc sin \frac{x}{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}{x^2+81}} dx \\ x &= 9 tan A \\ dx &= 9 sec^2 A dA \\ \int \frac{1}{x^2+81} dx &= \int \frac{9 sec^2 A}{(9 tan A)^2+81} dA \\ &= \int \frac{9 sec^2 A}{81 tan^2 A+81} dA \\ &= \int \frac{9 sec^2 A}{81 (tan^2 A+1)} dA \\ &= \int \frac{9 sec^2 A}{81 sec^2 A} dA \\ &= \int \frac{1}{9} dA \\ &= \frac{A}{9}+C \\ A &= arc tan \frac{x}{9} \\ &= \frac{arc tan \frac{x}{9}}{9}+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}{x\sqrt{x^2-49}} dx \\ x &= 7 sec A \\ dx &= 7 sec A tan A dA \\ \int \frac{1}{x\sqrt{x^2-49}} dx &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{(7 sec A)^2-49}} dA \\ &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{49 sec^2 A-49}} dA \\ &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{49(sec^2 A-1)}} dA \\ &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{49tan^2 A}} dA \\ &= \int \frac{7 sec A tan A}{49 sec A tan A} dA \\ &= \int \frac{1}{7} dA \\ &= \frac{A}{7}+C \\ A &= arc sec \frac{x}{7} \\ &= \frac{arc sec \frac{x}{7}}{7}+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]] cjlgso2hoeoczqs5b7qq3e21gfb26b7 108067 108066 2025-07-03T11:38:40Z Akuindo 8654 /* Integral lipat */ 108067 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_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. == 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-x^2}} = \arcsin {x \over a} + C</math> :<math>\int {dx \over {a^2+x^2}} = {1 \over a}\arctan {x \over a} + C</math> <math>\int {dx \over x\sqrt{x^2-a^2}} = {1 \over a}\arcsec {x \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 \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| \sec {x} \right|} + C</math> :<math>\int \cot{x} \, dx = -\ln{\left| \csc{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| \cosh x | + C</math> :<math>\int \coth x \, dx = \ln| \sinh x | + C</math> :<math>\int \mbox{sech}\,x \, dx = \arctan (\sinh x) + C</math> :<math>\int \mbox{csch}\,x \, dx = \ln\left| \tanh {x \over2}\right| + C</math> == Jenis integral == === integral biasa === Berikut contoh penyelesaian cara biasa. : <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> === integral substitusi === Berikut contoh penyelesaian cara substitusi. : <math>\int \frac{\ln(x)}{x}\,dx</math> : <math>t = \ln(x),\, dt = \frac{dx}{x}</math> Dengan menggunakan rumus di atas, : <math> \begin{aligned} &\; \int \frac{\ln(x)}{x}\,dx \\ =&\; \int t\,dt \\ =&\; \frac{1}{2} t^2 + C \\ =&\; \frac{1}{2} \ln^2 x + C \end{aligned} </math> === 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. : <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, : <math> \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> ;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. : <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, : <math> \begin{aligned} &\; \int x \sin(x)\,dx \\ =&\; (x)(-\cos(x)) - (1)(-\sin(x)) + C \\ =&\; -x \cos(x) + \sin(x) + C \end{aligned} </math> === integral pecahan parsial === Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional). : <math>\int \frac{dx}{x^2 - 4}</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{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 !! 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}}{4 x} + C \end{aligned} </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> == 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{25-x^2}} dx</math> * <math>\int \frac{1}{{x^2+81}} dx</math> * <math>\int \frac{1}{x\sqrt{x^2-49}} 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{25-x^2}} dx \\ x &= 5 sin A \\ dx &= 5 cos A dA \\ \int \frac{1}{\sqrt{25-x^2}} dx &= \int \frac{5 cos A}{\sqrt{25-(5 sin A)^2}} dA \\ &= \int \frac{5 cos A}{\sqrt{25-25 sin^2 A}} dA \\ &= \int \frac{5 cos A}{\sqrt{25(1-sin^2 A)}} dA \\ &= \int \frac{5 cos A}{\sqrt{25cos^2 A}} dA \\ &= \int \frac{5 cos A}{5 cos A} dA \\ &= \int dA \\ &= A+C \\ A &= arc sin \frac{x}{5} \\ &= arc sin \frac{x}{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}{x^2+81} dx \\ x &= 9 tan A \\ dx &= 9 sec^2 A dA \\ \int \frac{1}{x^2+81} dx &= \int \frac{9 sec^2 A}{(9 tan A)^2+81} dA \\ &= \int \frac{9 sec^2 A}{81 tan^2 A+81} dA \\ &= \int \frac{9 sec^2 A}{81 (tan^2 A+1)} dA \\ &= \int \frac{9 sec^2 A}{81 sec^2 A} dA \\ &= \int \frac{1}{9} dA \\ &= \frac{A}{9}+C \\ A &= arc tan \frac{x}{9} \\ &= \frac{arc tan \frac{x}{9}}{9}+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}{x\sqrt{x^2-49}} dx \\ x &= 7 sec A \\ dx &= 7 sec A tan A dA \\ \int \frac{1}{x\sqrt{x^2-49}} dx &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{(7 sec A)^2-49}} dA \\ &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{49 sec^2 A-49}} dA \\ &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{49(sec^2 A-1)}} dA \\ &= \int \frac{7 sec A tan A}{(7 sec A)\sqrt{49tan^2 A}} dA \\ &= \int \frac{7 sec A tan A}{49 sec A tan A} dA \\ &= \int \frac{1}{7} dA \\ &= \frac{A}{7}+C \\ A &= arc sec \frac{x}{7} \\ &= \frac{arc sec \frac{x}{7}}{7}+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]] 43gf47e5nmhf8zf4840x0szg6a2nmuc