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 108069 108056 2025-07-03T16:21:33Z Alfarq 799 /* Daftar Obat (2) */ 108069 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 | * Dexamethasone 0,5 mg * Arsenic trioxide (arsen anorganik) 0,5 mg |} # 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}}]] b4ggd46l7gf7v8zzc8ddrhegqvl2ma5 108070 108069 2025-07-03T16:31:57Z Alfarq 799 /* Daftar Obat (2) */ 108070 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 | * Dexamethasone 0,5 mg * Arsenic trioxide (arsen anorganik) 0,5 mg |- | Dexyl Syr | Komposisi per 5 ml: * Diphenhyramine HCl 12,5 mg * Ammonium chloride 135 mg * Sodium citrate 57,03 mg * Menthol 0,54 mg |} # 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}}]] b3zmoa35nqqexarv7tkd4vo422bwru2 0 23140 108071 108067 2025-07-04T01:23:16Z Akuindo 8654 /* Integral lipat */ 108071 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{50-288x^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{50-288x^2}} dx \\ x &= \frac{5\sqrt{2}}{12\sqrt{2}} sin A \\ dx &= \frac{5\sqrt{2}}{12\sqrt{2}} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5\sqrt{2}}{12\sqrt{2}} cos A}{\sqrt{50-288(\frac{5\sqrt{2}}{12\sqrt{2}} 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]] a84thbhmir64w5qg77r3rjjubf2rerb 108072 108071 2025-07-04T01:29:26Z Akuindo 8654 /* Integral lipat */ 108072 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{50-288x^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{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 5}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{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]] 2b1kqb1qb9kvhynlhvcdc57kp3nu6om 108074 108072 2025-07-04T01:35:06Z Akuindo 8654 /* Integral lipat */ 108074 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{50-288x^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{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 5}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{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]] fkrkl5nknwo9jn9t5eckr856mq17m3n 108075 108074 2025-07-04T01:38:51Z Akuindo 8654 /* Integral lipat */ 108075 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{50-288x^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{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{1}{12\sqrt{2}} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{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]] dmeirt6gr7uw8zkds7vmbj85pchlny9 108076 108075 2025-07-04T01:58:37Z Akuindo 8654 /* Integral lipat */ 108076 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{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+125x^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{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{1}{12\sqrt{2}} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+125x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+125x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+125(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{125 \cdot 4}{25} tan^2 A} 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]] id5cal6jl1v876t9dtvhwsi716hwk5f 108077 108076 2025-07-04T02:10:06Z Akuindo 8654 /* Integral lipat */ 108077 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{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^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{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{1}{12\sqrt{2}} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{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]] 6xc97m7xv9wemlv9rk44tl2o796iozb 108078 108077 2025-07-04T02:11:46Z Akuindo 8654 /* Integral lipat */ 108078 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{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^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{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{1}{12\sqrt{2}} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{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]] gs1a916x0snllqqv6g7ong26lj6ty1k 108079 108078 2025-07-04T03:26:57Z Akuindo 8654 /* Integral lipat */ 108079 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{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{1}{12\sqrt{2}} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{11}{7} sec A \\ dx &= \frac{11}{7} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{11}{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]] j09vwqj0r5pvhra489p4vh25z0fmjsi 108080 108079 2025-07-04T03:36:48Z Akuindo 8654 /* Integral lipat */ 108080 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{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{1}{12\sqrt{2}} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A)\sqrt{605(\frac{49}{121} sec^2 A)-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A)\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A)\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A)\sqrt{245tan^2 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]] l6s69efoqf2bnmt2qn75glzp7e9cmfe 108081 108080 2025-07-04T03:45:49Z Akuindo 8654 /* Integral lipat */ 108081 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{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{1}{12\sqrt{2}} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A\sqrt{605(\frac{49}{121} sec^2 A)-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{49\sqrt{5} sec A tan A}} dA \\ &= \int \frac{\sqrt{5}}{1.925} dA \\ &= \frac{A\sqrt{5}}{1.925}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{1.925}+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]] oeetsfsn8qxnrpsg1pilj4xzs3zxlqu 108082 108081 2025-07-04T03:52:55Z Akuindo 8654 /* Integral lipat */ 108082 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{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{1}{12\sqrt{2}} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A} {11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A\sqrt{605(\frac{49}{121} sec^2 A)-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{49\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{1.925} dA \\ &= \frac{A\sqrt{5}}{1.925}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{1.925}+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]] m7h4434pswsellef8c874ez818gr6an 108083 108082 2025-07-04T04:23:47Z Akuindo 8654 /* Integral lipat */ 108083 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{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{1}{12\sqrt{2}} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A} {11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A\sqrt{605(\frac{49}{121} sec^2 A)-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{7 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{\frac{7}{11} sec A tan A}{49\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> # 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]] fu00660lasw41y55x0l008fze4tnc1l 108084 108083 2025-07-04T04:45:32Z Akuindo 8654 /* Integral lipat */ 108084 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{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{1}{12\sqrt{2}} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{605(\frac{49}{121} sec^2 A)-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> # 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]] 1peu12k8x6ry3vg24viq5nhfw8i6im5 108085 108084 2025-07-04T04:47:28Z Akuindo 8654 /* Integral lipat */ 108085 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{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{605(\frac{49}{121} sec^2 A)-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> # 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]] d2n5gw582dwtm5ojc1oghgzts2kdf6v 108086 108085 2025-07-04T05:27:06Z Akuindo 8654 /* Integral lipat */ 108086 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{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{605(\frac{49}{121} sec^2 A)-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> # 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]] ee9q7a20zpebp4t7bpwkxvp65h8dz09 108094 108086 2025-07-04T07:20:04Z Akuindo 8654 /* Integral lipat */ 108094 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{50-288x^2}} dx</math> * <math>\int \frac{1}{{12+75x^2}} dx</math> * <math>\int \frac{1}{11x\sqrt{605x^2-245}} dx</math> ; jawaban ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{misalkan } u &= 5x^2-2x+7 \\ u &= 5x^2-2x+7 \\ du &= 10x-2 dx \\ &= 2(5x-1) dx \\ \frac{du}{2} &= (5x-1) dx \\ \int (5x-1)(5x^2-2x+7)^4 dx &= \int \frac{1}{2}u^4 du \\ &= \frac{1}{2}\frac{1}{5}u^5 + C \\ &= \frac{1}{10}(5x^2-2x+7)^5 + C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \text{gunakan integral parsial } \\ \int x^2 cos 3x dx &= \frac{x^2}{3} sin 3x-\frac{2x}{9} (-cos 3x)+\frac{2}{27} (-sin 3x)+C \\ &= \frac{x^2}{3} sin 3x+\frac{2x}{9} cos 3x-\frac{2}{27} sin 3x+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= sin x \\ u &= sin x \\ du &= cos x dx \\ \int sin^2 x cos x dx &= \int u^2 du \\ &= \frac{u^3}{3}+C \\ &= \frac{sin^3 x}{3}+C \\ * \text{cara 2} \\ \int sin^2 x cos x dx &= \int \frac{1-cos 2x}{2} cos x dx \\ &= \frac{1}{2} \int (1-cos 2x) cos x dx \\ &= \frac{1}{2} \int cos x-cos 2x cos x dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}(cos 3x+cox x) dx \\ &= \frac{1}{2} \int cos x-\frac{1}{2}cos 3x-\frac{1}{2}cos x dx \\ &= \frac{1}{2} \int \frac{1}{2}cos x-\frac{1}{2}cos 3x dx \\ &= \frac{1}{2} (\frac{1}{2}sin x-\frac{1}{6}sin 3x)+C \\ &= \frac{sin x}{4}-\frac{sin 3x}{12}+C \\ * \text{cara 3} \\ \int sin^2 x cos x dx &= \int sin x sin x cos x dx \\ &= \frac{1}{2} \int sin x sin 2x dx \\ &= \frac{1}{4} \int -(cos 3x-cos (-x)) dx \\ &= \frac{1}{4} \int -cos 3x+cos x dx \\ &= \frac{1}{4} (-\frac{sin 3x}{3}+sin x)+C \\ &= -\frac{sin 3x}{12}+\frac{sin x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} * \text{cara 1} \\ \text{misalkan } u &= cos x \\ u &= cos x \\ du &= -sin x dx \\ -du &= sin x dx \\ \int cos^2 x sin x dx &= -\int u^2 du \\ &= -\frac{u^3}{3}+C \\ &= -\frac{cos^3 x}{3}+C \\ * \text{cara 2} \\ \int cos^2 x sin x dx &= \int \frac{cos 2x+1}{2} sin x dx \\ &= \frac{1}{2} \int (cos 2x+1) sin x dx \\ &= \frac{1}{2} \int cos 2xsin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}(sin 3x-sin x)+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x-\frac{1}{2}sin x+sin x dx \\ &= \frac{1}{2} \int \frac{1}{2}sin 3x+\frac{1}{2}sin x dx \\ &= \frac{1}{2} (-\frac{1}{6}cos 3x-\frac{1}{2}cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ * \text{cara 3} \\ \int cos^2 x sin x dx &= \int cos x cos x sin x dx \\ &= \frac{1}{2} \int cos x sin 2x dx \\ &= \frac{1}{4} \int (sin 3x-sin (-x)) dx \\ &= \frac{1}{4} \int sin 3x+sin x dx \\ &= \frac{1}{4} (-\frac{cos 3x}{3}-cos x)+C \\ &= -\frac{cos 3x}{12}-\frac{cos x}{4}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int sec x dx &= \int sec x \frac{sec x+tan x}{sec x+tan x}dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ \text{misalkan } u &= sec x+tan x \\ u &= sec x+tan x \\ du &= sec xtan x+sec^2 x dx \\ &= \int \frac{sec^2 x+sec xtan x}{sec x+tan x}dx \\ &= \int \frac{1}{u}du \\ &= ln u+C \\ &= ln |sec x+tan x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int csc x dx &= \int csc x \frac{csc x+cot x}{csc x+cot x}dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ \text{misalkan } u &= csc x+cot x \\ u &= csc x+cot x \\ du &= -csc xcot x-csc^2 x dx \\ -du &= csc xcot x+csc^2 x dx \\ &= \int \frac{csc^2 x+csc xcot x}{csc x+cot x}dx \\ &= -\int \frac{1}{u}du \\ &= -ln u+C \\ &= -ln |csc x+cot x|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{2x^2-5x+3} dx &= \int \frac{1}{(2x-3)(x-1)} dx \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{A(x-1)+B(2x-3)}{(2x-3)(x-1)}) dx \\ &= \int (\frac{Ax-A+2Bx-3B}{2x^2-5x+3}) dx \\ &= \int (\frac{(A+2B)x+(-A-3B)}{2x^2-5x+3}) dx \\ \text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1} \\ &= \int (\frac{A}{(2x-3)}+\frac{B}{(x-1)}) dx \\ &= \int (\frac{2}{(2x-3)}+\frac{-1}{(x-1)}) dx \\ &= ln |2x-3|-ln |x-1|+C \\ &= ln |\frac{2x-3}{x-1}|+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{\sqrt{50-288x^2}} dx \\ x &= \frac{5}{12} sin A \\ dx &= \frac{5}{12} cos A dA \\ \int \frac{1}{\sqrt{50-288x^2}} dx &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-288(\frac{5}{12} sin A)^2}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-\frac{288 \cdot 25}{144} sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50-50 sin^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50(1-sin^2 A)}} dA \\ &= \int \frac{\frac{5}{12} cos A}{\sqrt{50cos^2 A}} dA \\ &= \int \frac{\frac{5}{12} cos A}{5cos A \sqrt{2}} dA \\ &= \int \frac{\sqrt{2}}{24} dA \\ &= \frac{\sqrt{2}}{24} \int dA \\ &= A+C \\ A &= arc sin \frac{12x}{5} \\ &= \frac{\sqrt{2}}{24} arc sin \frac{12x}{5}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{12+75x^2} dx \\ x &= \frac{2}{5} tan A \\ dx &= \frac{2}{5} sec^2 A dA \\ \int \frac{1}{12+75x^2} dx &= \int \frac{\frac{2}{5} sec^2 A}{12+75(\frac{2}{5} tan A)^2} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+\frac{75 \cdot 4}{25} tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12+12 tan^2 A} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 (1+tan^2 A)} dA \\ &= \int \frac{\frac{2}{5} sec^2 A}{12 sec^2 A} dA \\ &= \int \frac{1}{30} dA \\ &= \frac{A}{30}+C \\ A &= arc tan \frac{5x}{2} \\ &= \frac{arc tan \frac{5x}{2}}{30}+C \\ \end{aligned} </math> </div></div> ** <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{aligned} \int \frac{1}{11x\sqrt{605x^2-245}} dx \\ x &= \frac{7}{11} sec A \\ dx &= \frac{7}{11} sec A tan A dA \\ \int \frac{1}{11x\sqrt{605x^2-245}} dx &= \int \frac{\frac{7}{11} sec A tan A}{11(\frac{7}{11} sec A)\sqrt{605(\frac{7}{11} sec A)^2-245}} dA \\ &= \int \frac{sec A tan A} {11 sec A\sqrt{\frac{605 \cdot 49}{121} sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245sec^2 A-245}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245(sec^2 A-1)}} dA \\ &= \int \frac{sec A tan A}{11 sec A\sqrt{245tan^2 A}} dA \\ &= \int \frac{sec A tan A}{77\sqrt{5} sec A tan A} dA \\ &= \int \frac{\sqrt{5}}{385} dA \\ &= \frac{A\sqrt{5}}{385}+C \\ A &= arc sec \frac{11x}{7} \\ &= \frac{\sqrt{5}arc sec \frac{11x}{7}}{385}+C \\ \end{aligned} </math> </div></div> # 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]] 0ryyxcylnd7u2wr53ub5i4o9h5easub 100 25854 108068 2025-07-03T12:57:32Z Volstand 31387 ←Membuat halaman berisi '[[Berkas:Ketupat mie.jpg|jmpl|Ketupat mie yang dihidangkan dalam sebuah piring]] '''Ketupat mie''' merupakan makanan tradisional [[Indonesia]] yang berasal dari [[w:Martapura, Banjar|Martapura]], [[w:Kalimantan Selatan|Kalimantan Selatan]]. ==Bahan== * 2 buah ketupat * 2 keping mie keriting * 2 batang seledri * 2 sendok makan bawang merah goreng ==Bumbu== ===Bumbu halus=== * 5 butir bawang merah * 3 siung bawang putih * 1/3 buah pala * 1/2 sendok teh lada bubuk...' 108068 wikitext text/x-wiki [[Berkas:Ketupat mie.jpg|jmpl|Ketupat mie yang dihidangkan dalam sebuah piring]] '''Ketupat mie''' merupakan makanan tradisional [[Indonesia]] yang berasal dari [[w:Martapura, Banjar|Martapura]], [[w:Kalimantan Selatan|Kalimantan Selatan]]. ==Bahan== * 2 buah ketupat * 2 keping mie keriting * 2 batang seledri * 2 sendok makan bawang merah goreng ==Bumbu== ===Bumbu halus=== * 5 butir bawang merah * 3 siung bawang putih * 1/3 buah pala * 1/2 sendok teh lada bubuk * 1 cm Jahe ===Kuah ketupat=== * 1/4 ekor ayam * 1 liter air * 1/4 sendok teh garam * 1/2 sendok teh gula pasir * 1/4 sendok teh kaldu bubuk * 3 cm kayu manis * 2 buah cengkeh * 1/2 buah wortel * 1/4 buah kol * 2 lembar daun salam == Cara membuat == # Rebus ayam dengan daun salam dan jahe yang telah dipukul. # Haluskan semua bumbu halus menggunakan blender atau ulekan. # Tumis bumbu halus pada minyak panas hingga wangi, lalu tambahkan kayu manis dan cengkeh. # Tuangkan tumisan bumbu halus pada rebusan ayam yang telah mendidih. # Masukkan irisan wortel dan kol. # Tambahkan gula, garam, dan kaldu bubuk. # Rebus mie pada panci terpisah. # Tata potongan ketupat dan mie pada piring lalu sirah dengan kuah. # Taburkan irisan daun seledri dan bawang goreng sucukupnya.<ref>{{Cite web|date=2023-02-20|title=Resep Ketupat Mie Khas Martapura Kalsel oleh melda yanti|url=https://cookpad.com/id/resep/16804681|website=Cookpad|language=id|access-date=2025-07-03}}</ref> {{wikipedia|Ketupat mie}} == Referensi == <references /> [[Kategori:WikiBalalah II]] [[Kategori:Makanan tradisional]] clx2qdcbwiwyan4uyym4byzarreurlb 2 25855 108073 2025-07-04T01:33:15Z Etrin JP 40933 ←Membuat halaman berisi 'Halooooo perkenalkan saya Etrin. Mari berdiskusi dan berkolaborasi 🤝' 108073 wikitext text/x-wiki Halooooo perkenalkan saya Etrin. Mari berdiskusi dan berkolaborasi 🤝 gfdfpxgaixz36qg4tu1rcp4rrqb013d 0 25856 108087 2025-07-04T06:21:02Z Alemmwf29 41552 ←Membuat halaman berisi 'Penulis : Januardi Diterbitkan oleh : Dinas Perpustakaan dan Kearsipan Kota Padang Panjang Tahun terbit : 2020 Tebal : xii + 212 halaman Buku ini diterbitkan oleh Dinas Perpustakaan dan Kearsipan Kota Padang Panjang sebagai bagian dari penguatan literasi lokal dan dokumentasi identitas kota. Penulisnya, Januardi, adalah birokrat yang berpikir naratif—seorang perenca...' 108087 wikitext text/x-wiki Penulis : Januardi Diterbitkan oleh : Dinas Perpustakaan dan Kearsipan Kota Padang Panjang Tahun terbit : 2020 Tebal : xii + 212 halaman Buku ini diterbitkan oleh Dinas Perpustakaan dan Kearsipan Kota Padang Panjang sebagai bagian dari penguatan literasi lokal dan dokumentasi identitas kota. Penulisnya, Januardi, adalah birokrat yang berpikir naratif—seorang perencana pembangunan yang menyusun sejarah dan mimpi menjadi cerita yang hidup. Buku ini adalah manifestasi cinta dan kekhawatiran seorang anak negeri terhadap kota tempat ia dibesarkan, bertumbuh, dan mengabdi. dveo0irm81gn602o30m2ad7aj81bxdl 108088 108087 2025-07-04T06:25:06Z Alemmwf29 41552 108088 wikitext text/x-wiki Penulis : Januardi Diterbitkan oleh : Dinas Perpustakaan dan Kearsipan Kota Padang Panjang Tahun terbit : 2020 Tebal : xii + 212 halaman Buku ini diterbitkan oleh Dinas Perpustakaan dan Kearsipan Kota Padang Panjang sebagai bagian dari penguatan literasi lokal dan dokumentasi identitas kota. Penulisnya, Januardi, adalah birokrat yang berpikir naratif seorang perencana pembangunan yang menyusun sejarah dan mimpi menjadi cerita yang hidup. Buku ini adalah manifestasi cinta dan kekhawatiran seorang anak negeri terhadap kota tempat ia dibesarkan, bertumbuh, dan mengabdi. Buku dibuka dengan percakapan imajiner sepasang suami-istri urban Jakarta yang bosan berlibur ke Bali atau Lombok dan memutuskan mencari tempat eksotis yang berbeda yaitu Padang Panjang. Pilihan pembuka ini sangat cerdas. Ia seperti undangan lembut bagi pembaca agar membayangkan bahwa kota ini layak diperhitungkan dalam peta wisata nasional. Imajinasi ini bukanlah khayalan kosong, tetapi mimpi yang ditenagai oleh fakta dan potensi nyata. Padang Panjang digambarkan sebagai kota kecil dengan atmosfer sejuk seperti Malang, kuliner khas yang kaya rasa, potensi budaya yang hidup, dan sejarah panjang yang belum banyak diungkap ke publik luas. Prolog ini adalah jembatan antara mimpi dan realitas. Prolog ini sekaligus mengajak pembaca membayangkan kemungkinan baru: Padang Panjang sebagai destinasi nasional. lrjdwwlzffky8n1n6rbodj9b2mf5lks 108089 108088 2025-07-04T06:30:42Z Alemmwf29 41552 108089 wikitext text/x-wiki Penulis : Januardi Diterbitkan oleh : Dinas Perpustakaan dan Kearsipan Kota Padang Panjang Tahun terbit : 2020 Tebal : xii + 212 halaman Buku ini diterbitkan oleh Dinas Perpustakaan dan Kearsipan Kota Padang Panjang sebagai bagian dari penguatan literasi lokal dan dokumentasi identitas kota. Penulisnya, Januardi, adalah pegawai ASN yang berpikir naratif seorang perencana pembangunan yang menyusun sejarah dan mimpi menjadi cerita yang hidup. Buku ini adalah manifestasi cinta dan kekhawatiran seorang anak negeri terhadap kota tempat ia dibesarkan, bertumbuh, dan mengabdi. Buku dibuka dengan percakapan imajiner sepasang suami-istri urban Jakarta yang bosan berlibur ke Bali atau Lombok dan memutuskan mencari tempat eksotis yang berbeda yaitu Padang Panjang. Pilihan pembuka ini sangat cerdas. Ia seperti undangan lembut bagi pembaca agar membayangkan bahwa kota ini layak diperhitungkan dalam peta wisata nasional. Imajinasi ini bukanlah khayalan kosong, tetapi mimpi yang ditenagai oleh fakta dan potensi nyata. Padang Panjang digambarkan sebagai kota kecil dengan atmosfer sejuk seperti Malang, kuliner khas yang kaya rasa, potensi budaya yang hidup, dan sejarah panjang yang belum banyak diungkap ke publik luas. Prolog ini adalah jembatan antara mimpi dan realitas. Prolog ini sekaligus mengajak pembaca membayangkan kemungkinan baru: Padang Panjang sebagai destinasi nasional. nqrycqoewrmcbpvx03qec4ovs7vazpj 108090 108089 2025-07-04T06:31:10Z Alemmwf29 41552 108090 wikitext text/x-wiki Penulis : Januardi Diterbitkan oleh : Dinas Perpustakaan dan Kearsipan Kota Padang Panjang Tahun terbit : 2020 Tebal : xii + 212 halaman Buku ini diterbitkan oleh Dinas Perpustakaan dan Kearsipan Kota Padang Panjang sebagai bagian dari penguatan literasi lokal dan dokumentasi identitas kota. Penulisnya, Januardi, adalah pegawai ASN yang berpikir naratif seorang perencana pembangunan yang menyusun sejarah dan mimpi menjadi cerita yang hidup. Buku ini adalah manifestasi cinta dan kekhawatiran seorang anak negeri terhadap kota tempat ia dibesarkan, bertumbuh, dan mengabdi. Buku dibuka dengan percakapan imajiner sepasang suami-istri urban Jakarta yang bosan berlibur ke Bali atau Lombok dan memutuskan mencari tempat eksotis yang berbeda yaitu Padang Panjang. Pilihan pembuka ini sangat cerdas. Ia seperti undangan lembut bagi pembaca agar membayangkan bahwa kota ini layak diperhitungkan dalam peta wisata nasional. Imajinasi ini bukanlah khayalan kosong, tetapi mimpi yang ditenagai oleh fakta dan potensi nyata. Padang Panjang digambarkan sebagai kota kecil dengan atmosfer sejuk seperti Malang, kuliner khas yang kaya rasa, potensi budaya yang hidup, dan sejarah panjang yang belum banyak diungkap ke publik luas. Prolog ini adalah jembatan antara mimpi dan realitas. Prolog ini sekaligus mengajak pembaca membayangkan kemungkinan baru: Padang Panjang sebagai destinasi nasional. sqnz0s2ftlkoqmzqab54hlcxc3s4i5p 108091 108090 2025-07-04T06:42:23Z Alemmwf29 41552 108091 wikitext text/x-wiki Penulis : Januardi Diterbitkan oleh : Dinas Perpustakaan dan Kearsipan Kota Padang Panjang Tahun terbit : 2020 Tebal : xii + 212 halaman Buku ini diterbitkan oleh Dinas Perpustakaan dan Kearsipan Kota Padang Panjang sebagai bagian dari penguatan literasi lokal dan dokumentasi identitas kota. Penulisnya, Januardi, adalah pegawai ASN yang berpikir naratif seorang perencana pembangunan yang menyusun sejarah dan mimpi menjadi cerita yang hidup. Buku ini adalah manifestasi cinta dan kekhawatiran seorang anak negeri terhadap kota tempat ia dibesarkan, bertumbuh, dan mengabdi. Buku dibuka dengan percakapan imajiner sepasang suami-istri urban Jakarta yang bosan berlibur ke Bali atau Lombok dan memutuskan mencari tempat eksotis yang berbeda yaitu Padang Panjang. Pilihan pembuka ini sangat cerdas. Ia seperti undangan lembut bagi pembaca agar membayangkan bahwa kota ini layak diperhitungkan dalam peta wisata nasional. Imajinasi ini bukanlah khayalan kosong, tetapi mimpi yang ditenagai oleh fakta dan potensi nyata. Padang Panjang itu digambarkan sebagai kota kecil dengan atmosfer sejuk seperti Malang, kuliner khas yang kaya rasa, potensi budaya yang hidup, dan sejarah panjang yang belum banyak diungkap ke publik luas. Prolog ini adalah jembatan antara mimpi dan realitas. Prolog ini sekaligus mengajak pembaca membayangkan kemungkinan baru: Padang Panjang sebagai destinasi nasional. mp502ygax9g5vupjlzdl6sj8edn4bn1 108092 108091 2025-07-04T06:43:01Z Alemmwf29 41552 108092 wikitext text/x-wiki Penulis : Januardi Diterbitkan oleh : Dinas Perpustakaan dan Kearsipan Kota Padang Panjang Tahun terbit : 2020 Tebal : xii + 212 halaman Buku ini diterbitkan oleh Dinas Perpustakaan dan Kearsipan Kota Padang Panjang sebagai bagian dari penguatan literasi lokal dan dokumentasi identitas kota. Penulisnya, Januardi, adalah pegawai ASN yang berpikir naratif seorang perencana pembangunan yang menyusun sejarah dan mimpi menjadi cerita yang hidup. Buku ini adalah manifestasi cinta dan kekhawatiran seorang anak negeri terhadap kota tempat ia dibesarkan, bertumbuh, dan mengabdi. Buku dibuka dengan percakapan imajiner sepasang suami-istri urban Jakarta yang bosan berlibur ke Bali atau Lombok dan memutuskan mencari tempat eksotis yang berbeda yaitu Padang Panjang. Pilihan pembuka ini sangat cerdas. Ia seperti undangan lembut bagi pembaca agar membayangkan bahwa kota ini layak diperhitungkan dalam peta wisata nasional. Imajinasi ini bukanlah khayalan kosong, tetapi mimpi yang ditenagai oleh fakta dan potensi nyata. Padang Panjang itu digambarkan sebagai kota kecil dengan atmosfer sejuk seperti kota Malang, kuliner khas yang kaya rasa, potensi budaya yang hidup, dan sejarah panjang yang belum banyak diungkap ke publik luas. Prolog ini adalah jembatan antara mimpi dan realitas. Prolog ini sekaligus mengajak pembaca membayangkan kemungkinan baru: Padang Panjang sebagai destinasi nasional. a65rs43ytl60fzqosu5q6fcdt4624c0 108093 108092 2025-07-04T06:45:54Z Alemmwf29 41552 108093 wikitext text/x-wiki Penulis : Januardi Diterbitkan oleh : Dinas Perpustakaan dan Kearsipan Kota Padang Panjang Tahun terbit : 2020 Tebal : xii + 212 halaman Buku ini diterbitkan oleh Dinas Perpustakaan dan Kearsipan Kota Padang Panjang sebagai bagian dari penguatan literasi lokal dan dokumentasi identitas kota. Penulisnya, Januardi, adalah pegawai ASN yang berpikir naratif seorang perencana pembangunan yang menyusun sejarah dan mimpi menjadi cerita yang hidup. Buku ini adalah manifestasi cinta dan kekhawatiran seorang anak negeri terhadap kota tempat ia dibesarkan, bertumbuh, dan mengabdi. Buku dibuka dengan percakapan imajiner sepasang suami-istri urban Jakarta yang bosan berlibur ke Bali atau Lombok dan memutuskan mencari tempat eksotis yang berbeda yaitu Padang Panjang. Pilihan pembuka ini sangat cerdas. Ia seperti undangan lembut bagi pembaca agar membayangkan bahwa kota ini layak diperhitungkan dalam peta wisata nasional. Imajinasi ini bukanlah khayalan kosong, tetapi mimpi yang ditenagai oleh fakta dan potensi nyata. Padang Panjang itu digambarkan sebagai kota kecil dengan atmosfer sejuk seperti kota Malang, kuliner khas yang kaya rasa, potensi budaya yang hidup, dan sejarah panjang yang belum banyak diungkap ke publik luas. Prolog ini adalah jembatan antara mimpi dan realitas. Prolog ini sekaligus mengajak pembaca membayangkan kemungkinan baru yaitu di mana Padang Panjang sebagai destinasi nasional. g293j2jhow6r439d4ymum9rof6e3azs