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Kalkulus modul x integral

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  • mmm... bbutuh maklah nie tentang integral yang lengkap
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  • 1. 10 INTEGRAL10.1 ANTI TURUNAN (INTEGRAL TAK TENTU)• Integral adalah anti derivatif atau anti turunan.• Rumus Umum dari Integral Tak Tentu 1 ∫ x dx = n + 1 x n +1 n +C , n ≠ −1Contoh 10.1Cari anti turunan yang umum dari f(x) = x2Penyelesaian: 1 2+1 x3 ∫ x dx = 2 x = +C 2 +1 3Contoh 10.2Cari anti turunan yang umum dari f(x) = 20x4Penyelesaian: 20 4+1 ∫ 20x4 dx = x = 5x5 + C 4 +1Contoh 10.3Cari anti turunan yang umum dari fungsi y = x 2 3 1 ∫ x 2 dx = ( 3 2 + 1) x 2 + C 3 ( 3 +1) 1 = x( 2 + 2 ) + C 3 2 ( 3 2 + 2 2)       1 5 = x 2 +C 5 2 2 5 = x 2 +C 5Lukmanulhakim Almamalik II- 1 
  • 2. RUMUS UMUM INTEGRAL1. ∫ k dx = kx + C2. ∫ k f ( x) dx = k ∫ f (x) dx3. ∫ [ f ( x) ± g ( x)] dx = ∫ f ( x) dx ± ∫ g ( x) dx 1 ∫x dx = ∫ dx = ln | x | +C −14. x ∫e dx = e x + C x5.Contoh 10.3 ∫ ( 2x − 7 ) dx 5Cari anti turunan yang umum dari fungsiPenyelesaian ∫ (2x − 7 ) dx = ∫ 2 x 5 dx − ∫ 7 dx 5 = 2∫ x5 dx − ∫ 7 dx ⎛ 1 5+1 ⎞   = 2⎜ x ⎟ − 7x + C ⎝ 5 +1 ⎠ 1 6 = x − 7x + C 3Contoh 10.4 1 1 6 ∫ x dx = 5 + 1 x 5+11. 5 = x +C 62. ∫ x .dx = ∫ x½.dx = x = 2 x + C 3/ 2 3 3 2 3 1 1 x −2 13. ∫ .dx = ∫ x-3. dx = x −3+1 = =- +C x3 − 3 +1 −2 2x2 2 2m 34. ∫ 2m2.dm = m 2+1 = +C 2 +1 3 5 12 +1 10 λ35. ∫ 5 λ .dλ = 5λ 2 = λ = 1 +C 2 +1 1 36. ∫ 1 .dθ = ∫ θ-½.dθ = θ 1/ 2 = 2 θ +C θ 1/ 2 ⎛ 1 2x 2 ⎞ 37. ∫ ⎝ 3x 3 ⎟ dx ⎜ 2− ⎠Lukmanulhakim Almamalik II- 2 
  • 3. ⎛ 1 2x 2 ⎞ 3 3 1 2x 2 ∫ ⎝ 3x 3 ⎠ ⎜ 2− ⎟ dx = ∫ 2 dx − ∫ 3x 3 dx 1 2 3 = ∫ x −2 dx − ∫ x 2 dx 3 3 1⎛ 1 ⎞ 2⎛ 1 ⎞ = ⎜ x −2 +1 ⎟ − ⎜ x ( 2 +1) ⎟ + C 3 3 ⎝ −2 + 1 ⎠ 3 ⎜ ( 3 2 + 1) ⎝ ⎟ ⎠ 2⎛ 2 5 ⎞ = ( − x −1 ) − ⎜ x 2 ⎟ + C 1 3 3⎝ 5 ⎠ 5 1 4x 2 =− − +C 3 x 15 ⎛3 ⎞ ∫ ⎜ x + 2e + 5 ⎟ dx8. x ⎝ ⎠ ⎛3 ⎞ 1 ∫ ⎜ x + 2e + 5 ⎟ dx = 3∫ dx + 2 ∫ e x dx + 5∫ dx x ⎝ ⎠ x = 3ln | x | + 2 e x + 5 x + C du9. ∫ = ln u + c u10.Latihan1. ∫ (3x2 + 7x).dx 1 5 22. ∫( x + 2 + x + 4x3)dx x 3ATURAN PANGKAT YANG DIPERUMUM [ f ( x )] n+1 ∫ ⎡ f ( x) ⋅ ( f ( x)) ⎤ dx = +C n ⎣ ⎦ n +1Contoh 10.5 ∫ 6 x( x + 1) 2 dx 2CariPenyelesaian: Misalkan f(x) = x2+1 maka f′ (x) = 2x [ f ( x )] n+1 ∫ ⎡ f ( x) ⋅ ( f ( x)) ⎤ dx = +C nJadi menurut aturan ⎣ ⎦ n +1Lukmanulhakim Almamalik II- 3 
  • 4. [ x 2 + 1] 2+1 = +C 2 +1 = 1 / 3.( x 2 + 1)3 + CContoh 10.6 ∫x x 3 + 1 dx 2CarilahPenyelesaian:Jika kita ambil f ( x ) = x + 1 , sehingga kita misalkan u = x3 + 1 . 3 du duKita diferensiasikan u menjadi = 3x 2 → du = 3 x 2 dx → = dx dx 3x 2 duSekarang kita substitusikan x3 + 1 dengan u dan dx dengan , sehingga kita dapatkan 3x 2persamaan berikut. du ∫x x3 + 1 dx = ∫ x 2 u 2 3x 2Selanjutnya integralkan 1 =∫ u du 3 1 ∫ ( u ) du 1 = 2 3 1⎛ 2 3 ⎞ = ⎜ u 2 ⎟+C 3⎝ 3 ⎠ 2 3 = u 2 +C 9Sekarang kita harus substitusikan kembali u dengan x 3 + 1 untuk menemukan jawaban akhir. ∫ ( x + 1) 2 + C 2 3 3 x 2 x 3 + 1 dx = 9Lukmanulhakim Almamalik II- 4 
  • 5. 2 . ∫ x x 2 + 1 ⋅ dx ;misalkan u = x 2 + 1 du = 2 x ⋅ dx 1∫x x 2 + 1 ⋅ dx = ∫ 2 x ( x + 1) ⋅ dx 2 1/ 2 2 1 = ∫ u 1 / 2 du 2 1 2 3/2 = ⋅ u +C 2 3 1 = ( x 2 + 1) 3 / 2 + C 3Latihan ∫ (x + 6 x )5 (6 x 2 + 12)dx 31. ∫ (x + 4)10 x.dx 22. x23. ∫ ( + 3) 2 x 2 .dx 2Persamaan DiferensialRumus Umum ∫ f ( x)dx = F ( x) + CHal ini benar asalkan F′(x) = f (x). Jika F′(x) = f (x) maka ini setara dengan dF(x) = f(x) dx.Dengan demikian maka kita bisa tuliskan persamaan tersebut dalam bentuk ∫ f ( x)dx = F ( x) + CIntegral Sinus dan Kosinus∫ cos x.dx = sin x + C∫ sin x.dx = - cos x + C∫ tan x.dx = - ln (cos x) + CLukmanulhakim Almamalik II- 5 

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