Tugas 2 Kalkulus II

SOAL – SOAL 5.1

Jika anti turunan F(x) + C yang umum untuk masing – masing yang berikut :

1. F(x) = 4

Jawab : F’(x) = ∫ 4 dx

= 4x + C

3. F(x) = 3x2 + √2

Jawab : F’ (x) = ∫ 3x2 + √2 dx

= 3x3 + √2x 3 = x3 + √2x + C

5. F(x) = x 2/3 dx

Jawab : F’ (x) = ∫ x 2/3 dx

= 1 x2/3 +1

2/3+1

= 1 x5/3

5/3

= 3/5 x 5/3 + C

7. F(x) = 6x2 – 6x + 1 dx

Jawab : F’(x) = ∫ 6x2 – 6x + 1 dx

= 6 x3 – 6 x2 + x3 2

= 2x3 – 3x2 + x + C

9. F(x) = 18x8 – 25x4 + 3x2 dx

Jawab : F’(x) = ∫ 18x8 – 25x4 + 3x2 dx

= 18 x9 – 25 x5 + 3 x3 9 5 3

= 2x9 – 5x5 + x3 + C

11. F(x) = 4 - 3 dx x5 x4

Jawab : F’(x) = ∫4 - 3 dx x5 x4

= ∫ 4x-5 – 3x-4 dx

= 4 x-5+1 - 3 x-4+1 -5+1 -4+1

= 4 x-4 – 3 x-3 -4 -3

= -x-4 + x-3 + C

13. F(x) = 4x6 + 3x5 - 8dx x5

Jawab : F’(x) = ∫ 4x6 + 3x5 - 8dx x5

= ∫ 4x + 3 – 8 dx x5

= ∫ 4x + 3 – 8x-5 dx

= 4 x2 + 3x – 8 x-5+1

2 -5+1

= 2x2 + 3x + 2x-4 + C

Dalam Soal – soal 15 – 22, cari integaral tak tentu.

15. ∫ (x3 + √x) dx =

Jawab : ∫ x3 + x1/2 dx = 1 x4 + 1 x 3/2

4 3/2

= 1/4 x4 + 2/3 x 3/2 + C

17. ∫ (y2 + 4y)2 dy =

Jawab : ∫ (y2 + 4y)2 dy = ∫ (y4 + 8y3 + 16y2) dy

= 1/5 y5 + 8/4 y4 +16/3 y3 + C

= 1/5 y5 + 2y4 +16/3 y3 + C

19. ∫ x4 – 2x3 + 1 dx =

x2

Jawab : ∫ x4 – 2x3 + 1 dx = ∫ x2 – 2x + 1/x2 dx = ∫ x2 – 2x + x-2 dx

x2

= 1/3 x3 – x2 + -1 x-1

= 1/3 x3 – x2 -1 x-1 + C

21. ∫ (3 sin t – 2 cos t) dt =

Jawab : ∫ (3 sin t – 2 cos t) dt =

= - 3 cos t – 2 sin t + C

Dalam soal – soal 23 – 38, gunakan metode – metode pada contoh 5 dan 6 untuk mencari integral

tak tentu.

23. ∫ (3x + 1)4 3 dx =

Jawab : Andaikan g (x)= 3x + 1

g'(x) = 3

dengan demikian ∫ (3x + 1)4 3 dx = ∫(g(x))4 g’ dx

= (g(x))5 + C 5 = (3x + 1) 5 + C

5

25. ∫ (5x3 - 18)7 15x2 dx =Jawab : Andaikan g (x) = 5x3 – 18 g' (x) = 15x2

dengan demikian ∫ (5x3 - 18)7 15x2 dx = ∫ ∫(g(x))7 g’ dx

= (g(x))8 + C 8 = (5x3 – 18) 8 + C

8

27. ∫ 3x4(2x5 + 9)3 dx =Jawab : ∫ 3x4(2x5 + 9)3 dx = ( 2x5 + 9 )4 . 3x4

4 10x4

= ¼ (2x5 + 9 )4 . 3/10 = 3(2x5 + 9 )4 + C 40

29. ∫ (5x2 + 1) (5x3 + 3x - 8) dx =Jawab : Misalkan : V = 5x3 + 3x - 8 dv = 15x2 + 3Dengan demikian = ∫ 5x2 + 1. v6 dv

15x2 + 3

= ∫ 1/3 v6

= 1/3 (5x3 + 3x - 8)7 7 = (5x3 + 3x – 8)7 21

31. =

Jawab : ∫ 3t (2t2 – 11)⅓ dt t = 2t2 -11

∫ 3t (t∫⅓ 1/4t dt dt = 4t x du

3/4 ∫ t⅓ du = 1/4t x dt

3/4 x 3/4t ⅓

9/16 x t4/3

9/16 (2t2 – 11)4/3 + C

33. ∫ sin4 x cos x dx =

Jawab : Misalnya V = sin x

dv = cos x dx

Sehingga ∫ v4 dv = 1/5 v5 +C

35. ∫ (sin5 x2) (x cos x2) =

Jawab : 1/6 sin6 x2

1/12 sin6 x2 + C

37. ∫(x2 + 1)3 x2 dx =

Jawab : ∫(x2 + 1)3 x2 dx = ∫ (x6 + 3x4 + 3x2 +1). x2 dx

= x8 + 3x6 + 3x4 + x2

= 1/9 x9 + 3/7 x7 + 3/5 x5 + 1/3 x3 + C