Kumpulan Soal-soal Bab 5 Integral Beserta Jawaban Buku Kalkulus dan Geometri Analisis Jilid 1 Edisi Keempat

Soal-Soal 5.1

Carilah anti turunan F(x) + C yang umum untuk Soal-soal 1 – 14.

1.     f (x) = 4

2.     f (x) = 2x – 4

3.     f (x) = 3x² + √2

4.     f (x) = 5x⁴ + π

5.     f (x) = x^2/3

6.     f (x) = x^–3/4

7.     f (x) = 6x² – 6x + 1

8.     f (x) = 3x² + 10x – 7

9.     f (x) = 18x⁸ – 25x⁴ + 3x²

10.  f (x) = x²(20x⁷ – 7x⁴ + 6)

11.  f (x) = 4/x⁵ – 3/x⁴

12.  f (x) = 1/x³ + 6/x⁷

13.  f (x) = (4x⁶ + 3x⁵ – 8)/x⁵

14.  f (x) = (2x³ – 3x² + 1)/x²

 

Penyelesaian

1.     f(x) = 4

= ∫ 4 dx

= 4/1 x + c

= 4x + c

 

2.     f(x) = 2x – 4

= ∫ (2x – 4) dx

= 2/2 x² – 4/1 x + c

= x² – 4x + c

 

3.     f(x) = 3x² + √2

= ∫ (3x² + √2) dx

= 3/3 x³ + √2/1 x + c

= x³ + √2x + c

 

4.     f(x) = 5x⁴ + π

= ∫ (5x⁴ + π) dx

= 5/5 x⁵ + 1/1 πx + c

= x⁵ + πx + c

 

5.     f(x) = x²/3

= ∫ x^(2/3) dx

= 1/(5/3) x^5/3 + c

= 3/5 x^5/3 + c

 

6.     f(x) = x^–3/4

= ∫ x^–3/4 dx

= 1/(1/4) x^1/4 + c

= 4x^1/4 + c

 

7.     f(x) = 6x² – 6x + 1

= ∫ (6x² – 6x + 1) dx

= 6/3 x³ – 6/2 x² + 1/1 x + c

= 2x³ – 3x² + x + c

 

8.     f(x) = 3x² + 10x – 7

= ∫ (3x² + 10x – 7) dx

= 3/3x³ + 10/2x² – 7/1x + c

= x³ + 5x² – 7x + c

 

9.     f(x) = 18x⁸ – 25x⁴ + 3x²

= ∫ (18x⁸ – 25x⁴ + 3x²) dx

= 18/9 x⁹ – 25/5 x⁵ + 3/3 x³ + c

= 2x⁹ – 5x⁵ + x³ + c

 

10.  f(x) = x² (20x⁷ – 7x⁴ + 6)

= ∫ (20x⁹ – 7x⁶ + 6x²) dx

= 20/10 x¹⁰ – 7/7 x⁷ + 6/3 x³ + c

= 2x¹⁰ – x⁷ + 2x³ + c

 

11.  f(x) = 4/x⁵ – 3/x⁴

= ∫ (4x^–5 – 3x^–4) dx

= 4/(–4) x^-4 – 3/(–3) x^-3 + c

= –1/x⁴ + 1/x³ + c

 

12.  f(x) = 1/x³ + 6/x⁷

= ∫ (x^–3 + 6x^–7) dx

= 1/(–2) x^–2 + 6/(–6) x^–6 + c

= –1/2 x² – 1/x⁶ + c

 

13.  f(x) = (4x⁶ + 3x⁵ – 8)/x⁵

= ∫ (4x + 3 – 8x^–5) dx

= 4/2 x² + 3/1 x – 8/(–4) x^–4 + c

= 2x² + 3x + 2x^-4 + c

= 2x² + 3x + 2/x⁴ + c

 

14.  f(x) = (2x³ – 3x² + 1)/x²

= ∫ (2x – 3 + x^-2) dx

= 2/2 x² – 3/1 x + 1/(–1) x^–1 + c

= x² – 3x – 1/x + c

 

Dalam Soal-soal 15 - 22, carilah integral tak tentunya.

15.  ∫ (x³ + √x) dx

16.  ∫ (x² + 1)² dx

17.  ∫ (y² + 4y)² dy

18.  ∫ y²(y² – 3) dy

19.  ∫ (x⁴ – 2x³ + 1) dx

20.  ∫ (x³ – 3x² + 1)/√x dx

21.  ∫ (3 sin t – 2 cos t) dt

22.  ∫ (3t² – 2 sin t) dt

 

Penyelesaian

15.  ∫ (x³ + √x) dx

= ∫ (x³ + x^1/2) dx

= 1/4 x⁴ + 1/(3/2) x^3/2 + c

= 1/4 x⁴ + 2/3 x^3/2 + c

 

16.  ∫ (x² + 1)² dx

= ∫ (x⁴ + 2x² + 1) dx

= 1/5 x⁵ + 2/3 x³ + 1/1 x + c

= 1/5 x⁵ + 2/3 x³ + x + c

 

17.  ∫ (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

 

18.  ∫ (y² (y² – 3) dy

= ∫ (y⁴ – 3y²) dy

= 1/5 y⁵ – 3/3y³ + c

= 1/5 y⁵ – y³ + c

 

19.  ∫ (∫x⁴ – 2x³ + 1)/x² dx

= ∫ (x² – 2x + x^-2) dx

= 1/3 x³ – 2/2 x² + 1/(–1) x^–1 + c

= 1/3 x³ – x² – 1/x + c

 

20.  ∫ (x³ – 3x² + 1)/√x dx

= ∫ (x^5/2 – 3x^3/2 + x^–1/2) dx

= 1/(7/2) x^7/2 – 3/(5/2) x^5/2 + 1/(1/2) x^1/2 + c

= 2/7 x^7/2 – 6/5 x^5/2 +2√x + c

 

21.  ∫ (3 sin t – 2 cos t) dt

= – 3 cos t – 2 sin t + c

= – (3 cos t + 2 sin t) + c

 

22.  ∫ (3t² – 2 sin t) dt

= 3/3 t³ – (–2 cos t) + c

= t³ + 2 cos 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

24.  ∫ (x² – 4)³ 2x dx

25.  ∫ (5x³ – 18)⁷ 15x² dx

26.  ∫ (x² – 3x + 2)² (2x – 3) dx

27.  ∫ 3x⁴ (2x⁵ + 9)³ dx

28.  ∫ 3x √(3x² + 7) dx

29.  ∫ (5x² + 1) (5x³ + 3x – 8)⁶ dx

30.  ∫ (5x² + 1) √(5x³ + 3x – 2) dx

31.  ∫ 3t ∛(2t² – 11) dt

32.  ∫ 3y/√(2y² + 5) dy

33.  ∫ sin⁴ x cos x dx

34.  ∫ (cos⁴ 2x) (–2 sin 2x) dx

35.  ∫ (sin⁵ x²) (x cos x²) dx

36.  ∫ cos (3x + 1) sin (3x + 1) dx

37.  ∫ (x² + 1)³ x² dx

38.  ∫ (x⁴ – 1) x² dx

 

Penyelesaian

23.  ∫ (3x + 1)⁴ 3 dx

Jika u = (3x + 1), maka du = 3 dx;

= ∫ u⁴ du

= 1/5 u⁵ + c

= 1/5 (3x + 1)⁵ + c

 

24.  ∫ (x² – 4)³ 2x dx

Jika u = (x² – 4), maka du = 2x dx;

= ∫ u³ du

= 1/4 u⁴ + c

= 1/4 (x² – 4)⁴ + c

 

25.  ∫ (5x³ – 18)⁷ 15x² dx

Jika u = (5x³ – 18), maka du = 15x² dx;

= ∫ u⁷ du

= 1/8 u⁸ + c

= 1/8 (5x³ – 18)⁸ + c

 

26.  ∫ (x² – 3x + 2)² (2x – 3) dx

Jika u = (x² – 3x + 2), maka du = 2x – 3 dx;

= ∫ u² du

= 1/3 u³ + c

= 1/3 (x² – 3x + 2)³ + c

 

27.  ∫ 3x⁴ (2x⁵ + 9)³ dx

Jika u = (2x⁵ + 9), maka du = 10x⁴ dx;

= 3/10 ∫ u³ du

= 3/40 u⁴ + c

= 3/40 (2x⁵ + 9)⁴ + c

 

28.  ∫ 3x √(3x² + 7) dx

Jika u = (3x² + 7), maka du = 6x dx;

= 1/2 ∫ u^1/2 du

= 1/(2×3/2) u^3/2 + c

= 1/3 √((3x² + 7)³) + c

 

29.  ∫ (5x² + 1) (5x³ + 3x – 8)⁶ dx

Jika u = (5x³ + 3x – 8), maka du = 15x² + 3 dx;

=1/3 ∫ u⁶ du

= 1/(3×7) u⁷ + c

= 1/21 (5x³ + 3x – 8)⁷ + c

 

30.  ∫ (5x² + 1) √(5x³ + 3x – 2) dx

Jika u = (5x³ + 3x – 2), maka du = 15x² + 3 dx;

= 1/3 ∫ u^1/2 du

= 1/(3×3/2) u^3/2 + c

= 2/9 √((5x³ + 3x – 2)³) + c

 

31.  ∫ 3t ∛(2t² – 11) dt

Jika u = (2t² – 11), maka du = 4t dt;

= 3/4 ∫ u^1/3 du

= 3/(4×4/3) u^4/3 + c

= 9/16 ∛((2t² – 11)⁴) + c

 

32.  ∫ 3y/√(2y² + 5) dy

Jika u = (2y² + 5), maka du = 4y dy;

= 3/4 ∫ u^–1/2 du

= 3/(4×1/2) u^1/2 + c

= 3/2 √(2y² + 5) +c

 

33.  ∫ sin⁴ x cos x dx

Jika u = sin x, maka du = cos x dx;

= ∫ u⁴ du

= 1/5 u⁵ + c

= 1/5 sin⁵ + c

 

34.  ∫ (cos⁴ 2x) (–2 sin 2x) dx

Jika u = cos 2x, maka du = –2 sin 2x dx;

= ∫ u⁴ du

= 1/5 u⁵ + c

= 1/5 cos⁵ 2x + c

 

35.  ∫ (sin⁵ x²) (x cos x²) dx

Jika u = sin x², maka du = 2x cos x² dx;

= 1/2 ∫ u⁵ du

= 1/(2×6) u⁶ + c

= 1/12 sin⁶ x² + c

 

36.  ∫ cos (3x + 1) sin (3x + 1) dx

Jika u = cos (3x + 1), maka du = –3 sin (3x + 1) dx;

= –1/3 ∫ u du

= – 1/(3×2) u² + c

= –1/6 cos² (3x + 1) + c

 

37.  ∫ (x² + 1)³ x² dx

= c

 

38.  ∫ (x⁴ – 1) x² dx

= ∫ x⁶ – x² dx

= 1/7 x⁷ – 1/3 x³ + c

 

Dalam Soal-soal 39 – 44, f’’(x) diketahui. Cari f(x) melalui anti penurunan dua kali. Perhatikan bahwa dalam kasus ini jawab anda harus mengandung dua konstanta sebarang, dari masing-masing anti penurunan satu. Misalnya, jika f’’(x) = x, maka f’(x) = x²/2 + C₁ dan f(x) = x³/6 + C₁x + C₂. Konstanta C₁ dan C₂ tidak dapat digabungkan.

39.  f’’(x) = 3x + 1

40.  f’’(x) = –2x + 3

41.  f’’(x) = √x

42.  f’’(x) = x^4/3

43.  f’’(x) = (x⁴ + 1)/x³

44.  f’’(x) = 2 ∛(x + 1)

 

Penyelesaian

39.  f’’(x) = 3x + 1

f’(x) = 3/2 x² + 1/1 x + C₁

f’(x) = 3/2 x² + x + C₁

f(x) = 3/(2×3) x³ + 1/2 x² + C₁ + C₂

f(x) = 1/2 x³ + 1/2 x² + C₁ + C₂

 

40.  f’’(x) = –2x + 3

f'(x) = –2/2 x² + 3/1 x + C₁

f'(x) = –x² + 3x + C₁

f(x) = –1/3 x³ + 3/2 x² + C₁ + C₂

 

41.  f’’(x) = √x

f'(x) = 1/(3/2) x^3/2 + C­

f'(x) = 2/3 x^3/2 + C₁

f(x) = 2/(3×5/2) x^5/2 + C₁ + C₂

f(x) = 4/15 x^5/2 + C₁ + C₂

 

42.  f’’(x) = x^4/3

f'(x) = 1/(7/3) x^7/3 + C₁

f'(x) = 3/7 x^7/3 + C₁

f(x) = 3/(7×10/3) x^10/3 + C₁ + C₂

f(x) = 9/70 x^10/3 + C₁ + C₂

 

43.  f’’(x) = (x⁴ + 1)/x³

f’’(x) = x + x^-3

f’(x) = 1/2 x² + 1/(–2) x^–2 + C₁

f’(x) = 1/2 x² – 1/(2x²) + C₁

f(x) = 1/(2×3) x³ – 1/(2×(–1)) x^–1 + C₁ + C₂

f(x) = 1/6 x³ + 1/2x + C₁ + C₂

 

44.  f’’(x) = 2 ∛(x + 1)

f'(x) = 2/(4/3) (x + 1)^4/3 + C₁

f'(x) = 3/2 (x + 1)^4/3 + C₁

f(x) = 3/(2×7/3) (x + 1)^7/3 + C₁ + C₂

f(x) = 9/14 (x + 1)^7/3 + C₁ + C₂

 

Soal-Soal 5.6

Dalam Soal 1 – 14, gunakan Teorema Dasar Kalkulus untuk menghitung tiap integral tentu.

1.     ₀∫² x³ dx

2.     _-1∫² x⁴ dx

3.     -₁∫² (3x² – 2x + 3) dx

4.     ₁∫² (4x³ + 7) dx

5.     ₁∫⁴ 1/w² dw

6.     ₁∫³ 2/t³ dt

7.     ₀∫⁴ √t dt

8.     ₁∫⁸ ∛w dw

9.     _-4∫^-2 (y² + 1/y³) dy

10.  ₁∫⁴ (s⁴ – 8)/s² ds

11.  ₀∫^(π/2) cos x dx

12.  _(π/6)∫^(π/2) 2 sin t dt

13.  ₀∫¹ (2x⁴ – 3x² + 5) dx

14.  ₀∫¹ (x^4/3 – 2x^1/3) dx

 

Penyelesaian

1.     ₀∫² x³ dx

= [1/4 x⁴]₂

= 1/4 (2)⁴

= 16/4

= 4

 

2.     _–1∫² x⁴ dx

= [1/5 x⁵]₂ – [1/5 x⁵]_–1

= 2⁵/5 – (–1)⁵/5

= (32+1)/5

= 33/5

 

3.     _–1∫² (3x² – 2x + 3) dx

= [3/3 x³ – 2/2 x² + 3/1 x]₂ – [3/3 x³ – 2/2 x² + 3/1 x]_–1

= [x³ – x² + 3x]₂ – [x³ – x² + 3x]_–1

= [2³ – 2² + 3(2)] – [(–1)³ – (–1)² + 3(–1)]

= [8 – 4 + 6] – [(–1) – 1 – 3]

= 10 + 5

= 15

 

4.     ₁∫² (4x³ + 7) dx

= [4/4 x⁴ + 7/1 x]₂ – [4/4 x⁴ + 7/1 x]₁

= [x⁴ + 7x]₂ – [x⁴ + 7x]₁

= [2⁴ + 7(2)] – [1⁴ + 7(1)]

= [16 + 14] – [1 + 7]

= 30 – 8

= 22

 

5.     ₁∫⁴ 1/w² dw

= ₁∫⁴ w^–2 dw

= [1/(–1) w^–1]₄ – [1/(–1) w^–1]₁

= [–w^–1]₄ – [–w^–1]₁

= [–1/(4)] – [–1/1]

= 3/4

 

6.     ₁∫³ 2/t³ dt

= ₁∫³ 2t^-3 dt

= [2/(–2) t^–2]₃ – [2/(–2) t^–2]₁

= [–t^–2]₃ – [–t^–2]₃

= –1/3² – (–1/1²)

= –1/9 + 9/9

= 8/9

 

7.     ₀∫⁴ √t dt

= ₀∫⁴ t^1/2 dt

= [–1/(3/2) t^3/2]₄ – [–1/(3/2) t^3/2]₀

= [–2/3 t^3/2]₄ – [–2/3 t^3/2]₀

= 2/3 √(4³)

= 2/3 (8)

= 16/3

 

8.     ₁∫⁸ ∛w dw

= ₁∫⁸ w^1/3 dw

= [1/(4/3) w^4/3]₈ – [1/(4/3) w^4/3]₁

= [3/4 w^4/3]₈ – [3/4 w^4/3]₁

= 3/4 ∛(8⁴) – 3/4 ∛(1⁴)

= (48 – 3)/4

= 45/4

 

9.     _–4∫^–2 (y² + 1/y³) dy

= _–4∫^–2 (y² + y^–3) dy

= [1/3 y³ + (–1/2) y^–2]_–2 – [1/3 y³ + (–1/2) y^–2]_–4

= [1/3 (–2)³ + (–1/(2(–2)²))] – [1/3 (–4)³ + (–1/(2(–4)²))]

= [–8/3 – 1/8] – [–64/3 + (–1/32)]

= (–64 – 3)/24 + (2048 + 3)/96

= (–268)/96 + 2051/96

= 1783/96

 

10.  ₁∫⁴ (s⁴ – 8)/s² ds

= ₁∫⁴ s² – 8s^–2 ds

= [1/3 s³ – (8/(–1)) s^–1]₄ – [1/3 s³ – (8/(–1)) s^–1]₁

= [1/3 (4)³ + 8/4] – [1/3 (1)³ + 8/1]

= [64/3 + 8/4] – [(1 + 24)/3]

= (256 + 24)/12 – 100/12

= 180/12

= 15

 

11.  ₀∫^π/2 cos x dx

= [sin x]₉₀ – [sin x]₀

= sin(90) – sin(0)

= 1

 

12.  _π/6∫^π/2 2 sin t dt

= [–2 cos t]₉₀ – [–2 cos t]₃₀

= –2 cos (90) + 2 cos (30)

= –2(0) + 2(√3/2)

= √3

 

13.  ₀∫¹ (2x⁴ – 3x² + 5) dx

= [2/5 x⁵ – 3/3 x³ + 5/1 x]₁ – [2/5 x⁵ – 3/3 x³ + 5/1 x]₀

= 2/5 (1)⁵ – (1)³ + 5(1)

= (2 – 5 + 25)/5

= 22/5

 

14.  ₀∫¹ (x^4/3 – 2x^1/3) dx

= [1/(7/3) x^7/3 – 2/(4/3) x^4/3]₁ – [1/(7/3) x^7/3 – 2/(4/3) x^4/3]₀

= 3/7 (1) ^7/3 – 3/2 (1)^4/3

= (6 – 21)/14

= –15/14

 

Dalam Soal 15 – 30,  gunakan Teorema Dasar Kalkulus dikombinasikan dengan Aturan Pangkat Diperumum untuk menghitung integral tentu yang diberikan (lihat Contoh 7 – 10).

15.  ₀∫¹ (x² + 1)¹⁰ (2x) dx

16.  _-1∫⁰ √(x³ + 1) (3x²) dx

17.  _-1∫³ 1/(t + 2)² dt

18.  ₂∫¹⁰ √(y – 1) dy

19.  ₅∫⁸ √(3x + 1) dx

20.  ₁∫⁷ 1/√(2x + 2) dx

21.  _-3∫³ √(7 + 2t²) (8t) dt

22.  ₁∫³ (x² + 1)/ √(x³ + 3x) dx

23.  ₀∫^π/2 cos² x sin x dx

24.  ₀∫^π/2 sin² 3x cos 3x dx

25.  ₀∫^π/2 (2x + sin x) dx

26.  ₀∫^π/2 [4x + 3 + cos x) dx

27.  ₀∫⁴ [√x √(2x + 1)] dx

28.  _-4∫^-1 (1 – s⁴)/2s² ds

29.  ₀∫¹ (x² + 2x)² dx

30.  ₀∫^8a (a^1/3 – x^1/3)³ dx

 

Penyelesaian

15.  ₀∫¹ (x² + 1)¹⁰ (2x) dx

Jika u = (x² + 1), maka du = 2x dx;

= ₀∫¹ u¹⁰ du

= [1/11 (x² + 1)¹¹]₁ – [1/11 (x² + 1)¹¹]₀

= 1/11 [1² + 1]¹¹

= 2048/11

 

16.  _–1∫⁰ √(x³ + 1)(3x²) dx

Jika u = (x³ + 1), maka du = 3x² dx;

= _–1∫⁰ u^1/2 du

= [1/(3/2) (x³ + 1)^3/2]₀ – [1/(3/2) (x³ + 1)^3/2]_–1

= 2/3 √([(0³ + 1)]³) – 2/3 √([(–1)³ + 1]³

= 2/3(1) – 2/3(0)

= 2/3

 

17.  _–1∫³ 1/(t + 2)² dt

Jika u = (t + 2), maka du = dt;

= _–1∫³ u^-2 du

= [1/(–1) (t + 2)^–1]₃ – [1/(–1) (t + 2)^–1]_–1

= –1/((3 + 2)) – 1/((–1 + 2))

= –1/5 + 1

= 4/5

 

18.  ₂∫¹⁰ √(y – 1) dy

Jika u = (y – 1), maka du = dy;

= ₂∫¹⁰ u^1/2 du

= [1/(3/2) (y – 1)^3/2]₁₀ – [1/(3/2) (y – 1)^3/2]₂

= 2/3 √((10 – 1)³) – 2/3 √((2 – 1)³)

= 2/3(27) – 2/3(1)

= 52/3

 

19.  ₅∫⁸ √(3x + 1) dx

Jika u = (3x + 1), maka du = 3 dx;

= 1/3 ₅∫⁸ u^1/2 du

= [1/(3×3/2) (3x + 1)^3/2]₈ – [1/(3×3/2) (3x + 1)^3/2]₅

= 2/9 √([3(8) + 1]³) – 2/9 √([3(5) + 1]³)

= 2/9(125) – 2/9(64)

= (250 – 128)/9

= 122/9

 

20.  ₁∫⁷ 1/√(2x + 2) dx

Jika u = 2x + 2, maka du = 2 dx;

= 1/2 ₁∫⁷ u^–1/2 du

= [1/(2×1/2) (2x + 2)^1/2]₇ – [1/(2×1/2) (2x + 2)^1/2]₁

= √(2(7) + 2) – √(2(1) + 2)

= 4 – 2

= 2

 

21.  _–3∫³ √(7 + 2t²)(8t) dt

Jika u = (7 + 2t²), maka du = 4t dt

= 2 ∫ u^1/2 du

= [2/(3/2) (7 + 2t²)^3/2]₃ – [2/(3/2) (7 + 2t²)^3/2]_–3

= 4/3 √([7 + 2(3)²]³) – 4/3 √([7 + 2(–3)²]³)

= 4/3(125) – 4/3(125)

= 0

 

22.  ₁∫³ (x² + 1)/√(x³ + 3x) dx

Jika u = (x³ + 3x), maka du = (3x² + 3) dx

= 1/3 ₁∫³ u^–1/2 du

= [1/(3×1/2) (x³ + 3x) ^1/2]₃ – [1/(3×1/2) (x³ + 3x) ^1/2]₁

= 2/3 √(3³ + 3(3)) – 2/3 √(1³ + 3(1))

= 2/3(6) – 2/3(2)

= 8/3

 

23.  ₀∫^π/2 cos² x sin x dx

Jika u = cos x, maka du = –sin x dx

= – ₀∫⁹⁰ u² du

= [–1/3 cos³ x]₉₀ – [–1/3 cos³ x]₀

= –1/3 cos³ (90) + 1/3 cos³ (0)

= –1/3(0) + 1/3(1)

= 1/3

 

24.  ₀∫^π/2 sin² 3x cos 3x dx

Jika u = sin 3x, maka du = 3 cos 3x dx;

= 1/3 ₀∫^π/2 u² du

= [1/(3×3) sin³ 3x]₉₀ – [1/(3×3) sin³ 3x]₀

= 1/9 sin³ 3(90) – 1/9 sin³ 3(0)

= 1/9(–1) – 1/9(0)

= –1/9

 

25.  ₀∫^π/2 (2x + sin x) dx

= [2/2 x² + (–cos x)]₉₀ – [2/2 x² + (–cos x)]₀

= [x² – cos x)]₉₀ – [x² – cos x)]₀

= [(90)² – cos (90)] – [ (0)² – cos (0)]

= (8100 – 0) – (0 – 1)

= 8100 + 1

= 8101

 

26.  ₀∫^π/2 [4x + 3 + cos x] dx

= [4/2 x² + 3/1 x + sin x]₉₀ – [4/2 x² + 3/1 x + sin x]₀

= [2x² + 3x + sin x]₉₀ – [2x² + 3x + sin x]₀

= [2(90)² + 3(90) + sin (90)] – [2(0)² + 3(0) + sin (0)]

= (16200 + 270 + 1) – (0 + 0 + 0)

= 16471

 

27.  ₀∫⁴ [√x + √(2x + 1)] dx

Jika u = x, maka du = dx dan jika v = (2x + 1), maka dv = 2 dx;

= ₀∫⁴ u^1/2 du + 1/2 ₀∫⁴ v^1/2 dv

= [1/(3/2) x^3/2 + 1/(2×3/2) (2x + 1)^3/2]₄ – [1/(3/2) x^3/2 + 1/(2×3/2) (2x + 1)^3/2]₀

= [2/3 √(x³) + 1/3 √((2x + 1)³)]₄ – [2/3 √(x³) + 1/3 √((2x + 1)³)]₀

= [2/3 √((4)³) + 1/3 √((2(4) + 1)³)] – [2/3 √((0)³) + 2/3 √((2(0) + 1)³)]

= [2/3 (8) + 1/3 (27)] – [0 + 2/3]

= (16 + 27 – 2)/3

= 41

 

28.  _–4∫^–1 (1 – s⁴)/(2s²) ds

= _–4∫^–1 1/2(s^–2 – s²) ds

= [1/(2(–1)) s^–1 – 1/2(3) s³]_–1 – [ 1/(2(–1)) s^–1 – 1/2(3) s³]_–4

= [–1/2s – s³/6]_–1 – [–1/2s – s³/6]_–4

= [–1/2(–1) – ((–1)³)/6] – [(–1/2(–4) – ((–4)³)/6)]

= [1/2 + 1/6] – [1/8 + 64/6]

= (12 + 4 – 3 – 256)/24

= –243/24

= –81/8

 

29.  ₀∫¹ (x² + 2x)² dx

= ₀∫¹ x⁴ + 4x³ + 4x² dx

= [1/5 x⁵ + 4/4 x⁴ + 4/3 x³]₁ – [1/5 x⁵ + 4/4 x⁴ + 4/3 x³]₀

= 1/5 (1)⁵ + (1)⁴ + 4/3 (1)³

= (3 + 15 + 20)/15

= 38/15

 

30.  _a∫^8a (a^1/3 – x^1/3)³ dx
= _a∫^8a (a – 3a^2/3 x^1/3 + 3a^1/3 x^2/3 – x) dx

= [ax – 9/4 x^4/3 a^2/3 + 9/5 a^1/3 x^5/3 – 1/2 x²]_8a – [ax – 9/4 x^4/3 a^2/3 + 9/5 a^1/3 x^5/3 – 1/2 x²]_a

= [a(8a) – 9/4 (8a)^4/3 a^2/3 + 9/5 a^1/3 (8a)^5/3 – 1/2 (8a)²] – [a(a) – 9/4 (a)^4/3 a^2/3 + 9/5 a^1/3 (a)^5/3) – 1/2 (a)²]

= – 49/20 a²

 

Soal-Soal 5.8

Gunakan metode penggantian untuk mencari integral tak-tentu berikut.

1.     ∫ √(3x + 2) dx

2.     ∫ ∛(2x – 4) dx

3.     ∫ cos(3x + 2) dx

4.     ∫ sin(2x – 4) dx

5.     ∫ x √(x² + 4) dx

6.     ∫ x²(x³ + 5)⁹ dx

7.     ∫ x sin(x² + 4) dx

8.     ∫ x² cos(x³ + 5) dx

9.     ∫ (x sin√(x² + 4))/ √(x² + 4) dx

10.  ∫ x²(x³ + 5) cos[(x³ + 5)⁹] dx

11.  ∫ x cos(x² + 4) √(sin(x² + 4)) dx

12.  ∫ x² sin(x³ + 5) cos⁹(x³ + 5) dx

13.  ∫ ((√t + 4)³)/ √t dx

14.  ∫ (1 + 1/t)^-2 (1/t²) dt

 

Penyelesaian

1. ∫ √(3x + 2) dx

Jika u = (3x + 2), maka du = 3 dx

= 1/3 ∫ u^(1/2) du
= 1/(3×3/2) (3x + 2)^(3/2) + c
= 2/9 √((3x + 2)³) + c

 

2. ∫ ∛(2x – 4) dx

Jika u = (2x – 4), maka du = 2 dx

= 1/2 ∫ u^(1/3) du

= 1/(2×4/3) (2x – 4)^(4/3) + c

= 3/8 ∛((2x – 4)⁴) + c

 

3. ∫ cos(3x + 2) dx

Jika u = 3x + 2, maka du = 3 dx

= 1/3 ∫ cos u du

= 1/3 sin(3x + 2) + c

 

4. ∫ sin(2x – 4) dx

Jika u = 2x – 4, maka du = 2 dx

= 1/2 ∫ sin u du

= – 1/2 cos(2x – 4) + c

 

5. ∫ x √(x² + 4) dx

Jika u = x² + 4, maka du = 2x dx

= 1/2 ∫ u^(1/2) du
= 1/(2×3/2) (x² + 4)^(3/2) + c

= 1/3 √((x² + 4)³) + c

 

6. ∫ x² (x³ + 5)⁹ dx

Jika u = x³ + 5, maka du = 3x² dx

= 1/3 ∫ u⁹ du
= 1/(3×10) (x³ + 5)¹⁰ + c
= 1/30 (x³ + 5)¹⁰ + c

 

7. ∫ x sin(x² + 4) dx

Jika u = x² + 4, maka du = 2x dx;

= 1/2 ∫ sin u du

= – 1/2 cos (x² + 4) + c

 

8. ∫ x² cos(x³ + 5) dx

Jika u = x³ + 5, maka du = 3x² dx;

= 1/3 ∫ cos u du

= 1/3 sin(x³ + 5) + c

 

9. ∫ (x sin√(x² + 4))/√(x² + 4) dx

Jika u = (x² + 4)^(1/2), maka du = (x)(x² + 4)^(–1/2) dx

= ∫ sin u du

= – cos √(x² + 4) + c

 

10. ∫ x² (x³ + 5)⁸ cos[(x³ + 5)⁹] dx

Jika u = (x³ + 5)⁹, maka du = 27x² (x³ + 5)⁸ dx;

= 1/27 ∫ cos u du

= 1/27 sin[(x³ + 5)⁹] + c

 

11. ∫ x cos(x² + 4) √(sin(x² + 4)) dx

Jika u = sin(x² + 4), maka du = 2x cos(x² + 4) dx;

= 1/2 ∫ u^(1/2) du

= – 1/(2×3/2) sin(x² + 4)^(3/2) + c

= – 1/3 √(sin(x² + 4)³) + c

 

12. ∫ x² sin(x³ + 5) cos⁹ (x³ + 5) dx

Jika u = cos(x³ + 5), maka du = – 3x² sin(x³ + 5) dx;

= –1/3 ∫ u⁹ du

= –1/(3×10) cos¹⁰ (x³ + 5) + c

= –1/30 cos¹⁰ (x³ + 5) + c

 

13. ∫ (√t + 4)³/√t dt

Jika u = t^(1/2) + 4, maka du = 1/2 t^(–1/2) dt

= 2 ∫ u³ du

= 2/4 (t^(1/2) + 4)⁴ + c

= 1/2 (t^(1/2) + 4)⁴ + c

 

14. ∫ (1+1/t)^(–2) (1/t²) dt

Jika u = 1 + t^(–1) maka du = – t^(–2) dt;

= – ∫ u^^(–2) du

= –1/(–1) (1 + t^(–1))^{( –1)} + c

= 1/((1+1/t)) + c

 

Gunakan metode penggantian dalam integral tentu untuk menghitung masing-masing yang berikut (lihat Contoh 3 dan 4)

15.  ₀∫¹ (3x + 1)³ dx

16.  ₀∫⁴ √(2t + 1) dt

17.  ₀∫² t/√(t² + 9)² dt

18.  ₀∫^√5 √(9 – x²) x dx

19.  ₀∫¹ (x + 2)/(x² + 4x + 1)² dx

20.  ₀∫² x²/ (9 – x³)^3/2 dx

21.  ₀∫^π/6 sin³ θ cos θ dθ

22.  ₀∫^π/6 sin θ/ cos³ θ dθ

23.  ₀∫¹ cos(3x – 3) dx

24.  ₀∫^1/2 sin(2πx) dx

25.  ₀∫¹ x sin(πx²) dx

26.  ₀∫^π/4 (cos 2x + sin 2x) dx

27.  ₀∫^π/2 sin x sin(cos x) dx

28.  ₀∫¹ x cos³(x²) sin(x²) dx

29.  ₁∫⁴ 1/ √ t(√t + 1)³ dt

30.  ₁∫² (1 + 1/t)² (1/t²) dt

 

Penyelesaian

15. ₀∫¹ (3x + 1)³ dx

Jika u = (3x + 1) maka du = 3 dx;

= 1/3 ₀∫¹ u³ du

= [1/(3×4) (3x + 1)⁴]₁ – [1/(3×4) (3x + 1)⁴]₀

= 1/12 ([3(1) + 1]⁴ – [3(0) + 1]⁴)

= 1/12 (256 – 1)

= 255/12 = 85/4

 

16. ₀∫⁴ √(2t + 1) dt

Jika u = (2t + 1) maka du = 2 dt;

= 1/2 ₀∫⁴ u^(1/2) du

= [1/(2×3/2) (2t + 1)^(3/2)]₄ – [1/(2×3/2) (2t + 1)^(3/2)]₀

= 1/3 (√((2(4) + 1)³) – √((2(0) + 1)³))

= 1/3(27 – 1)

= 26/3

 

17. ₀∫² t/(t² + 9)² dt

Jika u = (t² + 9) maka du = 2t dt;

= 1/2 ₀∫² u^(–2) du

= [1/(2×(–1)) (t² + 9)^(-1)]₂­ – [1/(2×(–1)) (t² + 9)^(-1)]₀
= –1/2 (1/[(2)² + 9] – 1/[(0)² + 9])
= –1/2 (1/13 – 1/9)
= (–9 + 13)/234

= 4/234 = 2/117

 

18. ₀∫^(√5) √(9 – x²) x dx

Jika u = (9 – x²) maka du = – 2x dx;

= –1/2 ₀∫^(√5) u^(1/2) du

= [–1/(2×3/2) (9 – x²)^(3/2)]_√5 – [–1/(2×3/2) (9 – x²)^(3/2)]₀

= –1/3 (√([9 – (√5)²]³) – √([9 – (0)²]³))

= –1/3 (8 – 27)

= 19/3

 

19. ₀∫¹ (x + 2)/(x² + 4x + 1)² dx

Jika u = (x² + 4x + 1) maka du = (2x + 4) dx;

= 1/2 ₀∫¹ u^(-2) du

= [1/(2×(–1)) (x² + 4x + 1)^(–1)]₁ – [1/(2×(–1)) (x² + 4x + 1)^(–1)]₀

= –1/2 (1/[(1)² + 4(1) + 1] – 1/[(0)² + 4(0) + 1] )

= –1/2 (1/6 –1/1)

= (–1 + 6)/12

= 5/12

 

20. ₀∫² x²/(9 – x³)^(3/2) dx

Jika u = (9 – x³) maka du = – 3x² dx;

= –1/3 ₀∫² u^(–3/2) du
= [–1/(3×(–1/2)) (9 – x³)^(–1/2)]­₂ – [–1/(3×(–1/2)) (9 – x³)^(–1/2)]­₀
= 2/3 (1/√(9 – (2)³) – 1/√(9 – (0)³))
= 2/3 (1 – 1/3)
= (6 – 2)/9
= 4/9

 

21. ₀∫^(π/6) sin³ θ cos θ dθ

Jika u = sin θ maka du = cos θ dx;

= ₀∫^(π/6) u³ du
= [1/4 (sin θ)⁴]_(π/6) – [1/4 (sin θ)⁴]₀
= 1/4 (sin⁴ (π/6) – sin⁴ (0))
= 1/4 (1/16 – 0)
= 1/64

 

22. ₀∫^(π/6) sin θ/cos³ θ dθ

Jika u = cos θ maka du = – sin θ dθ;

= – ₀∫^(π/6) u^(–3) du
= [–1/2 (cos θ)^(-2)]_(π/6) – [–1/2 (cos θ)^(–2)]₀
= –1/2 (1/cos² (π/6) – 1/cos² 0)
= –1/2 (1/(3/4) – 1/1)
= (–4 + 4)/6

= –1/6

 

23. ₀∫¹ cos(3x – 3) dx

Jika u = 3x – 3 maka du = 3 dx;

= 1/3 ₀∫¹ cos u du
= [1/3 sin(3x – 3)]₁ – [1/3 sin(3x – 3)]₀
= 1/3 (sin[3(1) 3] – sin[3(0) – 3])
= sin(3)/3

 

24. ₀∫^(1/2) sin (2πx) dx

Jika u = 2πx maka du = 2π dx;

= 1/2π ₀∫^(1/2) sin u du
= [–1/2π cos 2πx]_(1/2) – [–1/2π cos 2πx]₀
= –1/2π (cos 2π(1/2) – cos 2π(0))
= –1/2π(–1–1)

= 1/π

 

25. ₀∫¹ x sin(πx²) dx

Jika u = πx² maka du = 2πx dx;

= 1/2π ₀∫¹ sin u du
= [–1/2π cos πx²]₁ – [–1/2π cos πx²]₀
= –1/2π (cos π(1)² – cos π(0)²)
= –1/2π(–1–1)

= 1/π

 

26. ₀∫^(π/4) (cos 2x + sin 2x) dx

Jika u = 2x maka du = 2 dx;

= 1/2 (₀∫^(π/4) cos u du + ₀∫^(π/4) sin u du)
= [1/2 (sin 2x – cos 2x)]_(π/4) – [1/2 (sin 2x – cos 2x)]_(π/4)
= 1/2 ([sin 2(π/4) – cos 2(π/4)] – [sin 2(0) – cos 2(0)])
= 1/2 ([1 – 0] – [0 – 1])

= 1/2 + 1/2
= 1

 

27. ₀∫^(π/2) sin x sin (cos x) dx

Jika u = cos x maka du = – sin x dx;

= – ₀∫^(π/2) sin u du
= [cos(cos x)]_(π/2) – [cos(cos x)]₀
= cos [cos (π/2) – cos (0)]

= cos (0 – 1)

= 1 – cos (1)

28. ₀∫¹ x cos³ (x²) sin(x²) dx

Jika u = cos(x²) maka du = –2x sin (x²) dx;

= –1/2 ₀∫¹ u³ du
= [–1/(2×4) cos⁴ (x²)]₁ – [–1/(2×4) cos⁴ (x²)]₀
= –1/8 (cos⁴ ([1]²) – cos⁴ ([0]²))

= – (cos⁴ (1)/8 – 1/8)

 

29. ₁∫⁴ 1/(√t (√t + 1)³) dt

Jika u = (t^(1/2) + 1) maka du = 1/2 t^(–1/2) dt;

= 2 ₁∫⁴ u^(–3) du
= [2/(–2) (t^(1/2) + 1)^(–2)]₄ – [2/(–2) (t^(1/2) + 1)^(–2)]₁
= –1/(√4 + 1)² + 1/(√1 + 1)²

= –1/9 + 1/4

= (–4 + 9)/36

= 5/36

 

30. ₁∫² (1 + 1/t)² (1/t²) dt

Jika u = (1 + t^(–1)), maka du = –t^(–2) dt;

= –₁∫² u² du
= [–1/3 (1 + t^(–1))³]₂ – [–1/3 (1 + t^(–1))³]₁
= –1/3 ((1 + 1/([2]))³ + (1 + 1/([1]))³)

= –1/3 (27/8 + 8)

= (–27 + 64)/24

= –37/24

 

Dalam Soal 31 – 38, gunakan simetri untuk membantu anda menghitung integral yang diberikan

31.  _–π∫^π (sin x + cos x) dx

32.  _–1∫¹ x³/ (1 + x²)⁴ dx

33.  _–π/2∫^π/2 sin x/ (1 + cos x) dx

34.  _–∛π∫^∛π x² cos(x³) dx

35.  _–π∫^π (sin x + cos x)² dx

 

Penyelesaian

31. _(–π)∫^π (sin x + cos x) dx

= [–cos x + sin x]_π – [–cos x + sin x]_–π

= (–cos π + sin π) – (–cos (–π) + sin (–π))

= (–(–1) + 0) – (– (–1) + 0)

= 1 – 1

= 0

 

32. _(–1)∫¹ x³/(1 + x²)⁴ dx

Jika u = (1 + x²), maka du = 2x dx;

= 1/2 _(–1)∫¹ (u – 1) u^(–4) du

= [1/2 (1/(–2) u^(–2) – 1/(–3) u^(–3))]₁ – [1/2 (1/(–2) u^(–2) – 1/(–3) u^(–3))] _–1

= (–1/2 [1/(2(1)²) – 1/(3(1)³)]) – (–1/2 [1/(2(–1)²) – 1/(3(–1)³)])

= (–1/4 + 1/6) – (–1/4 + 1/6)

= (–3 + 2 + 3 – 2)/12

= 0

 

33. _(–π/2)∫^(π/2) sin x/(1 + cos x) dx

Jika u = (1 + cos x), maka du = – sin x dx;

= –_(–π/2)∫^(π/2) u^(–1) du
= [ln|u|]_(π/2) – [ln|u|]_(–π/2)
= ln |1 + cos (π/2) | – ln |1 + cos (–π/2)|
= ln |1 + 0| – ln |1 + 0|
= 0

 

34. _(–∛π)∫^(∛π) x² cos (x³) dx

Jika u = x³, maka du = 3x² dx;

= 1/3 _(–∛π)∫^(∛π) cos u du

= [1/3 sin x³]_(∛π) – [1/3 sin x³]_(–∛π)

= (1/3 sin ([∛π]³)) – (1/3 sin ([–∛π]³))

= (1/3 (0)) – (1/3 (0))

= 0

 

35. _(–π)∫^π (sin x + cos x)² dx

= _(–π)∫^π 1 + sin 2x dx

= [x – 1/2 cos 2x]_(π) – [x – 1/2 cos 2x]_(–π)

= ((π) – 1/2 cos 2(π)) – ((–π) – 1/2 cos 2(–π))

= ((π) – 1/2) – ((–π) + 1/2)

= 2π