Solução_Calculo_Stewart_6e

F.
248 ¤ CHAPTER 5 INTEGRALS
TX.10
7.
x 4 − 1 2 x3 + 1 x − 2 dx = x5
4
5 − 1 x 4
2 4 + 1 x 2
4 2 − 2x + C = 1 5 x5 − 1 8 x4 + 1 8 x2 − 2x + C
9.
(1 − t)(2 + t 2 ) dt = (2 − 2t + t 2 − t 3 ) dt =2t − 2 t2 2 + t3 3 − t4 4 + C =2t − t2 + 1 3 t3 − 1 4 t4 + C
x 3 − 2 √
x x
3
11.
dx =
x
x − 2x1/2
x
13. (sin x +sinhx) dx = − cos x +coshx + C
15.
(θ − csc θ cot θ) dθ =
1
2 θ2 +cscθ + C
17. (1 + tan 2 α) dα = sec 2 αdα=tanα + C
dx = (x 2 − 2x −1/2 ) dx = x3
3 − 2 x1/2
1/2 + C = 1 3 x3 − 4 √ x + C
19.
cos x +
1
2 x dx =sinx + 1 4 x2 + C. The members of the family
in the figure correspond to C = −5, 0, 5,and10.
21.
23.
25.
27.
29.
2
0 (6x2 − 4x +5)dx = 6 · 1
3 x3 − 4 · 1
2 x2 +5x 2
= 2x 3 − 2x 2 +5x 2
=(16− 8+10)− 0=18
0 0
0
−1 (2x − ex ) dx = x 2 − e x 0
−1 =(0− 1) − 1 − e −1 = −2+1/e
2 (3u −2 +1)2 du = 2
−2 9u 2 +6u +1 du = 9 · 1
3 u3 +6· 1
2 u2 + u 2
= 3u 3 +3u 2 + u 2
−2 −2
4
1
−1
−2
= (24 + 12 + 2) − (−24 + 12 − 2) = 38 − (−14) = 52
√
4 4 t (1 + t) dt =
1 (t1/2 + t 3/2 2
) dt =
3 t3/2 + 2 5 t5/2 = 16
+ 64
3 5 − 2 + 2
3 5 =
14
+ 62 = 256
3 5 15
1
4y 3 + 2
−1
dy = 4 · 1 1
y 3 4 y4 +2·
−2 y−2 =
y 4 − 1 −1
=(1− 1) −
16 − 1
y 2 4 = −
63
4
1
31. x √ √
3 4
0 x + x
33.
4
1
dx = 1
0 (x4/3 + x 5/4 ) dx =
−2
−2
1
3
7 x7/3 + 4 9 x9/4 = 3
+ 4
7 9 − 0=
55
63
0
√ 4 5/x dx = 5
1 x−1/2 dx = √
5 2 √ 4
x = √ 5(2· 2 − 2 · 1) = 2 √ 5
1
35.
37.
π (4 sin θ − 3cosθ) dθ = − 4cosθ − 3sinθ π
=(4− 0) − (−4 − 0) = 8
0 0
π/4
0
1+cos 2 θ
cos 2 θ
dθ =
π/4
0
1
cos 2 θ + cos2 θ
dθ =
cos 2 θ
= tan θ + θ π/4
0
= tan π 4 + π 4
π/4
0
(sec 2 θ +1)dθ
− (0 + 0) = 1 +
π
4