Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 5 REVIEW ¤ 257
13.
9
1
√ u − 2u
2
du =
u
9
1
(u −1/2 − 2u) du =
2u 1/2 − u 2 9
=(6− 81) − (2 − 1) = −76
15. Let u = y 2 +1,sodu =2ydyand ydy = 1 du. Wheny =0, u =1;wheny =1, u =2.Thus,
2
1
0 y(y2 +1) 5 dy = 2
u5 1
du
= 1 1 u6 2
= 1 63
(64 − 1) = = 21 .
1 2 2 6 1 12 12 4
17.
5
1
dt
(t − 4) does not exist because the function f(t) = 1
has an infinite discontinuity at t =4;
2 (t − 4)
2
that is, f is discontinuous on the interval [1, 5].
19. Let u = v 3 ,sodu =3v 2 dv. Whenv =0, u =0;whenv =1, u =1.Thus,
1
0 v2 cos(v 3 ) dv = 1
cos u 1
du
0 3
= 1 1
3 sin u = 1 (sin 1 − 0) = 1 0 3 3
sin 1.
21.
23.
π/4
−π/4
t 4 tan t
2+cost dt =0by Theorem 5.5.7(b), since f(t) = t4 tan t
is an odd function.
2+cost
2 2 1 − x
1 1
dx =
x
x − 1 dx =
x − 2
2 x +1 dx = − 1 x − 2ln|x| + x + C
25. Let u = x 2 +4x. Thendu =(2x +4)dx =2(x +2)dx, so
x +2
√
x2 +4x dx = u −1/2 1
du = 1 · 2 2 2u1/2 + C = √ u + C = x 2 +4x + C.
1
27. Let u =sinπt. Thendu = π cos πt dt,so sin πt cos πt dt = u 1
π du = 1 π · 1
2 u2 + C = 1
2π (sin πt)2 + C.
29. Let u = √ x.Thendu = dx √
e
x
2 √ x ,so √ dx =2 x
e u du =2e u + C =2e √x + C.
31. Let u =ln(cosx). Thendu = − sin x dx = − tan xdx,so
cos x
tan x ln(cos x) dx = − udu= −
1
2 u2 + C = − 1 [ln(cos 2 x)]2 + C.
33. Let u =1+x 4 .Thendu =4x 3 dx,so
x 3
1+x dx = 1 1 4 4 u du = 1 ln|u| + C = 1 ln 1+x 4 + C.
4 4
35. Let u =1+secθ. Thendu =secθ tan θdθ,so
sec θ tan θ
1+secθ dθ = 1
1
1+secθ (sec θ tan θdθ)= du =ln|u| + C =ln|1+secθ| + C.
u
37. Since x 2 − 4 < 0 for 0 ≤ x 0 for 2