Solução_Calculo_Stewart_6e

F.
244 ¤ CHAPTER 5 INTEGRALS
TX.10
21.
23.
25.
27.
4 (5 − 2t 1 +3t2 ) dt = 5t − t 2 + t 3 4
=(20− 16 + 64) − (5 − 1+1)=68− 5=63
1
1 1
0 x4/5 5
dx =
9 x9/5 = 5 − 0= 5 9 9
0
2
1
3
2
t dt =3 4
1
t
t −4 −3 2
dt =3 = 3 2
1 1
= −1
−3
1
−3 t 3 1
8 − 1 = 7 8
2 x(2 + 0 x5 ) dx = 2
(2x + 0 x6 ) dx = x 2 + 1 x7 2
=
4+ 128
7 0 7 − (0 + 0) =
156
7
29.
9
1
x − 1
√
x
dx =
9
1
x√x
− 1 √
x
dx =
9
1
(x 1/2 − x −1/2 ) dx =
= 2 · 27 − 2 · 3 − 2
− 2 =12−
− 4 3 3 3 =
40
3
2
3 x3/2 − 2x 1/2 9
1
31.
33.
35.
37.
39.
π/4
sec 2 tdt= tan t π/4
=tan π 0 0 4
− tan 0 = 1 − 0=1
2
1 (1 + 2y)2 dy = 2
1 (1 + 4y +4y2 ) dy = y +2y 2 + 4 3 y3 2
1 = 2+8+ 32 3
9
1
√ 3/2
1/2
1
2x dx = 1 9
1
9
2 1 x dx = 1 ln |x| = 1 (ln 9 − ln 1) = 1 ln 9 − 2
2 2 0=ln91/2 =ln3
1
6
√ 3/2
√ dt =6 1 − t
2
1/2
1
√ dt =6 sin −1 t √
3/2
=6
1 − t
2 1/2
sin −1 √
3
2
1
−1 eu+1 du = e u+1 1
−1 = e2 − e 0 = e 2 − 1 [or start with e u+1 = e u e 1 ]
− 1+2+
4
3 =
62
− 13 = 49
3 3 3
− sin −1 1
=6 π
− π
2
3 6 =6 π
6 = π
sin x
41. If f(x) =
cos x
if 0 ≤ x