Solução_Calculo_Stewart_6e

F.
338 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
37.
Since I 1 is divergent, I is divergent.
0
−1
e 1/x
x 3
t
−
1 1
4
I 1 = lim
t→1 − 0 x − 1 + 4
t
1
dx = lim
t→0 − −1 x e1/x ·
x − 5
dx = lim
t→1 − − 1 4 ln |x − 1| + 1 4 ln |x − 5| t
0
= lim −
1
ln |t − 1| + 1 ln |t − 5|
t→1 − 4 4
− − 1 ln |−1| +
1
4 4
ln |−5|
= ∞, since lim
t→1 −
−
1
4 ln |t − 1| = ∞.
= lim
t→0 −
(u − 1)e
u −1
1/t
39. I = 2
0 z2 ln zdz= lim
1
dx = lim
x2
t→0 − 1/t
use parts
or Formula 96
−1
ue u (−du)
u =1/x,
du = −dx/x 2
1
= lim −2e −1 −
e
t→0 − t − 1 1/t
= − 2 e − lim (s − s→−∞ 1)es [s =1/t] = − 2 e − lim s − 1
= H − 2
s→−∞ e −s e − lim 1
s→−∞ −e −s
= − 2 e − 0=− 2 e . Convergent
2
t→0 + t
z
3
2 z2 ln zdz= lim
t→0 + 3 (3 ln z − 1) 2 t
integrate by parts
or use Formula 101
= lim 8 (3 ln 2 − 1) − 1
t→0 + 9 9 t3 (3 ln t − 1) = 8 ln 2 − 8 − 1 lim
3 9 9 t 3 (3 ln t − 1) = 8 ln 2 − 8 − 1 L.
t→0 + 3 9 9
Now L = lim
t→0 +
t 3 (3 ln t − 1) = lim
t→0 + 3lnt − 1
t −3
TX.10
H
= lim
t→0 + 3/t
−3/t 4 = lim
t→0 +
−t
3 =0.
Thus, L =0and I = 8 3 ln 2 − 8 9 .
Convergent
41. Area = 1
−∞ ex dx =
lim
t→−∞
e
x 1
= e − lim
t t→−∞ et = e
43. Area =
∞
−∞
2
∞
x 2 +9 dx =2· 2
t 1 x
=4 lim
t→∞ 3 tan−1 = 4 3
0
3 lim
t→∞
0
1
x 2 +9
t
dx = 4 lim
t→∞
0
1
x 2 +9 dx
tan −1 t
3 − 0
= 4 3 · π
2 = 2π 3
45. Area = π/2
sec 2 t
xdx= lim
0
t→(π/2) − 0 sec2 xdx=
= lim t − 0) = ∞
t→(π/2) −(tan
Infinite area
lim [tan x]t
t→(π/2) − 0