Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 7.8 IMPROPER INTEGRALS ¤ 341
61. (a) I = ∞
xdx= 0
xdx+ ∞
xdx,and ∞
t
xdx= lim xdx= lim 1 x2 t
= lim
1
−∞ −∞ 0 0 t→∞ 0 t→∞ 2 0 t→∞ 2 t2 − 0 = ∞,
so I is divergent.
(b) t
xdx= 1
x2 t
= 1
−t 2 −t 2 t2 − 1 t
2 t2 =0,so lim xdx=0. Therefore, ∞
t
xdx6= lim xdx.
t→∞ −t −∞ t→∞ −t
63. Volume =
65. Work =
∞
1
∞
R
2 1 t
dx
π dx = π lim
x
t→∞
1 x = π lim − 1 t
= π lim
1 − 1
= π0}.
s
(b) F (s)=
∞
0
= lim
n→∞
f(t)e −st dt =
∞
0
e
(1−s)n
1 − s − 1
1 − s
e t e −st dt = lim
n→∞
n − e−st
s
0
n
0
e
−sn
= lim
n→∞ −s + 1
. This converges to 1 only if s>0.
s
s
n 1
e t(1−s) dt = lim
n→∞ 1 − s et(1−s) 0
This converges only if 1 − s1,inwhichcaseF (s) = 1 with domain {s | s>1}.
s − 1