Solução_Calculo_Stewart_6e

F.
342 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
TX.10
(c) F (s) = ∞
f(t)e −st n
dt = lim
0 n→∞ 0 te−st dt. Use integration by parts: let u = t, dv = e −st dt ⇒ du = dt,
v = − e−st
s
.ThenF (s) = lim
n→∞
Therefore, F (s) = 1 and the domain of F is {s | s>0}.
s2
− t s e−st − 1 n −n
s 2 e−st = lim
n→∞
0
se − 1
sn s 2 e +0+ 1
= 1 only if s>0.
sn s 2 s2 73. G(s) = ∞
0
f 0 (t)e −st dt. Integrate by parts with u = e −st , dv = f 0 (t) dt ⇒ du = −se −st , v = f(t):
G(s) = lim
f(t)e
−st n
+ s ∞
f(t)e −st dt = lim
n→∞
0 0 n→∞ f(n)e−sn − f(0) + sF (s)
But 0 ≤ f(t) ≤ Me at ⇒ 0 ≤ f(t)e −st ≤ Me at e −st and lim
t→∞
Me t(a−s) =0for s>a. So by the Squeeze Theorem,
lim
t→∞ f(t)e−st =0for s>a ⇒ G(s) =0− f(0) + sF (s) =sF (s) − f(0) for s>a.
75. We use integration by parts: let u = x, dv = xe −x2 dx ⇒ du = dx, v = − 1 2 e−x2 .So
∞
0
x 2 e −x2 dx = lim
t→∞
− 1 2 xe−x2 t
0
+ 1 2
(The limit is 0 by l’Hospital’s Rule.)
∞
0
e −x2 dx = lim −
t
+ 1
t→∞ 2e t2 2
77. For the firstpartoftheintegral,letx =2tanθ ⇒ dx =2sec 2 θdθ.
1
2sec 2
√
x2 +4 dx = θ
2secθ dθ = sec θdθ=ln|sec θ +tanθ|.
∞
0
e −x2 dx = 1 2
From the figure, tan θ = x √
2 ,andsec θ = x2 +4
.So
2
∞
1
I = √
0 x2 +4 − C
√ dx = lim ln
x2 +4
x +2
t→∞ + x t 2 2 − C ln|x +2| 0
√
t2 +4+t
= lim ln
− C ln(t +2)− (ln 1 − C ln 2)
t→∞ 2
t + √ t
Now L = lim
2 +4 H 1+t/ √ t
= lim
2 +4
t→∞ (t +2) C t→∞ C (t +2) = 2
C−1 C lim (t +2) . C−1
t→∞
If C1, L =0and I diverges to −∞.
∞
0
e −x2 dx
√
t2 +4+t
= lim ln
+ln2 C t + √
t
=ln lim
2 +4
+ln2 C−1
t→∞ 2(t +2) C t→∞ (t +2) C
79. No, I = ∞
0
f(x) dx must be divergent. Since lim
x→∞ f(x) =1,theremustexistanN such that if x ≥ N,thenf(x) ≥ 1 2 .
Thus, I = I 1 + I 2 = N
f(x) dx + ∞
f(x) dx,whereI1 is an ordinary definite integral that has a finite value, and I2 is
0 N
improper and diverges by comparison with the divergent integral ∞
N
1
2 dx.