Solução_Calculo_Stewart_6e

F.
21.
∞
1
ln x
x
dx = lim
t→∞
(ln x)
2
2
t
1
TX.10
by substitution with
u =lnx, du = dx/x
(ln t) 2
= lim = ∞. Divergent
t→∞ 2
SECTION 7.8 IMPROPER INTEGRALS ¤ 337
23.
∞
−∞
Now
x 2 0
9+x dx = 6
∞
so 2
0
Convergent
x 2 dx
9+x 6
−∞
x 2 ∞
9+x dx + 6
u = x 3
du =3x 2 dx
x 2
t
dx = 2 lim
9+x6 t→∞
0
0
x 2 ∞
9+x dx =2 6
0
x 2
dx [since the integrand is even].
9+x6 1
3
=
du 1
u =3v
(3 dv)
3
=
9+u 2 du =3dv 9+9v = 1
dv
2 9 1+v 2
= 1 9 tan−1 v + C = 1 u
9 tan−1 + C = 1 x
3
3 9 tan−1 + C,
3
x 2
dx = 2 lim
9+x6 t→∞
1
9 tan−1 x
3
3
t
0
1 t
3
=2 lim
t→∞ 9 tan−1 = 2 3 9 · π
2 = π 9 .
25.
∞
e
1
t
dx = lim
x(ln x)
3 t→∞
e
1
dx = lim
x(ln x)
3 t→∞
= lim − 1
t→∞ 2(lnt) 2 + 1 2
ln t
1
u −3 du
u =lnx,
du = dx/x
=0+ 1 2 = 1 2 . Convergent
= lim − 1 ln t
t→∞ 2u 2 1
27.
1
0
3
dx = lim
x5 t→0 + 1
t
3x −5 dx = lim − 3 1
= − 3
t→0 + 4x 4 t
4 lim 1 − 1
= ∞. Divergent
t→0 + t 4
29.
14
−2
dx
4√ x +2
=
14
14
lim (x +2) −1/4 4
dx = lim
t→−2 + t→−2 + 3 (x +2)3/4
t
= 4 32
(8 − 0) = . Convergent
3 3
t
= 4
3 lim 16 3/4 − (t +2) 3/4
t→−2 +
31.
3
−2
dx 0
x = 4
−2
dx 3
x + 4
0
dx 0
x ,but 4
−2
dx
x 4
= lim
t→0 −
t
− x−3
3
−2
= lim − 1
t→0 − 3t − 1
= ∞. Divergent
3 24
33. There is an infinite discontinuity at x =1.
1
0 (x − 1)−1/5 dx = lim
t→1 − t
0 (x − 1)−1/5 dx = lim
33
1 (x − 1)−1/5 dx = lim
33
t→1 + t
33
0 (x − 1)−1/5 dx = 1
0 (x − 1)−1/5 dx + 33
1 (x − 1)−1/5 dx. Here
t→1 −
5
(x − 1) −1/5 dx = lim
5
t→1 +
Thus, 33
(x − 0 1)−1/5 dx = − 5 75
4
+20= . Convergent
4
5
t→1 −
4 (x − 1)4/5 t
0
= lim
33
(x − 4 1)4/5 = lim
t
(t − 4 1)4/5 − 5 = − 5 and
4 4
5
t→1 +
3
dx
3
35. I =
0 x 2 − 6x +5 = dx
1
0 (x − 1)(x − 5) = I 1 + I 2 =
0
1
Now
(x − 1)(x − 5) = A
x − 1 + B ⇒ 1=A(x − 5) + B(x − 1).
x − 5
Set x =5to get 1=4B,soB = 1 .Setx =1to get 1=−4A,soA = − 1 .Thus
4 4
dx
3
(x − 1)(x − 5) +
4 · 16 − 5 4 (t − 1)4/5
=20.
1
dx
(x − 1)(x − 5) .