Answers and Explanations 577
Answers
901–1,001
Therefore, it follows that
lim ( )
a
a a
→−∞
−
−
−











 = − − ∞
3 0 0
5
3
5
0 0
5
5
and you can conclude that the integral is divergent. Because the first integral is
divergent, you don’t need to evaluate the second integral;
3
4
−
(
)
−∞
∞
∫
x dx
is divergent.
972.
convergent, 3
2 3
e
− /
Begin by rewriting the integral using a limit:
e dx
e dx
x
a
x
a
−
∞
→∞
−
∫
∫
=
3
2
3
2
lim
To evaluate e dx
x
−
∫
3
, let u
x
= − 1
3
so that du
dx
= − 1
3
, or –3
 
du = dx. This gives you
e dx
e du
e C
e
C
x
u
u
x
−
−
∫
∫
= −
= −
+
= −
+
3
3
3
3
3
Because e dx
e
C
x
x
−
−
∫
= −
+
3
3
3
, it follows that
lim
lim
a
x
a
a
x
a
e dx
e
→∞
−
→∞
−
∫
=
−
(
)
3
2
3
2
3
Now simply evaluate the limit:
lim
lim
a
x
a
a
a
e
e
e
e
e
e
→∞
−
→∞
−
−
−∞
−
−
−
(
) = − +
(
)
= −
+
= +
3
3
3
3
3
0 3
3
2
3
23
3
23
2 3
= =
−
3
2 3
e
The answer is finite, so the integral is convergent.
973.
convergent, 1
Begin by writing the integral using a limit:
1
1
2
2
x x
dx
x x
dx
e
a
e
a
(ln )
lim
(ln )
∞
→∞
∫
∫
=
To evaluate
1
2
x x
dx
(ln )
∫
, use the substitution u = ln x so that du x
dx
= 1 :
1
1
1
2
2
1
u
du
u du
u
C
u
C
x
C
∫
∫
=
= −
+
= − +
= −
+
−
−
ln