Part II: The Answers
580
Answers
901–1,001
Notice that lim
a
a
e
→∞
−
−
(
) =
1
4
0
2
and that lim
a
a
ae
→∞
−
−
(
) = −∞
( )( )
1
2
0
2
, which is an indeterminate
form. To evaluate lim
a
a
ae
→∞
−
−
(
)
1
2
2
, use L’Hôpital’s rule:
lim
lim
lim
( )
a
a
a
a
a
a
ae
a
e
e
→∞
−
→∞
→∞
−
(
) = −
= −
= −
=
1
2
1
2
1
2
1
2
1
2
0
0
2
2
2
Using these values gives you
lim
( )
a
a
a
ae
e
e
e
→∞
−
−
− ( )
− ( )
−
−
− −
−
(
)
= + +
1
2
1
4
1
2
0
1
4
0 0 1
2
2
2
2 0
2 0
( ( )
0
1
4
1
4
2 0
2 0
e
e
− ( )
− ( )
+
=
The answer is finite, so the integral is convergent.
976.
divergent
Begin by splitting up the integral into two separate integrals, using limits to rewrite
each integral:
x e dx
x e dx
x e dx
x
a
x
a
b
x
b
4
4
0
4
0
5
5
5
−
−∞
∞
→−∞
−
→∞
−
∫
∫
∫
=
+
lim
lim
To evaluate x e dx
x
4
5
−
∫
, use the substitution u = –x
5
so that du = –5x
4
dx, or −
=
1
5
4
du x dx:
x e dx
e du
e C
e
C
x
u
u
x
4
5
5
1
5
1
5
1
5
−
−
∫
∫
= −
= −
+
= −
+
Because x e dx
e
C
x
x
4
5
5
1
5
−
−
∫
= −
+ , it follows that
lim
lim
a
x
a
a
x
a
x e dx
e
→−∞
−
→−∞
−
∫
=
−
4
0
0
5
5
1
5
Next, evaluate the limit:
lim
lim
a
x
a
a
a
e
e
e
→−∞
−
→−∞
−( )
−
−
=
−
− −
(
)
1
5
1
5
1
5
5
5
5
0
0
Notice that lim
a
a
e
→−∞
−
−
(
) = ∞
1
5
5
so that lim
a
x
a
x e dx
→−∞
−
∫
4
0
5
is divergent. Because this integral
is divergent, you don’t need to evaluate lim
b
x
b x e dx
→∞
−
∫
4
0
5
. You can conclude that
x e dx
x
4
5
−
−∞
∞
∫
is divergent.
580
Answers
901–1,001
Notice that lim
a
a
e
→∞
−
−
(
) =
1
4
0
2
and that lim
a
a
ae
→∞
−
−
(
) = −∞
( )( )
1
2
0
2
, which is an indeterminate
form. To evaluate lim
a
a
ae
→∞
−
−
(
)
1
2
2
, use L’Hôpital’s rule:
lim
lim
lim
( )
a
a
a
a
a
a
ae
a
e
e
→∞
−
→∞
→∞
−
(
) = −
= −
= −
=
1
2
1
2
1
2
1
2
1
2
0
0
2
2
2
Using these values gives you
lim
( )
a
a
a
ae
e
e
e
→∞
−
−
− ( )
− ( )
−
−
− −
−
(
)
= + +
1
2
1
4
1
2
0
1
4
0 0 1
2
2
2
2 0
2 0
( ( )
0
1
4
1
4
2 0
2 0
e
e
− ( )
− ( )
+
=
The answer is finite, so the integral is convergent.
976.
divergent
Begin by splitting up the integral into two separate integrals, using limits to rewrite
each integral:
x e dx
x e dx
x e dx
x
a
x
a
b
x
b
4
4
0
4
0
5
5
5
−
−∞
∞
→−∞
−
→∞
−
∫
∫
∫
=
+
lim
lim
To evaluate x e dx
x
4
5
−
∫
, use the substitution u = –x
5
so that du = –5x
4
dx, or −
=
1
5
4
du x dx:
x e dx
e du
e C
e
C
x
u
u
x
4
5
5
1
5
1
5
1
5
−
−
∫
∫
= −
= −
+
= −
+
Because x e dx
e
C
x
x
4
5
5
1
5
−
−
∫
= −
+ , it follows that
lim
lim
a
x
a
a
x
a
x e dx
e
→−∞
−
→−∞
−
∫
=
−
4
0
0
5
5
1
5
Next, evaluate the limit:
lim
lim
a
x
a
a
a
e
e
e
→−∞
−
→−∞
−( )
−
−
=
−
− −
(
)
1
5
1
5
1
5
5
5
5
0
0
Notice that lim
a
a
e
→−∞
−
−
(
) = ∞
1
5
5
so that lim
a
x
a
x e dx
→−∞
−
∫
4
0
5
is divergent. Because this integral
is divergent, you don’t need to evaluate lim
b
x
b x e dx
→∞
−
∫
4
0
5
. You can conclude that
x e dx
x
4
5
−
−∞
∞
∫
is divergent.