Part II: The Answers
588
Answers
901–1,001
986.
convergent, 2
Rewrite the integral using limits:
e dx
e dx
e dx
x
a
x
a
b
x
b
−
−∞
∞
→−∞
−
→∞
−
∫
∫
∫
=
+
lim
lim
0
0
On the interval (–∞, 0), you have − =
x x, and on the interval (0, ∞), you have
− =−
x
x. Therefore, the limit is
lim
lim
lim
lim
a
x
a
b
x
b
a
x
a
b
x
e dx
e dx
e dx
e dx
→−∞
−
→∞
−
→−∞
→ ∞
−
+
=
+
∫
∫
∫
0
0
0
0
b b
a
x
a
b
x
b
a
a
b
b
e
e
e e
e
∫
=
( ) + −
( )
=
−
(
) + −
→−∞
→ ∞
−
→−∞
→ ∞
−
lim
lim
lim
lim
0
0
0
− − −
( )
(
)
= − + +
=
e
0
1 0
0 1
2
(
) (
)
The answer is finite, so the integral is convergent.
987.
convergent, −14 2
3
Begin by rewriting the integral using a limit:
x
x
dx
x
x
dx
a
a
−
=
−
∫
∫
→
+
3
3
3
5
3
5
lim
To evaluate
x
x
dx
−
∫
3
, let u = x – 3 so that u – 3 = x and du = dx. This gives you
u
u
du
u
u
u
du
u
u
du
u
u
C
−
=
−






=
−
(
)
=
−
+
∫
∫
∫
−
3
3
3
3 2
3 1 2
1 2
1 2
1 2
1 2
1 2
3 2
1 2
= =
−
−
−
+
2
3
3
6
3
3 2
1 2
(
)
(
)
x
x
C
Because
x
x
dx
x
x
C
−
=
−
(
) −
−
(
) +
∫
3
2
3
3
6
3
3 2
1 2
, it follows that
x
x
dx
x
x
dx
x
x
a
a
a
a
−
=
−
=
−
(
) −
−
(
)




∫
∫
→
→
+
+
3
3
2
3
3
6
3
3
5
3
5
3
3 2
1 2
5
lim
lim
 
