Part II: The Answers
574
Answers
901–1,001
965.
divergent
Note that the function f x
x x
( ) = 1 has an infinite discontinuity when x = 0, which is
included in the interval of integration, so you have an improper integral. Begin by
rewriting the definite integral using limits; then integrate:
1
1
1 2
0
5
0
3 2
5
0
3 2
5
0
1 2
x x
dx
x
dx
x dx
x
a
a
a
a
a
a
∫
∫
∫
=
=
=
−
→
→
−
→
−
+
+
+
lim
lim
lim
5 5
0
5
0
2
1
2
1
5
1
= −
= −
−
→
→
+
+
lim
lim
a
a
a
x
a
Because lim
a
a
→
+
= ∞
0
1
, you can conclude that
1
0
5
x x
dx
∫
is divergent.
966.
divergent
Note that the function f x
x
( ) = 1 has an infinite discontinuity when x = 0, which is
included in the interval of integration, so you have an improper integral. Begin by
rewriting the definite integral using limits; then integrate:
1
1
2
0
2
0
2
0
2
0
x
dx
x
dx
x
a
a
a
a
a
a
∫
∫
=
=
( )
=
−
(
)
→
→
→
+
+
+
lim
lim ln
lim ln ln
Because lim ln
a
a
→
+
( )= −∞
0
, you know that 1
0
2
x
dx
∫
is divergent.
967.
divergent
Begin by rewriting the integral using a limit:
e dx
e dx
x
a
x
a
−
−∞
→−∞
−
∫
∫
=
4
1
4
1
lim
Then use elementary antiderivatives to get the following:
e dx
e dx
e
x
a
x
a
a
x
a
a
−
−∞
→−∞
−
→−∞
−
→−∞
∫
∫
=
=
−
=
4
1
4
1
4
1
1
4
lim
lim
lim − −
− −
(
)
−
−
1
4
1
4
4 1
4
e
e
a
( )
Because lim
a
a
e
→−∞
−
= ∞
4
, the integral is divergent.
574
Answers
901–1,001
965.
divergent
Note that the function f x
x x
( ) = 1 has an infinite discontinuity when x = 0, which is
included in the interval of integration, so you have an improper integral. Begin by
rewriting the definite integral using limits; then integrate:
1
1
1 2
0
5
0
3 2
5
0
3 2
5
0
1 2
x x
dx
x
dx
x dx
x
a
a
a
a
a
a
∫
∫
∫
=
=
=
−
→
→
−
→
−
+
+
+
lim
lim
lim
5 5
0
5
0
2
1
2
1
5
1
= −
= −
−
→
→
+
+
lim
lim
a
a
a
x
a
Because lim
a
a
→
+
= ∞
0
1
, you can conclude that
1
0
5
x x
dx
∫
is divergent.
966.
divergent
Note that the function f x
x
( ) = 1 has an infinite discontinuity when x = 0, which is
included in the interval of integration, so you have an improper integral. Begin by
rewriting the definite integral using limits; then integrate:
1
1
2
0
2
0
2
0
2
0
x
dx
x
dx
x
a
a
a
a
a
a
∫
∫
=
=
( )
=
−
(
)
→
→
→
+
+
+
lim
lim ln
lim ln ln
Because lim ln
a
a
→
+
( )= −∞
0
, you know that 1
0
2
x
dx
∫
is divergent.
967.
divergent
Begin by rewriting the integral using a limit:
e dx
e dx
x
a
x
a
−
−∞
→−∞
−
∫
∫
=
4
1
4
1
lim
Then use elementary antiderivatives to get the following:
e dx
e dx
e
x
a
x
a
a
x
a
a
−
−∞
→−∞
−
→−∞
−
→−∞
∫
∫
=
=
−
=
4
1
4
1
4
1
1
4
lim
lim
lim − −
− −
(
)
−
−
1
4
1
4
4 1
4
e
e
a
( )
Because lim
a
a
e
→−∞
−
= ∞
4
, the integral is divergent.