Part II: The Answers
582
Answers
901–1,001
x
xdx
x
x
x x
dx
x
x
x
dx
x
−
∫
∫
∫
= −
− −
= −
+
= −
4
3
3
3
4
3
1
3
1
3
1
1
3
1
3
1
3
ln
ln
ln
l ln
ln
ln
x
x dx
x
x
x
C
x
x
x
C
+
= −
+
−
+
= −
−
+
−
∫
1
3
1
3
1
3
1
3
1
3
1
9
4
3
3
3
3
Because x
xdx
x
x
x
C
−
∫
= −
−
+
4
3
3
1
3
1
9
ln
ln
, it follows that
lim ln
lim
ln
a
a
a
a
x
x
dx
x
x
x
→∞
→∞
∫
=
−
−
4
2
3
3
2
1
3
1
9
Evaluating the limit gives you
lim
ln
lim
ln
l
a
a
a
x
x
x
a
a
a
→∞
→∞
−
−
=
−
−
− − ( )
1
3
1
9
1
3
1
9
1
3 2
3
3
2
3
3
3
n n2
1
9 2
3
− ( )
Notice that lim ln
a
a
a
→∞ 3
3
has the indeterminate form ∞
∞
, so you can use L’Hôpital’s rule to
evaluate it:
lim ln
lim
lim
a
a
a
a
a
a
a
a
→∞
→∞
→∞
=
=
=
3
1
9
1
9
0
3
2
3
Because lim
a
a
→∞
=
1
9
0
3
, the limit becomes
lim
ln
ln
a
a
a
a
→∞
−
− − ( )
− ( )
= ( )
1
3
1
9
1
3 2
2
1
9 2
1
3 2
3
3
3
3
3 3
3
2
1
9 2
2
24
1
72
ln
ln
+ ( )
=
+
The answer is finite, so the integral is convergent.
979.
divergent
Begin by rewriting the integral using a limit:
x
x
dx
x
x
dx
a
a
2
0
2
0
4
4
+
=
+
∞
→∞
∫
∫
lim
582
Answers
901–1,001
x
xdx
x
x
x x
dx
x
x
x
dx
x
−
∫
∫
∫
= −
− −
= −
+
= −
4
3
3
3
4
3
1
3
1
3
1
1
3
1
3
1
3
ln
ln
ln
l ln
ln
ln
x
x dx
x
x
x
C
x
x
x
C
+
= −
+
−
+
= −
−
+
−
∫
1
3
1
3
1
3
1
3
1
3
1
9
4
3
3
3
3
Because x
xdx
x
x
x
C
−
∫
= −
−
+
4
3
3
1
3
1
9
ln
ln
, it follows that
lim ln
lim
ln
a
a
a
a
x
x
dx
x
x
x
→∞
→∞
∫
=
−
−
4
2
3
3
2
1
3
1
9
Evaluating the limit gives you
lim
ln
lim
ln
l
a
a
a
x
x
x
a
a
a
→∞
→∞
−
−
=
−
−
− − ( )
1
3
1
9
1
3
1
9
1
3 2
3
3
2
3
3
3
n n2
1
9 2
3
− ( )
Notice that lim ln
a
a
a
→∞ 3
3
has the indeterminate form ∞
∞
, so you can use L’Hôpital’s rule to
evaluate it:
lim ln
lim
lim
a
a
a
a
a
a
a
a
→∞
→∞
→∞
=
=
=
3
1
9
1
9
0
3
2
3
Because lim
a
a
→∞
=
1
9
0
3
, the limit becomes
lim
ln
ln
a
a
a
a
→∞
−
− − ( )
− ( )
= ( )
1
3
1
9
1
3 2
2
1
9 2
1
3 2
3
3
3
3
3 3
3
2
1
9 2
2
24
1
72
ln
ln
+ ( )
=
+
The answer is finite, so the integral is convergent.
979.
divergent
Begin by rewriting the integral using a limit:
x
x
dx
x
x
dx
a
a
2
0
2
0
4
4
+
=
+
∞
→∞
∫
∫
lim