Part II: The Answers
578
Answers
901–1,001
Because 1
1
2
u
du
x
C
∫
= −
+
ln
, it follows that
lim
(ln )
lim ln
a
e
a
a
e
a
x x
dx
x
→∞
→∞
∫
=
−
( )
1
1
2
Evaluating the limit gives you
1
1
1
1
2
2
x x
dx
x x
dx
x
e
a
e
a
a
e
a
a
(ln )
lim
(ln )
lim ln
lim
∞
→∞
→∞
→∞
∫
∫
=
=
−
( )
=
− l ln
ln
a
e
+
(
)
= +
( )
=
1
0 1
1
1
The answer is finite, so the integral is convergent.
974.
convergent, π
3
Begin by splitting the integral into two integrals, using limits to rewrite each integral:
x
x
dx
x
x
dx
x
x
dx
a
a
b
b
2
6
2
6
0
2
6
0
1
1
1
+
=
+
+
+
−∞
∞
→−
→
∞
∞
∫
∫
∫
lim
lim
To evaluate
x
x
dx
2
6
1 +
∫
, rewrite the integral as
x
x
dx
2
3
2
1 + ( )
∫
. Then use a substitution,
letting u = x
3
so that du = 3x
2 
dx, or 1
3
2
du x dx
=
. Using these substitutions gives you
x
x
dx
u
du
u C
x
C
2
3
2
2
1
1
3
1
1
3
1
1
1
3
1
3
+ ( )
=
+
=
+
=
( ) +
∫
∫
−
−
tan
tan
Therefore, as a approaches –∞, you have
lim
lim
tan
lim tan
a
a
a
a
a
x
x
dx
x
→−
→
∞
− ∞
−
→−∞
+
=
( )






=
∫
2
6
0
1
3
0
1
1
3
1
3
− −
−
( ) −
( )
(
)
=
− −
( )






=
1
3
1
3
0
1
3
0
2
6
tan a
π
π