Answers and Explanations 581
Answers
901–1,001
977.
convergent, 3
2
16
2
1
tan ( )
−
−
(
)
π
Begin by rewriting the integral using a limit:
3
1
3
1
4
4
4
4
x
x
dx
x
x
dx
a
a
+
=
+
−∞
−
→−∞
−
∫
∫
lim
To evaluate
3
1
4
x
x
dx
+
∫
, rewrite the integral as
3
1
2
2
x
x
dx
+ ( )
∫
. Then use a substitution
where u = x
2
so that du = 2x
 
dx, or 1
2
du x dx
=
:
3
1
3
2
1
1
3
2
3
2
2
2
2
1
1
2
x
x
dx
u
du
u C
x
+ ( )
=
+
=
+
=
( ) +
∫
∫
−
−
tan ( )
tan
C
Because
3
1
3
2
2
2
1
2
x
x
dx
x
C
+ ( )
=
( ) +
∫
−
tan
, it follows that
lim
lim
tan
a
a
a
a
x
x
dx
x
→−∞
−
→−∞
−
−
+
=
( )






∫
3
1
3
2
4
4
1
2
4
Now evaluate the limit:
lim
tan
lim tan ( ) tan
a
a
a
x
a
→−∞
−
−
→−∞
−
−
( )






=
−
( )
(
)
3
2
3
2
16
1
2
4
1
1
2
Notice that lim tan
a
a
→−∞
−
( ) =
1
2
2
π so that
3
2
16
3
2
16
2
1
1
2
1
lim tan
t an
tan ( )
a
a
→−∞
−
−
−
( )−
( )
(
)
=
−
(
)
π
The answer is finite, so the integral is convergent.
978.
convergent, ln2
24
1
72
+
Begin by rewriting the integral using a limit:
ln
lim ln
x
x
dx
x
x
dx
a
a
4
2
4
2
∞
→∞
∫
∫
=
To evaluate ln
ln
x
x
dx
x
xdx
4
4
=
−
∫
∫
, use integration by parts. If u = ln x, then du x
dx
= 1 ,
and if dv = x
−4 
dx, then v x
x
= −
= −
−3
3
3
1
3
: