Answers and Explanations 585
Answers
901–1,001
Because sin
s in( )
2
1
2
1
2
2
x dx
x
x
C
∫
=
−
(
) + , it follows that
lim sin
lim
sin( )
a
a
a
a
x dx
x
x
→∞
→∞
∫
=
−
(
)
2
0
0
1
2
1
2
2
Evaluating the limit gives you
lim
sin( )
lim
sin( )
s in ( )
a
a
a
x
x
a
a
→∞
→∞
−
(
)
=
−
− −
(
)
(
1
2
1
2
2
1
2
1
2
2
0 1
2
2 0
0
) )






Because lim
a
a
→∞
= ∞ and because sin(2a) is always between –1 and 1, the integral
is divergent.
983.
divergent
Notice that the function e
e
x
x
−1
has an infinite discontinuity at x = 0. Rewrite the integral
using limits:
e
e
dx
e
e
dx
e
e
dx
x
x
a
x
x
a
b
x
x
b
−
=
−
+
−
−
→
−
→
∫
∫
∫
+
−
1
1
1
1
1
0
1
0
1
lim
lim
To evaluate
e
e
dx
x
x
−
∫ 1
, use the substitution u = e
x
– 1 so that du = e
x 
dx to get
1
1
u
du
u C
e
C
x
∫
=
+
=
− +
ln
ln
Because
e
e
dx
e
C
x
x
x
−
=
− +
∫ 1
1
ln
, it follows that
lim
lim ln
a
x
x
a
a
x
a
e
e
dx
e
→
−
→
−
−
−
−
=
−
(
)
∫
0
1
0
1
1
1
Evaluating the limit gives you
lim ln
lim ln
ln
a
x
a
a
a
e
e
e
→
−
→
−
−
−
−
(
)
=
− −
−
(
)
0
1
0
1
1
1
1
However, because lim ln
a
a
e
→
−
− = −∞
0
1
, the integral lim
a
x
x
a
e
e
dx
→
−
−
−
∫
0
1
1
is divergent,
so you can conclude that
e
e
dx
x
x
−
−
∫
1
1
1
is divergent.
984.
convergent, 1
5
7
2
ln ( )
Begin by writing the integral using a limit:
1
6
1
6
2
4
2
4
x
x
dx
x
x
dx
a
a
+ −
=
+ −
∞
→∞
∫
∫
lim
To evaluate
1
6
2
x
x
dx
+ −
∫
, use a partial fraction decomposition:
1
6
1
3
2
3
2
2
x
x
x
x
A
x
B
x
+ −
=
+
−
= +
+ −
(
)(
)