Answers and Explanations 589
Answers
901–1,001
Substitute in the limits of integration and evaluate the resulting limit:
lim
lim
a
a
a
x
x
→
→
+
+
−
(
) −
−
(
)
=
−
(
) − −
(
)
3
3 2
1 2
5
3
3 2
1
2
3
3
6
3
2
3
5 3
6 5 3
2 2
3 2
1 2
3 2
1 2
2
3
3
6
3
2
3
2
6 2
0 0
2
3
2
−
−
(
) − −
(
)
(
)
= ( ) − ( ) − −
(
)
(
)
= ( )
a
a
3 3 2
1 2
1 2
6 2
2
4
3
6
14 2
3
− ( )
=
−
( )
= −
The answer is finite, so the integral is convergent.
988.
convergent, compare to
1
2
1 x
dx
∞
∫
Recall the comparison theorem: Suppose that f and g are continuous functions with
f
(x) ≥ g(x) ≥ 0 for x ≥ a.
✓ If f x dx
a
( )
∞
∫
is convergent, then g x dx
a
( )
∞
∫
is convergent.
✓ If g x dx
a
( )
∞
∫
is divergent, then f x dx
a
( )
∞
∫
is divergent.
The given improper integral is sin
2
2
1 1
x
x
dx
+
∞
∫
. Notice that on the interval [1, ∞], the
following inequalities are true: –1 ≤ sin x ≤ 1; therefore, 0 ≤ sin
2
x ≤ 1. Because
1 + x
2
≥ 0, you have 0 1
1
1
2
2
2
≤ +
≤ +
sin x
x
x
. However, 1
1
1
2
2
+
≤
x
x
on [1, ∞] so that
1
1
1
2
1
2
1
+
≤
∞
∞
∫
∫
x
dx
x
dx. An integral of the form
1
1 x
dx
p
∞
∫
converges if and only if p > 1,
so
1
2
1 x
dx
∞
∫
is convergent (you can also show this directly by using a limit to write
the integral and evaluating). Because 0
1
1
1
1
2
2
1
2
1
2
1
≤
+
≤
+
≤
∞
∞
∞
∫
∫
∫
sin x
x
dx
x
dx
x
dx, you can
conclude that sin
2
2
1 1
x
x
dx
+
∞
∫
is convergent.
989.
convergent, compare to
1
4
1 x
dx
∞
∫
The given improper integral is
dx
x e
x
4
3
1
+
∞
∫
. Notice that e
3x
≥ 0 on the interval
[1, ∞], so you have x
4
+ e
3x
≥ x
4
. It follows that
1
1
4
3
4
x e
x
x
+
≤
and that
0
1
1
4
3
1
4
1
≤
+
≤
∞
∞
∫
∫
x e
dx
x
dx
x
. An integral of the form
1
1 x
dx
p
∞
∫
converges if and only if
p > 1, so
1
4
1 x
dx
∞
∫
is convergent (you can also show this directly by using a limit to
write the integral and evaluating). Because 0
1
1
4
3
1
4
1
≤
+
≤
∞
∞
∫
∫
x e
dx
x
dx
x
, you know
that
dx
x e
x
4
3
1
+
∞
∫
also converges.
Answers
901–1,001
Substitute in the limits of integration and evaluate the resulting limit:
lim
lim
a
a
a
x
x
→
→
+
+
−
(
) −
−
(
)
=
−
(
) − −
(
)
3
3 2
1 2
5
3
3 2
1
2
3
3
6
3
2
3
5 3
6 5 3
2 2
3 2
1 2
3 2
1 2
2
3
3
6
3
2
3
2
6 2
0 0
2
3
2
−
−
(
) − −
(
)
(
)
= ( ) − ( ) − −
(
)
(
)
= ( )
a
a
3 3 2
1 2
1 2
6 2
2
4
3
6
14 2
3
− ( )
=
−
( )
= −
The answer is finite, so the integral is convergent.
988.
convergent, compare to
1
2
1 x
dx
∞
∫
Recall the comparison theorem: Suppose that f and g are continuous functions with
f
(x) ≥ g(x) ≥ 0 for x ≥ a.
✓ If f x dx
a
( )
∞
∫
is convergent, then g x dx
a
( )
∞
∫
is convergent.
✓ If g x dx
a
( )
∞
∫
is divergent, then f x dx
a
( )
∞
∫
is divergent.
The given improper integral is sin
2
2
1 1
x
x
dx
+
∞
∫
. Notice that on the interval [1, ∞], the
following inequalities are true: –1 ≤ sin x ≤ 1; therefore, 0 ≤ sin
2
x ≤ 1. Because
1 + x
2
≥ 0, you have 0 1
1
1
2
2
2
≤ +
≤ +
sin x
x
x
. However, 1
1
1
2
2
+
≤
x
x
on [1, ∞] so that
1
1
1
2
1
2
1
+
≤
∞
∞
∫
∫
x
dx
x
dx. An integral of the form
1
1 x
dx
p
∞
∫
converges if and only if p > 1,
so
1
2
1 x
dx
∞
∫
is convergent (you can also show this directly by using a limit to write
the integral and evaluating). Because 0
1
1
1
1
2
2
1
2
1
2
1
≤
+
≤
+
≤
∞
∞
∞
∫
∫
∫
sin x
x
dx
x
dx
x
dx, you can
conclude that sin
2
2
1 1
x
x
dx
+
∞
∫
is convergent.
989.
convergent, compare to
1
4
1 x
dx
∞
∫
The given improper integral is
dx
x e
x
4
3
1
+
∞
∫
. Notice that e
3x
≥ 0 on the interval
[1, ∞], so you have x
4
+ e
3x
≥ x
4
. It follows that
1
1
4
3
4
x e
x
x
+
≤
and that
0
1
1
4
3
1
4
1
≤
+
≤
∞
∞
∫
∫
x e
dx
x
dx
x
. An integral of the form
1
1 x
dx
p
∞
∫
converges if and only if
p > 1, so
1
4
1 x
dx
∞
∫
is convergent (you can also show this directly by using a limit to
write the integral and evaluating). Because 0
1
1
4
3
1
4
1
≤
+
≤
∞
∞
∫
∫
x e
dx
x
dx
x
, you know
that
dx
x e
x
4
3
1
+
∞
∫
also converges.