Answers and Explanations 591
Answers
901–1,001
993.
divergent, compare to 1
1 x
dx
∞
∫
The given improper integral is
5
1
+
−
∞
∫
e
x
dx
x
. Notice that on the interval [1, ∞), you have
e
–x
> 0 so that 5 + e
–x
> 5 and also 5
5 5 1
+
> =
−
e
x
x
x
x
. Therefore, it follows that
5
5 1
1
1
+
>
−
∞
∞
∫
∫
e
x
dx
x
dx
x
. The integral
1
1 x
dx
p
∞
∫
diverges if and only if p ≤ 1, so 5 1
1 x
dx
∞
∫
diverges. Because 5
5 1
1
1
+
>
−
∞
∞
∫
∫
e
x
dx
x
dx
x
, you know that 5
1
+
−
∞
∫
e
x
dx
x
also diverges.
994.
18.915
Recall that the trapezoid rule states the following:
f x
x f x
f x
f x
f x
f x
n
n
a
b ( )
...
≈
( )+ ( )+ ( )+ + ( )+ ( )
(
)
−
∫
∆
2
2
2
2
0
1
2
1
where ∆ x b a
n
= − and x i = a + iΔx.
You want to approximate
1
3
3
0
6
+
∫
x dx with n = 6. Using the trapezoid rule,
∆ x b a
n
= − = − =
6 0
6
1 so that
1
1
2
0 2 1 2 2 2 3 2 4 2 5
6
1
2
1
3
3
0
6
+
≈
+
+
+
+
+
+
)
(
=
∫
x dx
f
f
f
f
f
f
f
( )
( )
( )
( )
( )
( ) ( )
+ + +
+ +
+ +
+
+
+
+
+ + + )
(
≈
0
2 1 1 2 1 2
2 1 3
2 1 4
2 1 5
1 6
18 915
3
3
3
3
3
3
3
3
3
3
3
3
3
3
.
995.
0.105
You want to approximate
ln x
x
dx
1
2
1
2
+
∫
with n = 4. Using the trapezoid rule, you have
∆ x = − = =
2 1
4
1
4
0 25
. so that
ln
.
( )
( . )
( . )
( . ) ( )
.
x
x
dx
f
f
f
f
f
1
0 25
2
1 2 1 25 2 1 5 2 1 75
2
0
2
1
2
+
≈
+
+
+
+
)
(
=
∫
2 25
2
1
1 1
2
1 25
1 1 25
2
1 5
1 1 5
2
1 75
1
2
2
2
ln
ln( . )
( . )
ln( . )
( . )
ln( . )
+
+ +
+ +
+ + +
+ +

 

 
≈
( . )
ln
.
1 75
2
1 2
0 105
2
2
996.
–0.210
You want to approximate cos x
x
dx
1
3
∫
with n = 8. Using the trapezoid rule, you have
∆ x b a
n
= − = − =
3 1
8
0 25
. so that
cos
.
( )
( . )
( . )
( . )
( )
( .
x
x
dx
f
f
f
f
f
f
1
3
0 25
2
1 2 1 25 2 1 5 2 1 75 2 2
2 2
∫
≈
+
+
+
+
+
2 25 2 2 5 2 2 75
3
0 25
2
1
1
2
1 25
1
)
( . )
( . ) ( )
.
cos( )
cos( . )
.
+
+
+

 

 
=
+
f
f
f
2 25
2
1 5
1 5
2
1 75
1 75
2
2
2
2
2 25
2 25
2
+
+
+
+
+
cos( . )
.
cos( . )
.
cos( )
cos( . )
.
co os( . )
.
cos( . )
.
cos( )
.
2 5
2 5
2
2 75
2 75
3
3
0 210
+
+










≈ −