James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

534 Chapter 7 Techniques of Integration
Table 1
t
y t 2 0 e2x dx
1 0.7468241328
2 0.8820813908
3 0.8862073483
4 0.8862269118
5 0.8862269255
6 0.8862269255
Table 2
t y t fs1 1 1 e2x dyxg dx
2 0.8636306042
5 1.8276735512
10 2.5219648704
100 4.8245541204
1000 7.1271392134
10000 9.4297243064
Therefore, taking f sxd − e 2x and tsxd − e 2x 2 in the Comparison Theorem, we see that
y` 2
1 e2x dx is convergent. It follows that y` 2 0 e2x dx is convergent. n
In Example 9 we showed that y`
0 e2x 2 dx is convergent without computing its value. In
Exercise 72 we indicate how to show that its value is approximately 0.8862. In probability
theory it is important to know the exact value of this improper integral, as we will see in
Section 8.5; using the methods of multivariable calculus it can be shown that the exact
value is s y2. Table 1 illustrates the definition of an improper integral by showing how
the (computer-generated) values of y t 0 e2x 2 dx approach s y2 as t becomes large. In fact,
these values converge quite quickly because e 2x 2 l 0 very rapidly as x l `.
1 1 e
Example 10 The integral y`
2x
dx is divergent by the Comparison Theorem
1
because
x
1 1 e 2x
. 1 x x
and y`
1
s1yxd dx is divergent by Example 1 [or by (2) with p − 1]. n
Table 2 illustrates the divergence of the integral in Example 10. It appears that the
values are not approaching any fixed number.
;
1. Explain why each of the following integrals is improper.
(a) y 2 x
1 x 2 1 dx (b) 1
y`
0 1 1 x dx 3
(c) y`
x 2 e 2x 2 dx (d) y y4
cot x dx
2`
2. Which of the following integrals are improper? Why?
(a) y y4
tan x dx (b) y
tan x dx
0
(c) y 1 dx
(d) y`
e 2x 3 dx
21 x 2 2 x 2 2
0
3. Find the area under the curve y − 1yx 3 from x − 1 to x − t
and evaluate it for t − 10, 100, and 1000. Then find the total
area under this curve for x > 1.
4. (a) Graph the functions f sxd − 1yx 1.1 and tsxd − 1yx 0.9 in
the viewing rectangles f0, 10g by f0, 1g and f0, 100g
by f0, 1g.
(b) Find the areas under the graphs of f and t from x − 1
to x − t and evaluate for t − 10, 100, 10 4 , 10 6 , 10 10 ,
and 10 20 .
(c) Find the total area under each curve for x > 1, if it exists.
5–40 Determine whether each integral is convergent or divergent.
Evaluate those that are convergent.
5. y`
3
7. y 0 2`
1
3y2
dx
sx 2 2d 6. 1
y`
0
s 4 1 1 x dx
1
3 2 4x dx 8. 1
y`
1 s2x 1 1d dx 3
0
0
9. y`
2
11. y`
0
13. y`
e 25p dp 10. y 0 2`
2 r dr
x 2
s1 1 x 3 dx
2` xe2x 2 dx
15. y`
sin 2 d
0
17. y`
1
1
x 2 1 x dx
19. y 0 ze 2z dz
2`
ln x
21. y`
dx
1 x
23. y 0 z
2` z 4 1 4 dz
12. y`
2` sy3 2 3y 2 d dy
14. y`
1
e 21yx
x 2 dx
16. y`
sin 0 ecos d
18. dv
y`
2 v 2 1 2v 2 3
20. y`
ye 23y dy
ln x
22. y`
1 x dx 2
24. 1
y`
e xsln xd dx 2
dx
25. y`
0 e2sy dy 26. y`
1 sx 1 xsx
27. y 1 1
0 x dx 28. y 5 1
dx
0
s 3 5 2 x
29. y 14 dx
30. y 2 x
22
s 4 21
x 1 2 sx 1 1d dx 2
31. y 3 22
1
x dx 32. y 1
4 0
2
dx
s1 2 x 2
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