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Chapter 7. Integration Techniques, Solution Key
7.6 Improper Integrals
Review Questions
1. a. This is improper because there is an infinite discontinuity between the limits of integration at x = 3.
b. This is not improper.
c. This is improper because there is an infinite discontinuity at x = 0.
d. This is improper because the integral has an infinite interval of integration.
e. This is not improper.
2.
l
R∞ 1
−x1.001 1
dx
=
lim
l→∞
2.001
1.001 1 = 1.001
x
l
3.
Z−2
−∞
1
1
−
dx = lim [ln|x − 1|−ln|x + 1|]−2
l
l→−∞
x−1 x+1
= ln3 − 0 − lim [ln|l − 1|−ln|l + 1|]
l→−∞
l −1
= ln3 − ln l +1
l 1− 1
l = ln3 − ln l 1 + 1l = ln3 − 0
= ln3
4.
Z0
5x
1 5x 0
e
5
l
1
1
× l − e5l
5
5
e dx = lim
l→−∞
−∞
= lim
l→−∞
1
−0
5
1
=
5
=
5. The integral is divergent.
−π
6.
R2
tan x = 0 (Look at the symmetry of the graph on the interval.)
−π
2
245
7.6. Improper Integrals
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7.
Z1
0
1
√
dx = sin−1 x|10
1 − x2
= sin−1 1 − sin−1 0
π
= −0
2
π
=
2
8. a.
Z∞
(e−x )2 dx
V =π
0
Z∞
=π
e−2x dx
0
∞
−e−2x
=π
2
0
1
= π 0+
2
π
=
2
b.
Z∞
A = 2π
q
f (x) 1 + [ f 0 (x)]2 dx
0
Z∞
= 2π
e−x
q
1 + [−e−x ]2 dx
e−x
p
1 + e−2x dx
0
Z∞
= 2π
0
Let e−x = tan u. Then −e−x dx = sec2 u du.
246
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Chapter 7. Integration Techniques, Solution Key
Z∞
A = 2π
e−x
p
1 + e−2x dx
0
Z0
= 2π
∞
Z0
= 2π
sec2 u
p
1 + tan2 u du
sec3 u du
∞
1
1
sec u tan u + ln|sec u + tan u|
2
2
0 −2x 12 −x
−2x 21
−x = 2π (1 + e ) e + ln (1 + e ) + e ∞
√
= π[2 + ln( 2 + 1)]
= 2π
247