Concepts Review The integral diverges.
8.3 Concepts Review The integral diverges.
1. converge
b →∞ ∫ 0 cos x dx
3. ∫ – () f x dx ; () ∫ f x dx
11. ∫ ∞ e dx = [ln(ln )] x e =∞ –0 =∞
4. p>1
x ln x
The integral diverges.
Problem Set 8.3
dx ⎢ (ln ) x 2 ⎤ =∞ – =∞ In this section and the chapter review, it is understood
ln x
b The integral diverges.
that [ ( )] gx means lim [ ( )] gx and likewise for
a b →∞
1 1 1 similar expressions.
13. Let u = ln x, du = dx dv ,, = 2 dx v =− .
xx
1. e dx = ∫ ⎡⎤ e =∞ – =∞
∫ 2 2 dx = x lim b →∞ ∫ 2 2 dx
The integral diverges.
b →∞ ⎣ ⎢ x ⎥ b →∞ ∫ 2 2
dx ⎡ 1 ⎤
⎦ 1 e 14. xe – ∫ x 1 dx u = x, du = dx
3. ∫ 1 2 xe dx = ⎢ – e ⎥ = 0 – (– e –1 ) = ∞
4 4 dv = e dx v ,– = e ∫
⎦ 1 e The integral diverges.
1 1 The integral diverges.
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The integral diverges.
The integral diverges since both ∫ –
dx and ∫ 0 dx diverge.
– ∞ ( x 2 + 16) 2 ∫ – ∞ ( x 2 + 16) 2 ∫ 0 ( x 2 + 16) 2
dx
tan –1 +
by using the substitution x = 4 tan
19. ∫ 2 dx =
dx =
dx +
– ∞ x + 2 x + 10 ∫ – ∞ ( x + 1) 2 + 9 ∫ – ∞ ( x + 1) 2 + 9 ∫ 0 ( x + 1) 2 + 9
dx
1 1 –1 x + ∫ 1
dx = tan
by using the substitution x + 1 = 3 tan θ.
( x + 1) 2 + 9 3 3
–1 ∫ 1 dx =⎢ tan
∞ 2 dx x = ∫ – x dx + ∞ e –2 ∫ 0 e 2 x dx
For ∫ –
dx =
∫ – xe dx , use u = x, du = dx, dv = e dx v , = e 2 x ∞ . e ∞ 2
e ∫ 0 xe dx , use u = x, du = dx, dv = e dx v ,– = e 2 .
xe dx = ⎡ ⎢ – xe –2 x ⎤ +
–2 x
e –2 x dx = ⎡ – xe –2 x – e –2 x ⎤ = 0–0– ⎛ ⎞ ⎜ ⎟ =
– ∞ 2 x dx = – += 0
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21. ∫ – sech sech sech x dx = x dx =
25. The area is given by
∫ 0 x dx ∞ 2 ∞ ⎛ 1 1 ⎞
= [tan –1 (sinh )] x 0 + [tan –1 (sinh )] x ∞ ∫ 1 4 x 2 dx =
− 1 ∫ 1 – ⎜ ∞ 0 ⎝ 2–1 x 2 x + 1 ⎟ ⎠
22. ∫ 1 csch x dx = ∫ 1 dx = ∫ 1 x – x dx 2 ⎝ ⎝⎠ 3 ⎠ 2
sinh x
Note:. lim ln =
= 0 since
= ∫ 2 x dx x →∞ 2 x + 1 1
dx = ∫ e 2 du = ∫ e ⎜ –
⎝ u –1 u + 1 ⎟
⎠ 26. The area is
= [ln( – 1) – ln( u u + 1)]
≈ 0.7719 ⎣ ln x − ln x +⎤= 1 ⎦ 1 ⎢ ln
= 0 since lim
= 1 ⎟ ⎝ b →∞ b + 1 b →∞ b + 1 ⎠
27. The integral would take the form
k ∫ 3960 dx = [ k ln x
23. ∞− x
0 e cos x dx = ⎢ x (sin x − cos ) x
⎣ 2 e ⎦ 0 which would make it impossible to send anything out of the earth's gravitational field.
2 2 28. At x = 1080 mi, F = 165, so
(Use Formula 68 with a = –1 and b = 1.) k = 165(1080) 2 ≈ 1.925 10 × 8 . So the work done
⎡ in mi-lb is
(cos x + sin ) x ⎥
1.925 10 × 8 dx = 1.925 10 × 8 ⎡ − x − 1 ⎦ ⎤ 0 ∫ 1080 2
(Use Formula 67 with a = –1 and b = 1.)
29. FP =
100, 000 e − ⎢ t = 1,250,000
The present value is $1,250,000.
30. FP = ∞− 0.08 ∫
0 e t (100, 000 1000 ) + t dt
=− ⎡ ⎣ 1, 250, 000 e − t − 12,500 te − t − 156, 250 e − t ⎤
The present value is $1,406,250.
31. a.
() f x dx = ∫ 0 dx + ∫
dx + ∫ 0 dx
0 [] x a + 0 =
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x f x dx () −∞
x ⋅ 0 dx + ∫ a x dx +
b − a ∫ b x ⋅ 0 dx