Section 7.8 Improper Integrals

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Section 7.8 Improper Integrals

68. Improper Integrals that Are Both Type 1 and Type 2 The integralR∞

a f (x)dx is improper because the interval [a, ∞) is infinite. If f has an infinite discontinuity at a, then the integral is improper for a second reason. In this case we evaluate the integral by expressing it as a sum of improper integral of Type 2and Type 1 as follows:

Z ∞ a

f (x)dx = Z c

a

f (x)dx + Z ∞

c

f (x)dx c > a Evaluate the given integral if it is convergent.

Z ∞ 2

1 x√

x²− 4dx Solution:

716 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION 54. For 0   ≤ 1, sin²

√ ≤ 1

√. Now

 =

  0

√1

 = lim

→0⁺

 



^−12 = lim

→0⁺

2^12

 = lim

→0⁺

2 − 2√



= 2 − 0 = 2, so  is convergent, and by

comparison,

  0

sin²

√ is convergent.

55.

 _∞

0

√ 

 (1 + )=

 1 0

√ 

 (1 + )+

 _∞

1

√ 

 (1 + )= lim

→0⁺

 1



√ 

 (1 + )+ lim

→∞

  1

√ 

 (1 + ). Now

 

√ (1 + )=

 2 

(1 + ²)

 =√,  = ²,

 = 2 



= 2

 

1 + ² = 2 tan⁻¹ +  = 2 tan⁻¹√ + , so

 _∞

0

√ 

 (1 + ) = lim

→0⁺

2 tan⁻¹√

1

+ lim

→∞

2 tan⁻¹√

 1

= lim

→0⁺

2 4

− 2 tan⁻¹√

 + lim

→∞

2 tan⁻¹√

 − 2 4

= ^₂ − 0 + 2 2

−^2 = .

56.  ∞ 2



√

²− 4 =

 3 2



√

²− 4+

∞ 3



√

²− 4= lim

→2⁺

 3





√

²− 4+ lim

→∞

  3



√

²− 4. Now

 

√

²− 4=

 2 sec  tan  

2 sec  2 tan 

  = 2 sec , where 0≤   2 or  ≤   32



=¹₂ +  =¹₂sec⁻¹₁

2 + , so

 _∞

2



√

²− 4 = lim

→2⁺

1

2sec⁻¹1 2³

+ lim

→∞

1 2sec⁻¹1

2^

3=¹₂sec⁻¹3 2

− 0 +¹2

 2

−¹2sec⁻¹3 2

= ^₄.

57. If  = 1, then

 1 0



^ = lim

→0⁺

1





 = lim

→0⁺[ln ]¹_ = ∞. Divergent If  6= 1, then

 1 0



^ = lim

→0⁺

 1





^ [note that the integral is not improper if   0]

= lim

→0⁺

^−+1

− + 1

¹



= lim

→0⁺

1 1 − 

 1 − 1

^⁻¹



If   1, then  − 1  0, so 1

^⁻¹ → ∞ as  → 0⁺, and the integral diverges.

If   1, then  − 1  0, so 1

^⁻¹ → 0 as  → 0⁺and

 1 0



^ = 1 1 − 



→0lim⁺

1 − ¹^−

= 1

1 − . Thus, the integral converges if and only if   1, and in that case its value is 1

1 − .

58. Let  = ln . Then  =  ⇒  ∞





 (ln )^ =

 ∞ 1



^. By Example 4, this converges to 1

 − 1if   1 and diverges otherwise.

59. First suppose  = −1. Then

 1 0

^ln   =

 1 0

ln 

  = lim

→0⁺

 1



ln 

  = lim

→0⁺

1

2(ln )²¹

 = −¹2 lim

→0⁺(ln )²= −∞, so the integral

74. The average speed of molecules in an ideal gas is

¯ v = 4

√π

M

2RT

^3/2Z ∞ 0

v³e^{−M v}²^{/(2RT )}dv

where M is the molecular weight of the gas, R is the gas constant, T is the gas temperature, and v is the molecular speed. Show that

¯ v =

r8RT πM Solution:

718 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION So

^+1^− = −^+1^−−

−( + 1)^^− = −^+1^−+ ( + 1)

^^−and

lim→∞



0^+1^− = lim

→∞

−^+1^−

0+ ( + 1) lim

→∞



0^^−

= lim

→∞

−^+1^−+ 0+ ( + 1)! = 0 + 0 + ( + 1)! = ( + 1)!,

so the formula holds for  + 1. By induction, the formula holds for all positive integers. (Since 0! = 1, the formula holds for  = 0, too.)

61. (a)  =_∞

−∞  =0

−∞  +_∞

0  , and_∞

0   = lim

→∞



0  = lim

→∞

₁

2² 0= lim

→∞

₁

2²− 0

= ∞, so  is divergent.

(b)

−  =₁

2²

−=¹₂²−¹2²= 0, so lim

→∞



−  = 0. Therefore,_∞

−∞  6= lim_

→∞



− .

62.Let  = 

2 so that  = 4

√^32

 _∞

0

³^−². Let  denote the integral and use parts to integrate . Let  = ²,

 = ^−² ⇒  = 2 ,  = −1 2^−²:

 = lim

→∞



− 1

2²^−²

 0

+1



 _∞

0

^−²₀^= − 1 2 lim

→∞



²^−² +1

 lim

→∞



−1 2^−²



= −H 1

2· 0 − 1

2²(0 − 1) = 1 2² Thus,  = 4

√^32· 1

2² = 2

()^12 = 2

[ (2 )]^12 =2√ 2√

√ 

 =

8

.

63.Volume =

 _∞

1



1



2

 =  lim

→∞

  1



² =  lim

→∞



−1



 1

=  lim

→∞

 1 −1





=   ∞.

64.Work =

_∞



 

²  = lim

→∞

 



 

²  = lim

→∞ 

−1







=   lim

→∞

−1

 + 1





=  

 , where

 =mass of the earth = 598 × 10²⁴kg,  = mass of satellite = 10³kg,  = radius of the earth = 637 × 10⁶m, and

 =gravitational constant = 667 × 10⁻¹¹N·m²kg.

Therefore, Work = 667 × 10⁻¹¹· 598 × 10²⁴· 10³

637 × 10⁶ ≈ 626 × 10¹⁰J.

65.Work =

 _∞



  = lim

→∞

 





²  = lim

→∞

1

−1





= 

 . The initial kinetic energy provides the work, so ¹₂²₀= 

 ⇒ ⁰=

2

 . 66.() =

 



√ 2

²− ²() and () =¹₂( − )² ⇒

() = lim

→⁺

 



( − )²

√²− ²  = lim

→⁺

 



³− 2²+ ²

√²− ² 

= lim

→⁺

  



³

√²− ² − 2

 



²

√²− ² + ²





√ 

²− ²



= lim

→⁺

1− 2²+ ²3

= 

75. We know from Example 1 that the region R = {(x, y)|x ≥ 1, 0 ≤ y ≤ 1/x} has infinite area. Show that by rotating Rabout the x-axis we obtain a solid with finite volume.

Solution:

718 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION So

^+1^− = −^+1^−−

−( + 1)^^− = −^+1^−+ ( + 1)

^^−and

lim→∞



0^+1^− = lim

→∞

−^+1^−

0+ ( + 1) lim

→∞



0^^−

= lim

→∞

−^+1^−+ 0+ ( + 1)! = 0 + 0 + ( + 1)! = ( + 1)!,

so the formula holds for  + 1. By induction, the formula holds for all positive integers. (Since 0! = 1, the formula holds for  = 0, too.)

61. (a)  =_∞

−∞  =0

−∞  +_∞

0  , and_∞

0   = lim

→∞



0  = lim

→∞

₁

2² 0= lim

→∞

₁

2²− 0

= ∞, so  is divergent.

(b)

−  =₁

2²

−=¹₂²−¹2²= 0, so lim

→∞



−  = 0. Therefore,_∞

−∞  6= lim_

→∞



− .

62.Let  = 

2 so that  = 4

√^32

 _∞

0

³^−². Let  denote the integral and use parts to integrate . Let  = ²,

 = ^−² ⇒  = 2 ,  = −1 2^−²:

 = lim

→∞



− 1

2²^−²

 0

+1



 _∞

0

^−²₀^= − 1 2 lim

→∞



²^−² +1

 lim

→∞



−1 2^−²



= −H 1

2· 0 − 1

2²(0 − 1) = 1 2² Thus,  = 4

√^32· 1

2² = 2

()^12 = 2

[ (2 )]^12 =2√ 2√

√ 

 =

8

.

63.Volume =

 _∞

1



1



2

 =  lim

→∞

  1



² =  lim

→∞



−1



 1

=  lim

→∞

 1 −1





=   ∞.

64.Work =∞



 

²  = lim

→∞

 



 

²  = lim

→∞ 

−1







=   lim

→∞

−1

 + 1





=  

 , where

 =mass of the earth = 598 × 10²⁴kg,  = mass of satellite = 10³kg,  = radius of the earth = 637 × 10⁶m, and

 =gravitational constant = 667 × 10⁻¹¹N·m²kg.

Therefore, Work = 667 × 10⁻¹¹· 598 × 10²⁴· 10³

637 × 10⁶ ≈ 626 × 10¹⁰J.

65.Work =

 _∞



  = lim

→∞

 





²  = lim

→∞

1

−1





= 

 . The initial kinetic energy provides the work, so ¹₂²₀= 

 ⇒ ⁰=

2

 . 66.() =

 



√ 2

²− ²() and () =¹₂( − )² ⇒

() = lim

→⁺

 



( − )²

√²− ²  = lim

→⁺

 



³− 2²+ ²

√²− ² 

= lim

→⁺

  



³

√²− ² − 2

 



²

√²− ² + ²





√ 

²− ²



= lim

→⁺

1− 2²+ ²3

= 

80. As we saw in Section 3.8, a radioactive substance decays exponentially: The mass at time t is m(t) = m(0)e^kt, where m(0) is the initial mass and k is a negative constant. The mean life M of an atom in the substance is

M = −k Z ∞

0

te^ktdt

1

(2)

For the radioactive carbon isotope, ¹⁴C, used in radiocarbon dating, the value of k is −0.000121. Find the mean life of a¹⁴C atom.

Solution:

SECTION 7.8 IMPROPER INTEGRALS ¤ 719 For 1: Let  =√

²− ² ⇒ ²= ²− ², ²= ²+ ², 2  = 2 , so, omitting limits and constant of integration,

1=

 (²+ ²)

  =



(²+ ²)  = ¹₃³+ ² =¹₃(²+ 3²)

=¹₃√

²− ²(²− ²+ 3²) = ¹₃√

²− ²(²+ 2²) For 2: Using Formula 44, 2= 

2

√²− ²+²

2 ln +√

²− ².

For 3: Let  = ²− ² ⇒  = 2 . Then ³=1 2

 

√ =¹₂ · 2√

 =

²− ². Thus,

 = lim

→⁺



1 3

√²− ²(²+ 2²) − 2

 2

√²− ²+²

2 ln +√

²− ² + ²√

²− ²

^



= lim

→⁺



1 3

√²− ²(²+ 2²) − 2

 2

√²− ²+²

2 ln +√

²− ² + ²√

²− ²



− lim

→⁺



1 3

√²− ²

²+ 2²

− 2

 2

√²− ²+²

2 ln +√

²− ² + ²√

²− ²



=1 3

√²− ²(²+ 2²) − ²ln +√

²− ²

  −

−²ln ||

=¹₃√

²− ²(²+ 2²) − ²ln

 +√

²− ²





67.We would expect a small percentage of bulbs to burn out in the ﬁrst few hundred hours, most of the bulbs to burn out after close to 700 hours, and a few overachievers to burn on and on.

(a) (b) () = ⁰()is the rate at which the fraction  () of burnt-out bulbs increases as  increases. This could be interpreted as a fractional burnout rate.

(c)_∞

0 ()  = lim

→∞ () = 1, since all of the bulbs will eventually burn out.

68. =

 _∞

0

^ = lim

→∞

1

² ( − 1) ^

 0

[Formula 96, or parts] = lim

→∞

1

^− 1

²^



−



−1

²



.

Since   0 the ﬁrst two terms approach 0 (you can verify that the ﬁrst term does so with l’Hospital’s Rule), so the limit is equal to 1². Thus,  = − = −

1²

= −1 = −1(−0000121) ≈ 82645 years.

69.  =

 _∞

0

(1 − ^−)

 ^− = 

 lim

→∞

  0



^−− ⁽^{−−)}



=

 lim

→∞

 1

−^−− 1

− − ⁽^{−−)}

 0

= 

 lim

→∞

 1

−^+ 1

( + )^(+)−

 1

−+ 1

 + 



=



1

− 1

 + 



=



 +  − 

( + )



= 

( + )

2