Section 5.3 The Fundamental Theorem of Calculus

Section 5.3 The Fundamental Theorem of Calculus

58. Sketch the region enclosed by the given curves and calculate its area. y = 2x − x², y = 0.

Solution:

SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ¤ 511 46. Area = 1

0

³ =1 4⁴1

0= ¹₄− 0 = ¹4

47. Area = 2

−2(4 − ²)  =

4 −¹3³²

−2= 8 −⁸3

−

−8 +⁸3

= ³²₃

48. Area = 2

0 (2 − ²)  =

²− ¹3³2 0=

4 − ⁸3

− 0 = ⁴3

49. From the graph, it appears that the area is about 60. The actual area is

27

0 ^13 =

3 4^4327

0 = ³₄ · 81 − 0 = ²⁴³4 = 6075. This is ³₄ of the area of the viewing rectangle.

50. From the graph, it appears that the area is about ¹₃. The actual area is

 6 1

⁻⁴ =

⁻³

−3

6 1

=

−1 3³

6 1

= − 1

3 · 216 +1 3 = 215

648 ≈ 03318.

51. It appears that the area under the graph is about²₃ of the area of the viewing rectangle, or about²₃ ≈ 21. The actual area is



0 sin   = [− cos ]^0 = (− cos ) − (− cos 0) = − (−1) + 1 = 2.

52. Splitting up the region as shown, we estimate that the area under the graph is ^₃ + ¹₄

3 ·^3

≈ 18. The actual area is

3

0 sec²  = [tan ]^3₀ =√

3 − 0 =√

3 ≈ 173.

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68. Find the derivative of the function. g(x) =R1+2x

1−2x t sin tdt.

Solution:

512 ¤ CHAPTER 5 INTEGRALS 53. 2

−1³ =1 4⁴2

−1= 4 − ¹4 = ¹⁵₄ = 375

54. 2

6cos   = sin 2

6= 0 −¹2 = −¹2

55. () = ⁻⁴is not continuous on the interval [−2 1], so FTC2 cannot be applied. In fact,  has an infinite discontinuity at

 = 0, so1

−2⁻⁴does not exist.

56. () = 4

³ is not continuous on the interval [−1 2], so FTC2 cannot be applied. In fact,  has an infinite discontinuity at

 = 0, so 2

−1

4

³does not exist.

57. () = sec  tan is not continuous on the interval [3 ], so FTC2 cannot be applied. In fact,  has an infinite discontinuity at  = 2, so

3sec  tan  does not exist.

58. () = sec²is not continuous on the interval [0 ], so FTC2 cannot be applied. In fact,  has an infinite discontinuity at

 = 2, so

0 sec² does not exist.

59. () = 3

2

²− 1

²+ 1 =

 0 2

²− 1

²+ 1 +

 3

0

²− 1

²+ 1 = −

 2

0

²− 1

²+ 1 +

 3

0

²− 1

²+ 1 ⇒

⁰() = −(2)²− 1 (2)²+ 1· 

(2) + (3)²− 1 (3)²+ 1· 

(3) = −2 · 4²− 1

4²+ 1 + 3 ·9²− 1 9²+ 1

60. () = 1+2

1−2

 sin   =

 0 1−2

 sin   +

 1+2

0  sin   = −

 1−2

0

 sin   +

 1+2

0

 sin   ⇒

⁰() = −(1 − 2) sin(1 − 2) · 

(1 − 2) + (1 + 2) sin(1 + 2) · 

(1 + 2)

= 2(1 − 2) sin(1 − 2) + 2(1 + 2) sin(1 + 2)

61.  () = ²



^2 =

 0



^2 +

 ² 0

^2 = −

  0

^2 +

 ² 0

^2 ⇒

⁰() = −^2+ ^(2)2· 

(²) = −^2+ 2^4

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76. If f (x) =Rsin x 0

√1 + t²dt and g(y) =Ry

3 f (x)dx, find g⁰⁰(^π₆).

Solution:

SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ¤ 513 62.  () =

 2

√

arctan   =

 0

√

arctan   +

 2

0 arctan   = −

 ^√ 0

arctan   +

 2

0

arctan   ⇒

⁰() = − arctan√

 · 

(√

 ) + arctan 2 · 

(2) = − 1

2√arctan√

 + 2 arctan 2

63.  = sin  cos 

ln(1 + 2)  =

 0 cos 

ln(1 + 2)  +

 sin  0

ln(1 + 2) 

= −

 cos  0

ln(1 + 2)  +

 sin  0

ln(1 + 2)  ⇒

⁰= − ln(1 + 2 cos ) · 

cos  + ln(1 + 2 sin ) · 

sin  = sin  ln(1 + 2 cos ) + cos  ln(1 + 2 sin ) 64. () = 

0 (1 − ²)^2is increasing when ⁰() = (1 − ²)^2is positive.

Since ^2 0, ⁰()  0 ⇔ 1 − ² 0 ⇔ ||  1, so  is increasing on (−1 1).

65.  =  0

²

²+  + 2 ⇒ ⁰= ²

²+  + 2 ⇒

⁰⁰ = (²+  + 2)(2) − ²(2 + 1)

(²+  + 2)² = 2³+ 2²+ 4 − 2³− ²

(²+  + 2)² = ²+ 4

(²+  + 2)² = ( + 4) (²+  + 2)². The curve  is concave downward when ⁰⁰ 0; that is, on the interval (−4 0).

66. If  () =

1  () , then by FTC1, ⁰() =  (), and also, ⁰⁰() = ⁰().  is concave downward where ⁰⁰is negative; that is, where ⁰is negative. The given graph shows that  is decreasing (⁰ 0)on the interval (−1 1).

67.  () =

2 ^2 ⇒ ⁰() = ^2, so the slope at  = 2 is ²²= ⁴. The -coordinate of the point on  at  = 2 is

 (2) =2

2 ^2 = 0since the limits are equal. An equation of the tangent line is  − 0 = ⁴( − 2), or  = ⁴ − 2⁴. 68. () =

3  ()  ⇒ ⁰() =  (). Since () =sin  0

√1 + ², ⁰⁰() = ⁰() =

1 + sin² · cos , so ⁰⁰(^₆) =

1 + sin²(^₆) · cos^6 =

1 + (¹₂)²·^√2³ = ^√₂⁵ ·^√2³ = ^√₄¹⁵.

69. By FTC2,4

1 ⁰()  =  (4) − (1), so 17 = (4) − 12 ⇒ (4) = 17 + 12 = 29.

70. (a) erf() = 2

√

  0

^−2 ⇒

  0

^−2 =

√

2 erf()By Property 5 of definite integrals in Section 5.2,



0 ^−2 =

0 ^−2 +

 ^−2, so

 



^−2 =

  0

^−2 −

  0

^−2 =

√

2 erf() −

√

2 erf() = ¹₂√

 [erf() − erf()].

(b)  = ^2erf() ⇒ ⁰= 2^2erf() + ^2erf⁰() = 2 + ^2· 2

√^−2 [by FTC1] = 2 + 2

√.

71. (a) The Fresnel function () = 0 sin_

2²

has local maximum values where 0 = ⁰() = sin_

2² and

⁰changes from positive to negative. For   0, this happens when^₂² = (2 − 1) [odd multiples of ] ⇔

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78. Use l’Hospital’s Rule to evaluate the limit. lim

x→∞

1 x²

Rx

0 ln(1 + e^t)dt.

Solution: SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ¤ 553

78. lim

→∞

1

²

  0

ln(1 + ^)  = lim

→∞

  0

ln(1 + ^) 

²

form ^∞_∞ H

= lim

→∞

ln(1 + ^) 2

= limH

→∞

^ 1 + ^

2

= lim

→∞

^

2(1 + ^) = lim

→∞

^^

2(1 + ^)^ = lim

→∞

1 2

1

^ + 1

 = 1

2(0 + 1) = 1 2

79. By FTC2,4

1 ⁰()  =  (4) − (1), so 17 = (4) − 12 ⇒ (4) = 17 + 12 = 29.

80. (a) erf() = 2

√

  0

^−2 ⇒

  0

^−2 =

√

2 erf()By Property 5 of definite integrals in Section 5.2,



0 ^−2 =

0 ^−2 +

 ^−2, so

 



^−2 =

  0

^−2 −

  0

^−2 =

√

2 erf() −

√

2 erf() = ¹₂√

 [erf() − erf()].

(b)  = ^2erf() ⇒ ⁰= 2^2erf() + ^2erf⁰() = 2 + ^2· 2

√^−2 [by FTC1] = 2 + 2

√.

81. (a) The Fresnel function () = 0 sin

2²

has local maximum values where 0 = ⁰() = sin 2²and

⁰changes from positive to negative. For   0, this happens when ^₂²= (2 − 1) [odd multiples of ] ⇔

² = 2(2 − 1) ⇔  =√

4 − 2,  any positive integer. For   0, ⁰changes from positive to negative where



2²= 2 [even multiples of ] ⇔ ²= 4 ⇔  = −2√

. ⁰does not change sign at  = 0.

(b)  is concave upward on those intervals where ⁰⁰()  0. Differentiating our expression for ⁰(), we get

⁰⁰() = cos_

2²

2^₂

=  cos_

2²

. For   0, ⁰⁰()  0where cos(^₂²)  0 ⇔ 0  ^2² ^₂ or

2 −¹2

  ^₂²

2 +¹₂

,  any integer ⇔ 0    1 or√

4 − 1   √4 + 1,  any positive integer.

For   0, ⁰⁰()  0where cos(^₂²)  0 ⇔  2 −³2

  ^₂² 2 −¹2

,  any integer ⇔

4 − 3  ² 4 − 1 ⇔ √

4 − 3  || √

4 − 1 ⇒ √

4 − 3  − √

4 − 1 ⇒

−√

4 − 3    −√

4 − 1, so the intervals of upward concavity for   0 are

−√

4 − 1 −√

4 − 3,  any positive integer. To summarize:  is concave upward on the intervals (0 1),

−√

3 −1,√

3√ 5,

−√ 7 −√

5,

√7 3,    .

(c) In Maple, we use plot({int(sin(Pi*tˆ2/2),t=0..x),0.2},x=0..2);. Note that Maple recognizes the Fresnel function, calling it FresnelS(x). In Mathematica, we use

Plot[{Integrate[Sin[Pi*tˆ2/2],{t,0,x}],0.2},{x,0,2}]. In Derive, we load the utility file FRESNEL and plot FRESNEL_SIN(x). From the graphs, we see that

0 sin 2²

 = 02at  ≈ 074.

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93. Find a function f and a number a such that 6 +

Z x a

f (t)

t² dt = 2√

x for all x > 0

Solution:

SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ¤ 517 82. (a) If   0, then () =

0  ()  =

0 0  = 0.

If 0 ≤  ≤ 1, then () =

0  ()  =

0   =1 2²

0 = ¹₂². If 1   ≤ 2, then

() =

0  ()  =1

0  ()  +

1  ()  = (1) +

1(2 − ) 

= ¹₂(1)²+

2 − ¹2²

1 = ¹₂+

2 −¹2²

− 2 −¹2

= 2 −¹2²− 1.

If   2, then () =

0  ()  = (2) +

2 0  = 1 + 0 = 1. So

() =













0 if   0

1

2² if 0 ≤  ≤ 1 2 −¹2²− 1 if 1   ≤ 2

1 if   2

(b)

(c)  is not differentiable at its corners at  = 0, 1, and 2.  is differentiable on (−∞ 0), (0 1), (1 2) and (2 ∞).

is differentiable on (−∞ ∞).

83. Using FTC1, we differentiate both sides of 6 + 



 ()

²  = 2√

to get  ()

² = 2 1 2√

 ⇒ () = ^32. To find , we substitute  =  in the original equation to obtain 6 +

 



 ()

²  = 2√

 ⇒ 6 + 0 = 2√

 ⇒

3 =√

 ⇒  = 9.

84.  = 3 ⇒ 

0^ = 3

0 ^ ⇒ [^]^₀= 3 [^]^₀ ⇒ ^− 1 = 3(^− 1) ⇒ ^ = 3^− 2 ⇒

 = ln(3^− 2) 85. (a) Let  () =

0 () . Then, by FTC1, ⁰() =  () =rate of depreciation, so  () represents the loss in value over the interval [0 ].

(b) () = 1





 +

  0

 () 



=  +  ()

 represents the average expenditure per unit of  during the interval [0 ], assuming that there has been only one overhaul during that time period. The company wants to minimize average expenditure.

(c) () =1





 +

  0

 () 



. Using FTC1, we have ⁰() = −1

²



 +

  0

 () 

 +1

 ().

⁰() = 0 ⇒  () =  +

  0

 ()  ⇒ () = 1





 +

  0

 () 



= ().

86. (a) () = 1



  0

[ () + ()] . Using FTC1 and the Product Rule, we have

⁰() = 1

[ () + ()] − 1

²

  0

[ () + ()] . Set ⁰() = 0: 1

[ () + ()] − 1

²

  0

[ () + ()]  = 0 ⇒

[ () + ()] −1



  0

[ () + ()]  = 0 ⇒ [() + ()] − () = 0 ⇒ () =  () + ().

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