Section 5.5 The Substitution Rule

Section 5.5 The Substitution Rule

80. Evaluate the definite integralR16 1

x^1/2 1+x^3/4dx.

Solution:

SECTION 5.5 THE SUBSTITUTION RULE ¤ 571 72. 3

−3⁴sin   = 0by Theorem 7(b), since () = ⁴sin is an odd function.

73. Let  =  − 1, so  + 1 =  and  = . When  = 1,  = 0; when  = 2,  = 1. Thus,

 2 1

√

 − 1  =

 1 0

( + 1)√

  =

 1 0

(^32+ ^12)  =

2

5^52+²₃^321

0= ²₅+ ²₃ = ¹⁶₁₅. 74. Let  = 1 + 2, so  = ¹₂( − 1) and  = 2 . When  = 0,  = 1; when  = 4,  = 9. Thus,

 4 0

√ 

1 + 2 =

 9 1

1 2( − 1)

√



2 = ¹₄

 9 1

(^12− ^−12)  = ¹₄

2

3^32− 2^129 1= ¹₄· ²3

^32− 3^129 1

= ¹₆[(27 − 9) − (1 − 3)] = ²⁰6 = ¹⁰₃

75. Let  = ln , so  = 

. When  = ,  = 1; when  = ⁴;  = 4. Thus,

 ⁴





√ ln  =

 4 1

^−12 = 2

^124

1 = 2(2 − 1) = 2.

76. Let  = ( − 1)², so  = 2( − 1) . When  = 0,  = 1; when  = 2,  = 1. Thus,

 2

0 ( − 1)^(−1)2 =

 1 1

^1 2

= 0since the limits are equal.

77. Let  = ^+ , so  = (^+ 1) . When  = 0,  = 1; when  = 1,  =  + 1. Thus,

 1 0

^+ 1

^+  =

 +1 1

1

 =

ln ||+1

1 = ln | + 1| − ln |1| = ln( + 1).

78. Let  = √. Then ²= , 2  = ,  = 1

2√, and 1

√ = 2 . When  = 1,  = 1; when  = 4,  = 2.

Thus,  4

1

1 ( + 1)√

 =

 2 1

1

²+ 1(2 ) = 2

arctan 2

1= 2(arctan 2 − arctan 1)

= 2

arctan 2 − 4

= 2 arctan 2 −  2 79. Let  = 1 + √, so  = 1

2√

 ⇒ 2√  =  ⇒ 2( − 1)  = . When  = 0,  = 1; when  = 1,

 = 2. Thus,

 1 0



(1 +√

 )⁴ =

 2 1

1

⁴ · [2( − 1) ] = 2

 2 1

 1

³ − 1

⁴



 = 2



− 1 2² + 1

3³

2 1

= 2

−¹8+ ₂₄¹

−

−¹2+ ¹₃

= 21 12

= ¹₆

80. Let  = 1 + ^34. Then ^34=  − 1,  = ³4^−14, and ^−14 = ⁴₃. When  = 1,  = 2; when  = 16,  = 9.

Thus,

 16 1

^12

1 + ^34  =

 16 1

^34· ^−14 1 + ^34  =

 9 2

 − 1



4 3



= 4 3

 9 2

 1 −1





 = 4 3

 − ln ||9 2

= ⁴₃[(9 − ln 9) − (2 − ln 2)] = ⁴3(7 − ln 9 + ln 2) = ⁴3

7 + ln²₉

° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

83. Evaluate R2

−2(x + 3)√

4 − x²dx by writing it as a sum of two integrals and interpreting one of those integrals in terms of an area.

Solution:

SECTION 5.5 THE SUBSTITUTION RULE ¤ 531 73. Let  = 1 + √, so  = 1

2√

 ⇒ 2√

  =  ⇒ 2( − 1)  = . When  = 0,  = 1; when  = 1,

 = 2. Thus,

 1 0



(1 +√

 )⁴ =

 2 1

1

⁴ · [2( − 1) ] = 2

 2 1

 1

³ − 1

⁴



 = 2



− 1 2² + 1

3³

2 1

= 2

−¹8+ ₂₄¹

−

−¹2+¹₃

= 2₁

12

= ¹₆

74. If () = sin√³

, then (−) = sin√³

− = sin(−√³

 ) = − sin√³

 = −(), so  is an odd function. Now

 =3

−2sin√³

  =2

−2sin√³

  +3 2 sin√³

  = 1+ 2. 1= 0by Theorem 7(b). To estimate 2, note that 2 ≤  ≤ 3 ⇒ √³

2 ≤√³

 ≤√³

3 [≈ 144] ⇒ 0 ≤√³

 ≤ ^2 [≈ 157] ⇒ sin 0 ≤ sin√³

 ≤ sin^2 [since sine is increasing on this interval] ⇒ 0 ≤ sin√³

 ≤ 1. By Property 5.2.8, 0(3 − 2) ≤ ²≤ 1(3 − 2) ⇒ 0 ≤ ²≤ 1 ⇒ 0 ≤  ≤ 1.

75. From the graph, it appears that the area under the curve is about

1 +a little more than ¹₂· 1 · 07, or about 14. The exact area is given by

 =1 0

√2 + 1 . Let  = 2 + 1, so  = 2 . The limits change to

2 · 0 + 1 = 1 and 2 · 1 + 1 = 3, and

 =3 1

√1 2

= ¹₂

2 3^323

1= ¹₃ 3√

3 − 1

=√

3 −¹3 ≈ 1399.

76. From the graph, it appears that the area under the curve is almost ¹₂·  · 26, or about 4. The exact area is given by

 =

0(2 sin  − sin 2)  = −2 cos 

0 −

0 sin 2 

= −2(−1 − 1) − 0 = 4

Note: 

0 sin 2  = 0since it is clear from the graph of  = sin 2 that

2sin 2  = −2

0 sin 2 .

77. First write the integral as a sum of two integrals:

 =2

−2( + 3)√

4 − ² = 1+ 2=2

−2√

4 − ² +2

−23√

4 − ². 1 = 0by Theorem 7(b), since

 () = √

4 − ²is an odd function and we are integrating from  = −2 to  = 2. We interpret ²as three times the area of a semicircle with radius 2, so  = 0 + 3 · ¹2

 · 2²= 6.

78. Let  = ². Then  = 2  and the limits are unchanged (0² = 0and 1² = 1), so

 =1 0 √

1 − ⁴ =¹₂1 0

√1 − ². But this integral can be interpreted as the area of a quarter-circle with radius 1.

So  = ¹₂· ¹4

 · 1²

= ¹₈.

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

94. If f is continuous and R9

0 f (x)dx = 4, findR3

0 xf (x²)dx.

Solution:

SECTION 5.5 THE SUBSTITUTION RULE ¤ 533

85.  30 0

()  =

 30 0



 0^{−}  = 0

 ^{−30}

1 (−)

  = ^{−}

 =−^^{−}



= 0



−^{−30}

1 = 0(−^{−30} + 1) The integral30

0 () represents the total amount of urea removed from the blood in the first 30 minutes of dialysis.

86. Number of calculators = (4) − (2) =4 2 5000

1 − 100( + 10)⁻²



= 5000

 + 100( + 10)⁻¹4

2= 5000

4 +¹⁰⁰₁₄

−

2 +¹⁰⁰₁₂

≈ 4048 87. Let  = 2. Then  = 2 , so2

0  (2)  =4 0  ()₁

2

= ¹₂4

0  ()  = ¹₂(10) = 5.

88. Let  = ². Then  = 2 , so3

0  (²)  =9 0  ()1

2

= ¹₂9

0  ()  = ¹₂(4) = 2.

89. Let  = −. Then  = −, so



 (−)  =_−

− ()(−) =_−

−  ()  =_−

−  () 

From the diagram, we see that the equality follows from the fact that we are reflecting the graph of , and the limits of integration, about the -axis.

90. Let  =  + . Then  = , so



 ( + )  =+

+ ()  =+

+  () 

From the diagram, we see that the equality follows from the fact that we are translating the graph of , and the limits of integration, by a distance .

91. Let  = 1 − . Then  = 1 −  and  = −, so

1

0 ^(1 − )^ =0

1 (1 − )^^(−) =1

0 ^(1 − )^ =1

0 ^(1 − )^.

92. Let  =  − . Then  = −. When  = ,  = 0 and when  = 0,  = . So



0  (sin )  = −0

( − ) (sin( − ))  =

0( − ) (sin ) 

= 

0  (sin )  −

0   (sin )  = 

0  (sin )  −

0   (sin )  ⇒ 2

0   (sin )  = 

0  (sin )  ⇒ 

0   (sin )  = ^₂

0  (sin ) .

93.  sin 

1 + cos² =  · sin 

2 − sin² =   (sin ), where () = 

2 − ². By Exercise 92,

  0

 sin  1 + cos² =

  0

  (sin )  =  2

  0

 (sin )  = 2

  0

sin  1 + cos²

Let  = cos . Then  = − sin  . When  = ,  = −1 and when  = 0,  = 1. So

 2

  0

sin 

1 + cos² = − 2

 ₋₁

1



1 + ² =  2

 1

−1



1 + ² =  2

tan⁻¹¹

−1

= 

2[tan⁻¹1 − tan⁻¹(−1)] = 2

  4−

− 4



= ² 4

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

98. If f is continuous on [0, π], use the substitution u = π − x to show that Z π

0

xf (sin x)dx = π 2

Z π 0

f (sin x)dx Solution:

SECTION 5.5 THE SUBSTITUTION RULE ¤ 533 85.

 30 0

()  =

 30 0



 0^{−} = 0

 ^{−30}

1 (−)

  = ^{−}

 =−^^{−}



= 0

−^{−30}

1 = 0(−^{−30} + 1) The integral30

0 () represents the total amount of urea removed from the blood in the first 30 minutes of dialysis.

86. Number of calculators = (4) − (2) =4 2 5000

1 − 100( + 10)⁻²



= 5000

 + 100( + 10)⁻¹4

2= 5000

4 +¹⁰⁰₁₄

−

2 +¹⁰⁰₁₂

≈ 4048 87. Let  = 2. Then  = 2 , so2

0  (2)  =4 0  ()₁

2

= ¹₂4

0  ()  = ¹₂(10) = 5.

88. Let  = ². Then  = 2 , so3

0  (²)  =9 0  ()1

2

= ¹₂9

0  ()  = ¹₂(4) = 2.

89. Let  = −. Then  = −, so



 (−)  =_−

−  ()(−) =_−

−  ()  =_−

−  () 

From the diagram, we see that the equality follows from the fact that we are reflecting the graph of , and the limits of integration, about the -axis.

90. Let  =  + . Then  = , so



 ( + )  =+

+  ()  =+

+ () 

From the diagram, we see that the equality follows from the fact that we are translating the graph of , and the limits of integration, by a distance .

91. Let  = 1 − . Then  = 1 −  and  = −, so

1

0 ^(1 − )^ =0

1 (1 − )^^(−) =1

0 ^(1 − )^ =1

0 ^(1 − )^.

92. Let  =  − . Then  = −. When  = ,  = 0 and when  = 0,  = . So



0  (sin )  = −0

( − ) (sin( − ))  =

0( − ) (sin ) 

= 

0  (sin )  −

0   (sin )  = 

0  (sin )  −

0   (sin )  ⇒ 2

0   (sin )  = 

0  (sin )  ⇒ 

0   (sin )  = ^₂ 

0  (sin ) .

93.  sin 

1 + cos² =  · sin 

2 − sin²=   (sin ), where () = 

2 − ². By Exercise 92,

  0

 sin  1 + cos² =

  0

  (sin )  =  2

  0

 (sin )  =  2

  0

sin  1 + cos²

Let  = cos . Then  = − sin  . When  = ,  = −1 and when  = 0,  = 1. So

 2

  0

sin 

1 + cos² = − 2

 ₋₁

1



1 + ² =  2

 1

−1



1 + ² =  2

tan⁻¹¹

−1

= 

2[tan⁻¹1 − tan⁻¹(−1)] =  2

  4 −

− 4



= ² 4

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

99. Use Exercise 98 to evaluate the integralRπ 0

x sin x 1+cos²xdx.

Solution:

SECTION 5.5 THE SUBSTITUTION RULE ¤ 533

85.  30 0

()  =

 30 0



 0^{−}  = 0

 ^{−30}

1 (−)

  = ^{−}

 =−^^{−}



= 0



−^{−30}

1 = 0(−^{−30} + 1) The integral30

0 () represents the total amount of urea removed from the blood in the first 30 minutes of dialysis.

86. Number of calculators = (4) − (2) =4 2 5000

1 − 100( + 10)⁻²



= 5000

 + 100( + 10)⁻¹⁴

2= 5000

4 +¹⁰⁰₁₄

−

2 +¹⁰⁰₁₂

≈ 4048 87. Let  = 2. Then  = 2 , so2

0  (2)  =4 0  ()1

2

= ¹₂4

0  ()  = ¹₂(10) = 5.

88. Let  = ². Then  = 2 , so3

0  (²)  =9 0  ()1

2

= ¹₂9

0  ()  = ¹₂(4) = 2.

89. Let  = −. Then  = −, so



 (−)  =_−

− ()(−) =_−

−  ()  =_−

−  () 

From the diagram, we see that the equality follows from the fact that we are reflecting the graph of , and the limits of integration, about the -axis.

90. Let  =  + . Then  = , so



 ( + )  =+

+ ()  =+

+  () 

From the diagram, we see that the equality follows from the fact that we are translating the graph of , and the limits of integration, by a distance .

91. Let  = 1 − . Then  = 1 −  and  = −, so

1

0 ^(1 − )^ =0

1 (1 − )^^(−) =1

0 ^(1 − )^ =1

0 ^(1 − )^.

92. Let  =  − . Then  = −. When  = ,  = 0 and when  = 0,  = . So



0  (sin )  = −0

( − ) (sin( − ))  =

0( − ) (sin ) 

= 

0  (sin )  −

0   (sin )  = 

0  (sin )  −

0   (sin )  ⇒ 2

0   (sin )  = 

0  (sin )  ⇒ 

0   (sin )  = ^₂

0  (sin ) .

93.  sin 

1 + cos² =  · sin 

2 − sin² =   (sin ), where () = 

2 − ². By Exercise 92,

  0

 sin  1 + cos² =

  0

  (sin )  =  2

  0

 (sin )  = 2

  0

sin  1 + cos²

Let  = cos . Then  = − sin  . When  = ,  = −1 and when  = 0,  = 1. So

 2

  0

sin 

1 + cos² = − 2

 ₋₁

1



1 + ² =  2

 1

−1



1 + ² =  2

tan⁻¹¹

−1

= 

2[tan⁻¹1 − tan⁻¹(−1)] = 2

  4−

− 4

= ² 4

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