Explanation Formula used: Chain Rule: d y d x = d y d u ⋅ d u d x . Theorem: Substitution Rule for Indefinite Integrals. Let u = g ( x ) and g ′ ( x ) be continuous function on an interval and let f be continuous function on the corresponding range of g on that interval ∫ f ( g ( x ) ) g ′ ( x ) d x = ∫ f ( u ) d u . Explanation: The given integral is ∫ 0 1 x x + a p d x , a > 0 . The substitution function is, u = x + a p can be expressed as u p − a = x . If u = x + a p , then by the chain rule, u p = x + a p u p − 1 d u = d x Since, the integrand is changed in terms of the variable u , the limits of integration must also be expressed in terms of u . Use the substitution u = x + a p and find the corresponding limits of u as follows, Lower limit: if x = 0 then u = a p . Upper limit: if x = 1 then u = 1 + a p . Therefore, the given ∫ 0 1 x x + a p d x can be expressed and calculated as, ∫ 0 1 x x + a p d x = ∫ a p 1 + a p ( u p − a ) ⋅ u ⋅ p u p − 1 d u = p ∫ a p 1 + a p ( u 2 p − a u p ) d u = p ( u 2 p + 1 2 p + 1 − a u p + 1 p + 1 ) | a p 1 + a p = p ( ( ( 1 + a p ) 2 p + 1 2 p + 1 − a ( 1 + a p ) p + 1 p + 1 ) − ( ( a p ) 2 p + 1 2 p + 1 − a ( 1 + a p ) p + 1 p + 1 ) ) Can be simplified further as, ∫ 0 1 x x + a p d x = p ( 1 + a ) 2 1 + a p 2 p + 1 − a p ( 1 + a ) 1 + a p p + 1 − p a 2 a p 2 p + 1 + p a 2 a p p + 1 . Therefore the value of the definite integral is ∫ 0 1 x x + a p d x using the substitution u = x + a p is p ( 1 + a ) 2 1 + a p 2 p + 1 − a p ( 1 + a ) 1 + a p p + 1 − p a 2 a p 2 p + 1 + a 2 p a p p + 1 . The calculation of the ∫ 0 1 x x + a p d x , a > 0 using the substitution u = x + a is as follows, As u = x + a , by the chain rule, d u = d x

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 5.5, Problem 100E

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To find: The value of ∫01xx+ap dx,a>0 by using the substitutions u=x+ap and u=x+a .

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Chapter 5.5, Problem 99E

Chapter 5.5, Problem 101E

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Calculus: Early Transcendentals (2nd Edition)

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The value of ∫ 0 1 x x + a p d x , a > 0 by using the substitutions u = x + a p and u = x + a .

Solution Summary: The author explains how to find the value of displaystyle 'underset' by using the substitutions u=sqrt[p]x+adx and

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To find: The value of ∫01xx+ap dx,a>0 by using the substitutions u=x+ap and u=x+a .

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Chapter 5 Solutions

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The value of ∫ 0 1 x x + a p d x , a &gt; 0 by using the substitutions u = x + a p and u = x + a .

Solution Summary: The author explains how to find the value of displaystyle 'underset' by using the substitutions u=sqrt[p]x+adx and

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To find: The value of ∫01xx+ap dx,a>0 by using the substitutions u=x+ap and u=x+a .

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Chapter 5 Solutions

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The value of ∫ 0 1 x x + a p d x , a > 0 by using the substitutions u = x + a p and u = x + a .