Explanation Formula: The expression for the average value of f on the interval f ave [ a , b ] is, f ave [ a , b ] = 1 b − a ∫ a b f ( x ) d x (1) Given information: The average value of the function, a < c < b is the average value of f on interval [ a , b ] and f ave [ a , b ] = 1 b − a ∫ a b f ( x ) d x . Calculation: Since a < c < b , modify integral limit a to b by the addition of integral limits a to c and c to b . Modify equation (1). f ave [ a , b ] = 1 b − a ∫ a c f ( x ) d x + 1 b − a ∫ c b f ( x ) d x Multiply and divide by terms ( c − a ) and ( b − c ) . f ave [ a , b ] = c − a b − a [ 1 c − a ∫ a c f ( x ) d x ] + b − c b − a [ 1 b − c ∫ c b f ( x ) d x ] (2) Express the function f ave [ a , c ] by the use of equation (1)

Bundle: Single Variable Calculus: Early Transcendentals, Loose-leaf Version, 8th + Webassign Printed Access Card For Calculus, Multi-term Courses, Life Of Edition

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ISBN: 9780357008034

Author: Stewart

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Chapter 6.5, Problem 26E

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To show: The expression fave[a,b]=(c−ab−a)fave[a,c]+(b−cb−c)fave[c,b].

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Chapter 6.5, Problem 25E

Chapter 6, Problem 1RCC

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Bundle: Single Variable Calculus: Early Transcendentals, Loose-leaf Version, 8th + Webassign Printed Access Card For Calculus, Multi-term Courses, Life Of Edition

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The expression f ave [ a , b ] = ( c − a b − a ) f ave [ a , c ] + ( b − c b − c ) f ave [ c , b ] .

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To show: The expression fave[a,b]=(c−ab−a)fave[a,c]+(b−cb−c)fave[c,b].

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Chapter 6 Solutions