Explanation Given: The function f ( x ) = x 2 − 2 x − 1 , and the interval [ 0 , 3 ] . Concept Used: The average value of an integrable functionf on the interval [ a , b ] , is given by 1 b − a ∫ a b f ( x ) d x . The area that is bounded by the graph of f ( x ) above the x-axis is positive. The area that is bounded by the graph of f ( x ) below the x-axis is negative. Power rule of integration: ∫ x n d x = x n + 1 n + 1 Calculation: Observe that the average value of an integrable function f on the interval [ a , b ] , is given by 1 b − a ∫ a b f ( x ) d x , so using this concept the exact average value of f ( x ) = x 2 − 2 x − 1 on the interval [ 0 , 3 ] is given by, f avg = 1 3 − 0 ∫ 0 3 ( x 2 − 2 x − 1 ) d x =

14th Edition

ISBN: 9781464151088

Author: Associate Professor Laura Taalman, Peter Kohn

Publisher: W.H. Freeman & Company

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Chapter 4.6, Problem 61S

To determine

The exact average value of f(x)=x2−2x−1 on the interval [0,3] using definite integrals and the Fundamental Theorem of Calculus.

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Chapter 4.6, Problem 60S

Chapter 4.6, Problem 62S

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The exact average value of f ( x ) = x 2 − 2 x − 1 on the interval [ 0 , 3 ] using definite integrals and the Fundamental Theorem of Calculus.

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The exact average value of f(x)=x2−2x−1 on the interval [0,3] using definite integrals and the Fundamental Theorem of Calculus.

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Chapter 4 Solutions