Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.8 Average Value Of A Function And It's Applications - Exercises Set 4.8 - Page 335: 7

Answer

$${f_{{\text{ave}}}} = \frac{1}{{21}}$$

Work Step by Step

$$\eqalign{ & {\text{Let }}f\left( x \right) = \frac{x}{{{{\left( {5{x^2} + 1} \right)}^2}}}{\text{ on the interval }}\left[ {0,2} \right] \cr & {\text{The average value of the function is}} \cr & {f_{{\text{ave}}}} = \frac{1}{{b - a}}\int_a^b {f\left( x \right)dx} \cr & {\text{Then}} \cr & {f_{{\text{ave}}}} = \frac{1}{{2 - 0}}\int_0^2 {\frac{x}{{{{\left( {5{x^2} + 1} \right)}^2}}}dx} \cr & {f_{{\text{ave}}}} = \frac{1}{{2\left( {10} \right)}}\int_0^2 {{{\left( {5{x^2} + 1} \right)}^{ - 2}}\left( {10x} \right)dx} \cr & {f_{{\text{ave}}}} = \frac{1}{{20}}\left[ {\frac{{{{\left( {5{x^2} + 1} \right)}^{ - 1}}}}{{ - 1}}} \right]_0^2 \cr & {f_{{\text{ave}}}} = - \frac{1}{{20}}\left[ {\frac{1}{{5{x^2} + 1}}} \right]_0^2 \cr & {\text{Evaluating}} \cr & {f_{{\text{ave}}}} = - \frac{1}{{20}}\left[ {\frac{1}{{5{{\left( 2 \right)}^2} + 1}} - \frac{1}{{5{{\left( 0 \right)}^2} + 1}}} \right] \cr & {f_{{\text{ave}}}} = - \frac{1}{{20}}\left[ { - \frac{{20}}{{21}}} \right] \cr & {f_{{\text{ave}}}} = \frac{1}{{21}} \cr} $$
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