Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - 6.2 Setting Up Integrals: Volume, Density, Average Value - Exercises - Page 298: 46



Work Step by Step

The average is given as follows $$ \frac{1}{3-0} \int_{0}^{3}\frac{x}{(x^2+16)^{3/2}} d x=\frac{1}{6}\int_{0}^{3}2x(x^2+16)^{-3/2} d x\\=-\left.\frac{1}{3}(x^2+16)^{-1/2}\right|_{0 } ^{3}=-\frac{1}{3}\left( \frac{1}{5}-\frac{1}{4}\right)=\frac{1}{60}. $$
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