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: 28


$4.95 \times 10^{-5}.$

Work Step by Step

The total charge is given by $$ \int_{0}^{10} \rho(x) d x=10^{-4} \int_{0}^{10} \frac{x}{\left(x^{2}+1\right)^{2}} d x\\ =\left.10^{-4}\left(-\frac{1}{2}\left(x^{2}+1\right)^{-1}\right)\right|_{0} ^{10}\\ =5 \times 10^{-5}\left(1-\frac{1}{101}\right)=4.95 \times 10^{-5}. $$
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