Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 15 - Multiple Integrals - Review - Exercises - Page 1102: 23



Work Step by Step

$\int\int_{D} \frac{y}{1+x^{2}}dA=\int_{0}^{1}\int_{0} ^{\sqrt x} \frac{y}{1+x^{2}}dydx$ $=\int_{0}^{1}[\frac{y^{2}}{2}\cdot \frac{1}{1+x^{2}}]_{0} ^{\sqrt x\sqrt x}dx$ $=\frac {1}{4}\int_{0} ^{1}\frac{2x}{1+x^{2}}dx$ Put $u=1+x^{2}$ and $2xdx=du$ $=\frac{1}{4}\int_{1} ^{2}\frac{du}{u}$ $=\frac{1}{4}[lnu]_{1} ^{2}$ $=\frac{ln2}{4}$
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