Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 15 - Review - Exercises - Page 1062: 24



Work Step by Step

$\int\int_{D} \frac{1}{1+x^{2}}dA=\int_{0}^{1}\int_{x} ^{1} \frac{1}{1+x^{2}}dydx$ $=\int_{0}^{1} \frac{y}{1+x^{2}}]_{x} ^{1}dx$ $=\int_{0} ^{1}\frac{1-x}{1+x^{2}}dx$ $=\int_{0} ^{1}\frac{1}{1+x^{2}}dx-\int_{0} ^{1}\frac{x}{1+x^{2}}dx$ $=[tan^{-1}x]_{0} ^{1}-\frac{1}{2}\int_{0} ^{1}\frac{2x}{1+x^{2}}dx$ $=[tan^{-1}(1)-tan^{-1}(0)]-\frac{1}{2}[ln{(1+x^{2})}]_{0} ^{1}$ $=\frac{\pi}{4}-\frac{1}{2}ln2$
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