Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Section 15.3 - Area by Double Integration - Exercises 15.3 - Page 887: 15



Work Step by Step

$\int^{\pi/4}_{0}\int^{cos x}_{sinx}dydx$ =$\int^{\pi/4}_{0}(cosx-sinx)dx=[sinx+cosx]^{\pi/4}_0$ =$(\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2})-(0+1)$ =$\sqrt{2}-1$
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