Answer
$\sqrt{2}-1$
Work Step by Step
$\int^{\pi/4}_{0}\int^{cos x}_{sinx}dydx$
=$\int^{\pi/4}_{0}(cosx-sinx)dx=[sinx+cosx]^{\pi/4}_0$
=$(\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2})-(0+1)$
=$\sqrt{2}-1$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
