Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 16 - Multiple Integration - 16.1 Integration in Two Variables - Exercises - Page 847: 30

Answer

$$\frac{1}{2}-\frac{1}{\sqrt{2}}$$

Work Step by Step

\begin{align*} \int_{0}^{\pi / 4} \int_{\pi / 4}^{\pi / 2} \cos (2 x+y) d y d x&=\int_{0}^{\pi / 4} \sin (2 x+y) \bigg|_{\pi / 4}^{\pi / 2} d x\\ &= \int_{0}^{\pi / 4} \sin (2 x+\frac{\pi}{2})- \sin (2 x+\frac{\pi}{4})d x\\ &=\frac{-1}{2}\cos (2x+\pi/2)+\frac{1}{2}\cos (2x+\pi/4)\bigg|_{0}^{\pi/4}\\ &=\frac{1}{2}-\frac{1}{\sqrt{2}} \end{align*}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After 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.