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: 17



Work Step by Step

$\int^{0}_{-1} \int^{1-x}_{-2x}dydx+\int^{2}_{0} \int^{1-x}_{-x/2}dydx$ =$\int^{0}_{-1}(1+x)dx+\int^{2}_{0}(1-\frac{x}{2})dx$ =$[x+\frac{x^2}{2}]^0_{-1}+[x-\frac{x^2}{4}]$ =$(-1+\frac{1}{2})+(2-1)=\frac{3}{2}$
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.