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 15.1 - Double and Iterated Integrals over Rectangles - Exercises 15.1 - Page 874: 2



Work Step by Step

$\int^2_0\int^1_{-1}(x-y)dydx$ =$\int^2_0[xy-\frac{1}{2}y^2]^1_{-1}dx$ =$\int^2_02xdx$ =$[x^2]^2_0$ =$4$
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.