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: 8



Work Step by Step

$\int^4_1\int^4_0(\frac{x}{2}+\sqrt y)dxdy$ =$\int^4_1[\frac{1}{4}x^2+x\sqrt y]^4_0dy$ =$\int^4_1(4+4y^\frac{1}{2})dy$ =$[4y+\frac{8}{3}y^\frac{3}{2}]^4_1$ =$\frac{92}{3}$
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.