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: 12



Work Step by Step

$\int^{1}_{0} \int^{\sqrt{x}}_{-x}1dydx +\int^{4}_{0} \int^{\sqrt{x}}_{x-2} 1dydx $ =$\int^{1}_{0} [y]^{\sqrt{x}}_{-x}dx+\int^{4}_{0} [y]^{\sqrt{x}}_{x-2}dx$ =$\int^{1}_{0} (\sqrt{x}+x)dx+\int^{4}_{0} (\sqrt{x}-x+2)dx$ =$[\frac{2}{3}x^{3/2}+\frac{1}{2}x^2]^1_0+[\frac{2}{3}x^{3/2}-\frac{1}{2}x^2+2x]^4_1$ =$\frac{13}{3}$
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.