University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 14 - Section 14.1 - Double and Iterated Integrals over Rectangles - Exercises - Page 759: 4



Work Step by Step

Re-arrange the given integral as follows: $\int_{0}^{1}[x-\dfrac{x^3}{6}-\dfrac{y^2}{2}]_{0}^{1}]dy=\int_{0}^{1}[(\dfrac{5}{6} -\dfrac{y^2}{2} ) dy$ This implies that $\int_{0}^{1}[(\dfrac{5}{6} -\dfrac{y^2}{2} ) dy= (\dfrac{5y}{6}-\dfrac{y^3}{6})_{0}^{1}$ Thus, we have, $\dfrac{5}{6}-\dfrac{1}{6}=\dfrac{2}{3}$
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.