Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 409: 49



Work Step by Step

We can evaluate the integral to find the area of the region: $\int_{0}^{2}(2y-y^2)~dy$ $=(y^2-\frac{y^3}{3})~\vert_{0}^{2}$ $=(2^2-\frac{2^3}{3})-(0^2-\frac{0^3}{3})$ $=(4-\frac{8}{3})-(0)$ $= \frac{4}{3}$
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.