Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 400: 48

Answer

$= \frac{4}{3} $

Work Step by Step

$\int^{2}_0 2x-x^{2} dx$ $= 2\frac{x^{2}}{2} - \frac{x^{3}}{3} |^{2}_0$ $= x^{2} - \frac{x^{3}}{3} |^{2}_0$ $= [2^{2} - \frac{2^{3}}{3}] - [0^{2} - \frac{0^{3}}{3}]$ $= [4 - \frac{8}{3}] - 0$ $= [4 - \frac{8}{3}] $ $= \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.