Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.2 The Definite Integral - Exercises - Page 245: 37

Answer

$-\frac{2}{3}$

Work Step by Step

$\int^{1}_{0}(u^{2}-2u)du=\int^{1}_{0}u^{2}du-2\int^{1}_{0}udu $ $=[\frac{u^{3}}{3}]^{1}_{0}-2[\frac{u^{2}}{2}]^{1}_{0}$ $=(\frac{1^{3}}{3}-\frac{0^{3}}{3})-2(\frac{1^{2}}{2}-\frac{0^{2}}{2})=-\frac{2}{3}$
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.