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 $
$=[\frac{u^{3}}{3}]^{1}_{0}-2[\frac{u^{2}}{2}]^{1}_{0}$
$=(\frac{1^{3}}{3}-\frac{0^{3}}{3})-2(\frac{1^{2}}{2}-\frac{0^{2}}{2})=-\frac{2}{3}$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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