Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Review - Exercises - Page 422: 14



Work Step by Step

$\int^1_{0}(1-x)^9dx$ $=[-1\times\frac{1}{10}(1-x)^{10}] ^1_{0}$ $=[-\frac{1}{10}(1-x)^{10}] ^1_{0}$. Sub in the bounds and subtract the lower bound (0) from the upper bound (1): $=[-\frac{1}{10}(1-(1))^{10}]- [-\frac{1}{10}(1-(0))^{10}] $ $=0- (-\frac{1}{10}) $ $=0.1$.
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