Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - Review - Exercises - Page 349: 14



Work Step by Step

We use substitution to allow us to integrate the term $(1-x)^9$ as we cannot reasonably integrate the current term otherwise. Using the substitution $u = 1-x$ and $du = -dx$ we can rewrite the statement as $-\int_{-1}^{0}u^9du$, which evaluates to $-\frac{u^{10}}{10}\Big|_{-1}^0$ We now substitute $1-x$ into $u$ and get $-\frac{(1-x)^{10}}{10}\Big|_{0}^1$. Substituting $1 $ and $0$ into $x$ results in $-\frac{(1-1)^{10}}{10}$ - $(-\frac{(1)^{10}}{10}) = 0$ - ($-\frac{1}{10}) $ = $\frac{1}{10}$
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