Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises: 9

Answer

$$\int(1-2x)^9dx=-\frac{(1-2x)^{10}}{20}+C$$

Work Step by Step

$$A=\int(1-2x)^9dx$$ Let $u=1-2x$ Then $du=(1-2x)'dx=-2dx$. So $dx=-\frac{1}{2}du$ Substitute into $A$, we have $$A=\int u^9(-\frac{1}{2})du$$ $$A=-\frac{1}{2}\int u^9du$$ $$A=-\frac{1}{2}\frac{u^{10}}{10}+C$$ $$A=-\frac{u^{10}}{20}+C$$ $$A=-\frac{(1-2x)^{10}}{20}+C$$
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