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: 16


$$\int e^{-5r}dr=-\frac{e^{-5r}}{5}+C$$

Work Step by Step

$$A=\int e^{-5r}dr$$ Let $u=-5r$. Then $du=-5 dr$, so $dr=-\frac{1}{5}du$ Substitute into $A$, we have $$A=-\frac{1}{5}\int e^udu$$ $$A=-\frac{1}{5}e^u+C$$ $$A=-\frac{e^{-5r}}{5}+C$$
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