Calculus 8th Edition

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

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 347: 68


$$\int e^{-5r}dr=-\frac{1}{5}e^{-5r}+c$$

Work Step by Step

To evaluate the integral $$\int e^{-5r}dr$$ we will use substitution $-5r=t$ which gives us $-5dr=dt\Rightarrow dr=-\frac{dt}{5}$. Putting this into the integral we get: $$\int e^{-5r}dr=\int e^t \Big(-\frac{dt}{5}\Big)=-\frac{1}{5}\int e^tdt=-\frac{1}{5}e^t+c$$ where $c$ is arbitrary constant. Now we have to express solution in terms of $x$: $$\int e^{-5r}dr=-\frac{1}{5}e^t+c=-\frac{1}{5}e^{-5r}+c$$
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