Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 419: 16

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

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