University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 27


$$\int r^2\Big(\frac{r^3}{18}-1\Big)^5dr=\Big(\frac{r^3}{18}-1\Big)^6+C$$

Work Step by Step

$$A=\int r^2\Big(\frac{r^3}{18}-1\Big)^5dr$$ We set $u=\frac{r^3}{18}-1$ Then $$du=\frac{3r^2}{18}dr=\frac{r^2}{6}dr$$ That means $$r^2dr=6du$$ Therefore, $$A=6\int u^5du=\frac{6u^6}{6}+C=u^6+C$$ $$A=\Big(\frac{r^3}{18}-1\Big)^6+C$$
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