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: 28

Answer

$$\int r^4\Big(7-\frac{r^5}{10}\Big)^3dr=-\frac{1}{2}\Big(7-\frac{r^5}{10}\Big)^4+C$$

Work Step by Step

$$A=\int r^4\Big(7-\frac{r^5}{10}\Big)^3dr$$ We set $u=7-\frac{r^5}{10}$ Then $$du=-\frac{5r^4}{10}dr=-\frac{r^4}{2}dr$$ That means $$r^4dr=-2du$$ Therefore, $$A=-2\int u^3du=-\frac{2u^4}{4}+C=-\frac{u^4}{2}+C$$ $$A=-\frac{1}{2}\Big(7-\frac{r^5}{10}\Big)^4+C$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.