Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 28

Answer

-$\frac{1}{2}$$(7-\frac{r^{5}}{10})^{4}$

Work Step by Step

u=7- $\frac{r^{5}}{10}$ =$\gt$ du= - $\frac{5r^{4}}{10}$dr =$\gt$ -2du=$r^{4}$dr substition : $\int$(7-$\frac{r^{5}}{10}$)$^{3}$$r^{4}$dr= $\int$$u^{3}$(-2du)=-2$\int$$u^{3}$du -2$\int$$4^{3}$=-2($\frac{u^{3+1}}{3+1}$)+C= -$\frac{1}{2}$$u^{4}$+C So, u=7-$\frac{r^{5}}{10}$ we get -$\frac{1}{2}$$(7-\frac{r^{5}}{10})^{4}$
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