University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 337: 7


$$\int^{1}_{-1}\frac{5r}{(4+r^2)^2}dr=0$$ $$\int^{1}_{0}\frac{5r}{(4+r^2)^2}dr=\frac{1}{8}$$

Work Step by Step

Part a) $$A=\int^{1}_{-1}\frac{5r}{(4+r^2)^2}dr$$ Set $u=4+r^2$, we then have $$du=2rdr$$ $$rdr=\frac{1}{2}du$$ For $r=-1$, we have $u=5$ and for $r=1$, we have $u=5$. Therefore, $$A=\frac{1}{2}\int^{5}_5\frac{5}{u^2}du=0$$ Part b) $$A=\int^{1}_{0}\frac{5r}{(4+r^2)^2}dr$$ We carry out a similar substitution as in part a). For $r=0$, we have $u=4$ and for $r=1$, we have $u=5$. Therefore, $$A=\frac{1}{2}\int^{5}_4\frac{5}{u^2}du$$ $$A=\frac{5}{2}\Big(-\frac{1}{u}\Big)\Big]^5_4$$ $$A=-\frac{5}{2}\Big(\frac{1}{5}-\frac{1}{4}\Big)$$ $$A=-\frac{5}{2}\Big(-\frac{1}{20}\Big)=\frac{1}{8}$$
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