University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 343: 99



Work Step by Step

$$A=\int^{\ln5}_{0}e^r(3e^r+1)^{-3/2}dr$$ We set $u=3e^r+1$, which means $$du=3e^rdr$$ $$e^rdr=\frac{1}{3}du$$ For $r=\ln5$, $$u=3e^{\ln5}+1=3\times5+1=16$$ For $r=0$, $$u=3e^{0}+1=3\times1+1=4$$ Therefore, $$A=\frac{1}{3}\int^{16}_{4}u^{-3/2}du$$ $$A=\frac{1}{3}\times\frac{u^{-1/2}}{-\frac{1}{2}}\Big]^{16}_{4}$$ $$A=-\frac{2}{3\sqrt u}\Big]^{16}_{4}$$ $$A=-\Big(\frac{2}{3\sqrt{16}}-\frac{2}{3\sqrt4}\Big)$$ $$A=-\Big(\frac{1}{6}-\frac{1}{3}\Big)$$ $$A=\frac{1}{6}$$
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