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: 80



Work Step by Step

$$A=\int^{1}_0\frac{dr}{\sqrt[3]{(7-5r)^2}}=\int^{1}_0\frac{dr}{(7-5r)^{2/3}}$$ We set $a=7-5r$, which means $$da=-5dr$$ $$dr=-\frac{1}{5}da$$ For $r=0$, $a=7$ and for $r=1$, $a=2$ Therefore, $$A=-\frac{1}{5}\int^{2}_7\frac{da}{a^{2/3}}=-\frac{1}{5}\int^{2}_7a^{-2/3}da$$ $$A=\Big(-\frac{1}{5}\times\frac{a^{1/3}}{\frac{1}{3}}\Big)\Big]^2_7$$ $$A=\Big(-\frac{3}{5}\sqrt[3]a\Big)\Big]^2_7$$ $$A=-\frac{3}{5}(\sqrt[3]2-\sqrt[3]7)$$
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