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: 81



Work Step by Step

$$A=\int^{1}_{1/8}x^{-1/3}(1-x^{2/3})^{3/2}dx$$ We set $a=1-x^{2/3}$, which means $$da=-\frac{2}{3}x^{-1/3}dx$$ $$x^{-1/3}dx=-\frac{3}{2}da$$ For $x=1/8$, $a=1-(1/8)^{2/3}=1-1/4=3/4$ and for $x=1$, $a=0$ Therefore, $$A=-\frac{3}{2}\int^0_{3/4}a^{3/2}da$$ $$A=\Big(-\frac{3}{2}\times\frac{a^{5/2}}{\frac{5}{2}}\Big)\Big]^0_{3/4}=-\frac{3}{5}a^{5/2}\Big]^0_{3/4}$$ $$A=-\frac{3}{5}\Big(0-\Big(\frac{3}{4}\Big)^{5/2}\Big)$$ $$A=-\frac{3}{5}\Big(-\frac{\sqrt{243}}{32}\Big)$$ $$A=\frac{3\sqrt{243}}{160}=\frac{27\sqrt3}{160}$$
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