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: 8


$$\int^{1}_{0}\frac{10\sqrt v}{(1+v^{3/2})^2}dv=\frac{10}{3}$$ $$\int^{4}_{1}\frac{10\sqrt v}{(1+v^{3/2})^2}dv=\frac{70}{27}$$

Work Step by Step

Part a) $$A=\int^{1}_{0}\frac{10\sqrt v}{(1+v^{3/2})^2}dv$$ Set $u=1+v^{3/2}$, we then have $$du=\frac{3}{2}v^{1/2}dv=\frac{3}{2}\sqrt vdv$$ $$\sqrt vdv=\frac{2}{3}du$$ For $v=0$, we have $u=1$ and for $v=1$, we have $u=2$. Therefore, $$A=\frac{2}{3}\int^{2}_1\frac{10}{u^2}du$$ $$A=\frac{20}{3}\Big(-\frac{1}{u}\Big)\Big]^2_1$$ $$A=-\frac{20}{3}\Big(\frac{1}{2}-1\Big)$$ $$A=-\frac{20}{3}\Big(-\frac{1}{2}\Big)=\frac{10}{3}$$ Part b) $$A=\int^{4}_{1}\frac{10\sqrt v}{(1+v^{3/2})^2}dv$$ We carry out a similar substitution as in part a). For $v=4$, we have $u=9$ and for $v=1$, we have $u=2$. Therefore, $$A=\frac{2}{3}\int^{9}_2\frac{10}{u^2}du=\frac{20}{3}\Big(-\frac{1}{u}\Big)\Big]^9_2$$ $$A=-\frac{20}{3}\Big(\frac{1}{9}-\frac{1}{2}\Big)$$ $$A=-\frac{20}{3}\Big(-\frac{7}{18}\Big)=\frac{70}{27}$$
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