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 338: 46



Work Step by Step

$$A=\int^{3}_{0}\frac{ydy}{\sqrt{5y+1}}$$ We set $u=\sqrt{5y+1}$, which means $u^2=5y+1$ and $y=\frac{u^2-1}{5}$ $$du=\frac{(5y+1)'}{2\sqrt{5y+1}}dy=\frac{5}{2\sqrt{5y+1}}dy$$ $$\frac{dy}{\sqrt{5y+1}}=\frac{2}{5}du$$ - For $y=3$, we have $u=\sqrt{5\times3+1}=4$ - For $y=0$, we have $u=\sqrt{5\times0+1}=1$ Therefore, $$A=\frac{2}{5}\int^{4}_{1}\frac{u^2-1}{5}du=\frac{2}{25}\int^{4}_{1}(u^2-1)du$$ $$A=\frac{2}{25}\Big(\frac{u^3}{3}-u\Big)\Big]^{4}_{1}$$ $$A=\frac{2}{25}\Big((\frac{64}{3}-4)-(\frac{1}{3}-1)\Big)$$ $$A=\frac{2}{25}\Big(\frac{52}{3}-(-\frac{2}{3})\Big)$$ $$A=\frac{2}{25}\times\frac{54}{3}=\frac{36}{25}$$
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