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: 111



Work Step by Step

$$A=\int^{2/3}_{\sqrt2/3}\frac{dy}{|y|\sqrt{9y^2-1}}=\int^{2/3}_{\sqrt2/3}\frac{dy}{|y|\sqrt{(3y)^2-1^2}}$$ Since here we consider $y\in[\sqrt2/3,2/3]$, we know that $y\gt0$, so $|y|=y$ Set $u=3y$, which means $$du=3dy$$ $$dy=\frac{1}{3}du$$ Also, we would have $y=1/3u$ For $y=2/3$, we have $u=2$ and for $y=\sqrt2/3$, we have $u=\sqrt2$ Therefore, $$A=\int^{2}_{\sqrt2}\frac{\frac{1}{3}du}{\frac{1}{3}u\sqrt{u^2-1^2}}=\int^{2}_{\sqrt2}\frac{du}{u\sqrt{u^2-1^2}}$$ We have $$\int\frac{dx}{x\sqrt{x^2-a^2}}=\frac{1}{a}\sec^{-1}\Big|\Big(\frac{x}{a}\Big)\Big|+C$$ Therefore, $$A=\frac{1}{1}\sec^{-1}|u|\Big]^2_{\sqrt2}$$ $$A=\sec^{-1}2-\sec^{-1}\sqrt2$$ $$A=\frac{\pi}{3}-\frac{\pi}{4}$$ $$A=\frac{\pi}{12}$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.