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: 16


$$\int^4_1\frac{dy}{2\sqrt y(1+\sqrt y)^2}=\frac{1}{6}$$

Work Step by Step

$$A=\int^4_1\frac{dy}{2\sqrt y(1+\sqrt y)^2}$$ We set $u=1+\sqrt y$, which means $$du=\frac{dy}{2\sqrt y}$$ For $y=1$, we have $u=2$ For $y=4$, we have $u=3$ Therefore, $$A=\int^3_2\frac{1}{u^2}du=\Big(-\frac{1}{u}\Big)\Big]^3_2$$ $$A=-\Big(\frac{1}{3}-\frac{1}{2}\Big)$$ $$A=\frac{1}{6}$$
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.