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: 56


$$\int^e_{1}\frac{\sqrt{\ln x}}{x}dx=\frac{2}{3}$$

Work Step by Step

$$A=\int^e_{1}\frac{\sqrt{\ln x}}{x}dx$$ We set $a=\ln x$, which means $$da=\frac{1}{x}dx$$ For $x=1$, $a=0$ and for $x=e$, $a=1$ Therefore, $$A=\int^{1}_{0} \sqrt ada$$ $$A=\int^{1}_{0} a^{1/2}da$$ $$A=\Big(\frac{a^{3/2}}{\frac{3}{2}}\Big)\Big]^{1}_0=\Big(\frac{2a^{3/2}}{3}\Big)\Big]^{1}_0$$ $$A=\frac{2\times1^{3/2}}{3}-\frac{2\times0^{3/2}}{3}$$ $$A=\frac{2}{3}-0=\frac{2}{3}$$
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.