University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 30


$$\int^{4}_{1}\frac{2^{\sqrt x}}{\sqrt x}dx=\frac{4}{\ln2}$$

Work Step by Step

$$A=\int^{4}_{1}\frac{2^{\sqrt x}}{\sqrt x}dx$$ We set $\sqrt x=u$, which means $$\frac{1}{2\sqrt x}dx=du$$ $$\frac{1}{\sqrt x}dx=2du$$ - For $x=4$, we have $u=2$ and for $x=1$, we have $u=1$ Therefore, $$A=2\int^{2}_{1}2^udu$$ $$A=2\times\frac{2^u}{\ln2}\Big]^{2}_{1}$$ $$A=\frac{2}{\ln2}(2^2-2^1)$$ $$A=\frac{2\times2}{\ln2}=\frac{4}{\ln2}$$
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.