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: 34


$$\int^{2}_{1}\frac{2^{\ln x}}{x}dx=\frac{2^{\ln2}-1}{\ln2}$$

Work Step by Step

$$A=\int^{2}_{1}\frac{2^{\ln x}}{x}dx$$ We set $u=\ln x$, which means $$du=\frac{1}{x}dx$$ - For $x=2$, we have $u=\ln2$ - For $x=1$, we have $u=\ln1=0$ Therefore, $$A=\int^{\ln2}_02^udu=\frac{2^u}{\ln2}\Big]^{\ln2}_0$$ $$A=\frac{2^{\ln2}-1}{\ln2}$$
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