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: 38


$$\int^4_1\frac{\log_{2}x}{x}dx=\ln 4$$

Work Step by Step

$$A=\int^4_1\frac{\log_{2}x}{x}dx=\int^4_1\frac{\frac{\ln x}{\ln2}}{x}dx$$ $$A=\frac{1}{\ln2}\int^4_1\frac{\ln x}{x}dx$$ We set $u=\ln x$, which means $$du=\frac{1}{x}dx$$ - For $x=4$, we have $u=\ln4$ - for $x=1$, we have $u=\ln1=0$ Therefore, $$A=\frac{1}{\ln2}\int^{\ln4}_0 udu$$ $$A=\frac{1}{\ln2}\times\frac{u^2}{2}\Big]^{\ln4}_0$$ $$A=\frac{1}{2\ln2}((\ln4)^2-0^2)$$ $$A=\frac{(\ln4)^2}{2\ln2}=\ln4$$
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