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: 42


$$\int^{10}_{1/10}\frac{\log_{10}(10x)}{x}dx=\ln 100$$

Work Step by Step

$$A=\int^{10}_{1/10}\frac{\log_{10}(10x)}{x}dx=\int^{10}_{1/10}\frac{\frac{\ln(10x)}{\ln10}}{x}dx$$ $$A=\frac{1}{\ln10}\int^{10}_{1/10}\frac{\ln10x}{x}dx$$ We set $u=\ln10x$, which means $$du=\frac{(10x)'}{10x}dx=\frac{10}{10x}dx=\frac{1}{x}dx$$ - For $x=10$, we have $u=\ln100$ - For $x=1/10$, we have $u=\ln1=0$ Therefore, $$A=\frac{1}{\ln10}\int^{\ln100}_0udu=\frac{1}{\ln10}\times\frac{u^2}{2}\Big]^{\ln100}_0$$ $$A=\frac{(\ln100)^2-0^2}{2\ln10}$$ $$A=\frac{(\ln100)^2}{2\ln10}=\ln 100$$
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