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: 40



Work Step by Step

$$A=\int^e_1\frac{2\ln10\log_{10}x}{x}dx=\int^e_1\frac{2\ln10\times\frac{\ln x}{\ln10}}{x}dx$$ $$A=2\int^e_1\frac{\ln x}{x}dx$$ We set $u=\ln x$, which means $$du=\frac{1}{x}dx$$ - For $x=e$, we have $u=\ln e=1$ - for $x=1$, we have $u=\ln1=0$ Therefore, $$A=2\int^{1}_0 udu$$ $$A=2\times\frac{u^2}{2}\Big]^{1}_0=u^2\Big]^{1}_0$$ $$A=1^2-0^2=1$$
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