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: 41



Work Step by Step

$$A=\int^2_0\frac{\log_2(x+2)}{x+2}dx=\int^2_0\frac{\frac{\ln(x+2)}{\ln2}}{x+2}dx$$ $$A=\frac{1}{\ln2}\int^2_0\frac{\ln(x+2)}{x+2}dx$$ We set $u=\ln(x+2)$, which means $$du=\frac{(x+2)'}{x+2}dx=\frac{1}{x+2}dx$$ - For $x=2$, we have $u=\ln4$ - For $x=0$, we have $u=\ln2$ Therefore, $$A=\frac{1}{\ln2}\int^{\ln4}_{\ln2}udu=\frac{1}{\ln2}\times\frac{u^2}{2}\Big]^{\ln4}_{\ln2}$$ $$A=\frac{(\ln4)^2-(\ln2)^2}{2\ln2}$$ $$A=\frac{(\ln4-\ln2)(\ln4+\ln2)}{2\ln2}$$ $$A=\frac{(\ln\frac{4}{2})(\ln(4\times2))}{2\ln2}$$ $$A=\frac{\ln2\ln8}{2\ln2}=\frac{\ln8}{2}$$
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