University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 343: 104



Work Step by Step

$$A=\int^{e}_1\frac{8\ln3\log_3\theta}{\theta}d\theta$$ We set $u=\log_3\theta$, which means $$du=\frac{1}{\theta\ln3}d\theta$$ $$\frac{1}{\theta}d\theta=\ln3du$$ For $\theta=1$, we have $u=\log_31=0$ For $\theta=e$, we have $u=\log_3e$ Therefore, $$A=\int^{log_3e}_{0}8\ln3u(\ln3du)=\int^{log_3e}_{0}8(\ln3)^2udu$$ $$A=\frac{8(\ln3)^2}{2}u^2\Big]^{log_3e}_{0}=4(\ln3)^2u^2\Big]^{log_3e}_{0}$$ $$A=4(\ln3)^2(\log_3e)^2$$ $$A=4\Big(\ln3\times\frac{1}{\ln3}\Big)^2$$ $$A=4\times1^2=4$$
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