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: 100



Work Step by Step

$$A=\int^{\ln9}_0e^{\theta}(e^\theta-1)^{1/2}d\theta$$ We set $u=e^\theta-1$, which means $$du=e^\theta d\theta$$ For $\theta=\ln9$, we have $$u=e^{\ln9}-1=9-1=8$$ For $\theta=0$, we have $$u=e^0-1=1-1=0$$ Therefore, $$A=\int^8_0u^{1/2}du$$ $$A=\frac{2u^{3/2}}{3}\Big]^8_0$$ $$A=\frac{2}{3}(8^{3/2}-0^{3/2})$$ $$A=\frac{2\times16\sqrt2}{3}$$ $$A=\frac{32\sqrt2}{3}$$
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