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: 98



Work Step by Step

$$A=\int^{0}_{-\ln2}e^{2w}dw$$ We set $u=2w$, which means $$du=2dw$$ $$dw=\frac{1}{2}du$$ For $w=0$, $u=-(-1+1)=0$ and for $w=-\ln2$, $u=-2\ln2=\ln2^{-2}=\ln\frac{1}{4}$ Therefore, $$A=\frac{1}{2}\int^0_{\ln1/4}e^udu$$ $$A=\frac{1}{2}e^u\Big]^0_{\ln1/4}$$ $$A=\frac{1}{2}(e^0-e^{\ln1/4})$$ $$A=\frac{1}{2}(1-\frac{1}{4})$$ $$A=\frac{1}{2}\times\frac{3}{4}=\frac{3}{8}$$
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