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: 28



Work Step by Step

$$A=\int^0_{-2}5^{-\theta}d\theta$$ We set $-\theta=u$, which means $$-d\theta=du$$ $$d\theta=-du$$ - For $\theta=-2$, we have $u=2$ and for $\theta=0$, we have $u=0$ Therefore, $$A=-\int^{0}_{2}5^udu$$ $$A=-\frac{5^u}{\ln5}\Big]^{0}_{2}$$ $$A=-\frac{1}{\ln5}(5^{0}-5^2)$$ $$A=-\frac{1-25}{\ln5}$$ $$A=\frac{24}{\ln5}$$
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