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: 27

Answer

$$\int^1_02^{-\theta}d\theta=\frac{1}{2\ln2}$$

Work Step by Step

$$A=\int^1_02^{-\theta}d\theta$$ We set $-\theta=u$, which means $$-d\theta=du$$ $$d\theta=-du$$ - For $\theta=1$, we have $u=-1$ and for $\theta=0$, we have $u=0$ Therefore, $$A=-\int^{-1}_02^udu$$ $$A=-\frac{2^u}{\ln2}\Big]^{-1}_0$$ $$A=-\frac{1}{\ln2}(2^{-1}-2^0)$$ $$A=-\frac{\frac{1}{2}-1}{\ln2}$$ $$A=\frac{\frac{1}{2}}{\ln2}=\frac{1}{2\ln2}$$
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