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: 36



Work Step by Step

$$A=\int^{e}_{1}x^{(\ln2)-1}dx$$ $$A=\frac{x^{\ln2}}{\ln2}\Big]^{e}_{1}$$ $$A=\frac{(e^{\ln2}-1^{\ln2})}{\ln2}$$ $$A=\frac{2-1}{\ln2}=\frac{1}{\ln2}$$
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