Answer
$$\int^{e}_{1}x^{(\ln2)-1}dx=\frac{1}{\ln2}$$
Work Step by Step
$$A=\int^{e}_{1}x^{(\ln2)-1}dx$$ $$A=\frac{x^{\ln2}}{\ln2}\Big]^{e}_{1}$$ $$A=\frac{(e^{\ln2}-1^{\ln2})}{\ln2}$$ $$A=\frac{2-1}{\ln2}=\frac{1}{\ln2}$$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
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
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