University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 34


$[\frac{1}{\pi\cdot ln\pi}]\cdot(1-\frac{1}{\pi})\approx0.1896$

Work Step by Step

$\int_{-1}^{0} ({\pi^{x-1}})dx=[\frac{\pi^{x-1}}{ln\pi}]_{-1}^{0}=[\frac{\pi^{0-1}}{ln\pi}]-[\frac{\pi^{-1-1}}{ln\pi}]=[\frac{1}{\pi\cdot ln\pi}]-[\frac{1}{\pi^2\cdot ln\pi}]=[\frac{1}{\pi\cdot ln\pi}]\cdot(1-\frac{1}{\pi})\approx0.1896$
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