Answer
$ln2+e^{-2}-e^{-1}\approx0.4606$
Work Step by Step
$\int_{1}^{2} (\frac{1}{x}-e^{-x}) dx=[ln|x|-(-e)^{-x}]_{1}^{2}=[ln|x|+e^{-x}]_{1}^{2}=[ln|2|+e^{-2}]-[ln|1|+e^{-1}]=ln2+e^{-2}-0-e^{-1}=ln2+e^{-2}-e^{-1}\approx0.4606$
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 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 30
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