Answer
$\ln 2+\frac{1}{e^2}-\frac{1}{e}$
Work Step by Step
$\int^{2}_{1}(\frac{1}{x}-e^{-x})dx$
=$(ln x+e^{-x})|^2_1$
=$(ln2+e^{-2})-(ln 1+e^{-1})$
=$\ln 2+\frac{1}{e^2}-\frac{1}{e}$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 286: 30
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