Thomas' Calculus 13th Edition

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


$\ln 2+\frac{1}{e^2}-\frac{1}{e}$

Work Step by Step

$\int^{2}_{1}(\frac{1}{x}-e^{-x})dx$ =$(ln x+e^{-x})|^2_1$ =$(ln2+e^{-2})-(ln 1+e^{-1})$ =$\ln 2+\frac{1}{e^2}-\frac{1}{e}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.