Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises: 85



Work Step by Step

$\int_0^2\frac{dx}{e^{\pi x}}$ $=\int_0^2 e^{-\pi x}dx$ Let $u=-\pi x$. Then $du=-\pi dx$, and $\frac{1}{-\pi}du=dx$. $=\int_{-\pi*0}^{-\pi*2}e^u* \frac{1}{-\pi} du$ $=\frac{1}{-\pi}\int_0^{-2\pi} e^u du$ $=-\frac{1}{\pi}(e^{-2\pi}-e^0)$ $=-\frac{1}{\pi}(e^{-2\pi}-1)$ $=\frac{1}{\pi}(1-e^{-2\pi})$
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.