Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 387: 67


$$\int e^{9-2x}dx=-\frac{1}{2}\ e^{9-2x}+c.$$

Work Step by Step

Assume that $ u=9-2x $, then $ du=-2dx $ and hence $$\int e^{9-2x}dx=-\frac{1}{2}\int e^udu=-\frac{1}{2}\ e^u+c=-\frac{1}{2}\ e^{9-2x}+c.$$
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