Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 78


$$-\frac{1}{12} e^{9-12t}+c. $$

Work Step by Step

We have $$\int e^{9-12t} dt=e^9\int e^{-12t}dt=\frac{e^9}{-12} e^{-12t}+c\\ =-\frac{1}{12} e^{9-12t}+c. $$
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