Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 59

Answer

$$ - \frac{{{3^{ - 2x}}}}{{2\ln 3}} + C$$

Work Step by Step

$$\eqalign{ & \int {{3^{ - 2x}}} dx \cr & {\text{substitute }}u = - 2x,{\text{ }}du = - 2dx \cr & = - \frac{1}{2}\int {{3^u}} du \cr & {\text{find the antiderivative}} \cr & = - \frac{1}{2}\left( {\frac{{{3^u}}}{{\ln 3}}} \right) + C \cr & {\text{with }}u = - 2x \cr & = - \frac{1}{2}\left( {\frac{{{3^{ - 2x}}}}{{\ln 3}}} \right) + C \cr & = - \frac{{{3^{ - 2x}}}}{{2\ln 3}} + C \cr} $$
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