Calculus: Early Transcendentals (2nd Edition)

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

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 103

Answer

$$\frac{{{e^{2x}} - {e^{ - 2x}}}}{2}$$

Work Step by Step

$$\eqalign{ & \int {\frac{{{e^{2x}} - {e^{ - 2x}}}}{2}dx} \cr & {\text{split the numerator}} \cr & = \int {\left( {\frac{{{e^{2x}}}}{2} - \frac{{{e^{ - 2x}}}}{2}} \right)dx} \cr & = \int {\frac{{{e^{2x}}}}{2}dx} - \int {\frac{{{e^{ - 2x}}}}{2}dx} \cr & = \frac{1}{2}\int {{e^{2x}}dx} - \frac{1}{2}\int {{e^{ - 2x}}dx} \cr & \operatorname{integrate} \cr & = \frac{1}{2}\left( {\frac{1}{2}{e^{2x}}} \right) - \frac{1}{2}\left( { - \frac{1}{2}{e^{ - 2x}}} \right) + C \cr & {\text{simplify}} \cr & = \frac{1}{4}{e^{2x}} + \frac{1}{4}{e^{ - 2x}} + C \cr & \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dx}}\left( {\frac{1}{4}{e^{2x}} + \frac{1}{4}{e^{ - 2x}} + C} \right) \cr & {\text{ = }}\frac{1}{4}{e^{2x}}\left( 2 \right) + \frac{1}{4}{e^{ - 2x}}\left( 2 \right) + 0 \cr & {\text{ = }}\frac{1}{2}{e^{2x}} - \frac{1}{2}{e^{ - 2x}} \cr & = \frac{{{e^{2x}} - {e^{ - 2x}}}}{2} \cr} $$
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