Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 355: 126

Answer

$$A = - \frac{1}{2}\left( {{e^{ - 6}} - {e^2}} \right)$$

Work Step by Step

$$\eqalign{ & {\text{From the graph we can see that the region is given by:}} \cr & A = \int_{ - 1}^3 {{e^{ - 2x}}} dx \cr & {\text{Integrate and evaluate}} \cr & A = \left[ { - \frac{1}{2}{e^{ - 2x}}} \right]_{ - 1}^3 \cr & A = - \frac{1}{2}\left[ {{e^{ - 2x}}} \right]_{ - 1}^3 \cr & A = - \frac{1}{2}\left[ {{e^{ - 2\left( 3 \right)}} - {e^{ - 2\left( { - 1} \right)}}} \right] \cr & A = - \frac{1}{2}\left( {{e^{ - 6}} - {e^2}} \right) \cr & A \approx 3.69328 \cr} $$
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