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: 127

Answer

$$A = - 2\left[ {{e^{ - 3/2}} - 1} \right]$$

Work Step by Step

$$\eqalign{ & {\text{From the graph we can see that the region is given by:}} \cr & A = \int_0^{\sqrt 6 } {x{e^{ - {x^2}/4}}} dx \cr & {\text{Integrate and evaluate}} \cr & A = - 2\int_0^{\sqrt 6 } {\left( { - \frac{1}{2}x} \right){e^{ - {x^2}/4}}} dx \cr & A = - 2\left[ {{e^{ - {x^2}/4}}} \right]_0^{\sqrt 6 } \cr & A = - 2\left[ {{e^{ - {{\left( {\sqrt 6 } \right)}^2}/4}} - {e^{ - {{\left( 0 \right)}^2}/4}}} \right] \cr & A = - 2\left[ {{e^{ - 3/2}} - {e^0}} \right] \cr & A = - 2\left[ {{e^{ - 3/2}} - 1} \right] \cr & A \approx 1.5537 \cr} $$
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