Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 8 - Further Techniques and Applications of Integration - 8.4 Improper Integrals - 8.4 Exercises - Page 452: 26

Answer

$-25$

Work Step by Step

\[\begin{align} & \int_{-\infty }^{0}{x{{e}^{0.2x}}}dx \\ & \text{By the definition of improper integrals} \\ & \int_{-\infty }^{b}{f\left( x \right)}dx=\underset{a\to -\infty }{\mathop{\lim }}\,\int_{a}^{b}{f\left( x \right)}dx \\ & \text{We obtain,} \\ & \int_{-\infty }^{0}{x{{e}^{0.2x}}}dx=\underset{a\to -\infty }{\mathop{\lim }}\,\int_{a}^{0}{x{{e}^{0.2x}}}dx \\ & \text{Integrating }\int{x{{e}^{0.2x}}dx}\text{ by parts we obtain }5x{{e}^{0.2x}}-25{{e}^{0.2x}}+C \\ & \underset{a\to -\infty }{\mathop{\lim }}\,\int_{a}^{0}{x{{e}^{0.2x}}}dx=\underset{a\to -\infty }{\mathop{\lim }}\,\left[ 5x{{e}^{0.2x}}-25{{e}^{0.2x}} \right]_{a}^{0} \\ & =\underset{a\to -\infty }{\mathop{\lim }}\,\left[ \left( 5\left( 0 \right){{e}^{0}}-25{{e}^{0}} \right)-\left( 5a{{e}^{a}}-25{{e}^{a}} \right) \right] \\ & =\underset{a\to -\infty }{\mathop{\lim }}\,\left[ -25-5a{{e}^{a}}+25{{e}^{a}} \right] \\ & \text{Evaluate when }a\to -\infty \\ & =-25-5\left( -\infty \right){{e}^{-\infty }}+25{{e}^{-\infty }} \\ & =-25-0+0 \\ & =-25 \\ \end{align}\]
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