Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 7 - Principles Of Integral Evaluation - 7.2 Integration By Parts - Exercises Set 7.2 - Page 498: 4

Answer

$$\frac{{{x^2}}}{2}{e^{ - 2x}} - \frac{x}{2}{e^{ - 2x}} + \frac{1}{4}{e^{ - 2x}} + C$$

Work Step by Step

$$\eqalign{ & \int {{x^2}{e^{ - 2x}}} dx \cr & {\text{substitute }}u = {x^2},{\text{ }}du = 2xdx \cr & dv = {e^{ - 2x}},{\text{ }}v = - \frac{1}{2}{e^{ - 2x}} \cr & {\text{ integration by parts}}{\text{, we have}} \cr & \int {{x^2}{e^{ - 2x}}} dx = - \frac{{{x^2}}}{2}{e^{ - 2x}} - \int {\left( { - \frac{1}{2}{e^{ - 2x}}} \right)\left( {2xdx} \right)} \cr & \int {{x^2}{e^{ - 2x}}} dx = - \frac{{{x^2}}}{2}{e^{ - 2x}} + \int {x{e^{ - 2x}}dx} \cr & {\text{substitute }}u = x,{\text{ }}du = dx \cr & dv = {e^{ - 2x}},{\text{ }}v = - \frac{1}{2}{e^{ - 2x}} \cr & {\text{ integration by parts}}{\text{, we have}} \cr & \int {{x^2}{e^{ - 2x}}} dx = - \frac{{{x^2}}}{2}{e^{ - 2x}} - \frac{x}{2}{e^{ - 2x}} + \int {\left( { - \frac{1}{2}{e^{ - 2x}}} \right)} dx \cr & \int {{x^2}{e^{ - 2x}}} dx = - \frac{{{x^2}}}{2}{e^{ - 2x}} - \frac{x}{2}{e^{ - 2x}} - \frac{1}{2}\int {{e^{ - 2x}}} dx \cr & {\text{integrating}} \cr & \int {{x^2}{e^{ - 2x}}} dx = - \frac{{{x^2}}}{2}{e^{ - 2x}} - \frac{x}{2}{e^{ - 2x}} - \frac{1}{2}\left( { - \frac{1}{2}{e^{ - 2x}}} \right) + C \cr & \int {{x^2}{e^{ - 2x}}} dx = - \frac{{{x^2}}}{2}{e^{ - 2x}} - \frac{x}{2}{e^{ - 2x}} + \frac{1}{4}{e^{ - 2x}} + C \cr} $$
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