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: 3

Answer

$${x^2}{e^x} - 2x{e^x} + 2{e^x} + C$$

Work Step by Step

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