Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 418: 2


$$\int xe^{-x^2}dx=-\frac{e^{-x^2}}{2}+C$$

Work Step by Step

$$A=\int xe^{-x^2}dx$$ Let $u=-x^2$ Then $du=(-x^2)'dx=-2xdx$. So $xdx=-\frac{1}{2}du$. Substitute into $A$, we have $$A=\int e^u(-\frac{1}{2})du$$ $$A=-\frac{1}{2}\int e^udu$$ $$A=-\frac{1}{2}e^u+C$$ $$A=-\frac{e^{-x^2}}{2}+C$$
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