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 419: 60

Answer

$\frac{1}{2}-\frac{1}{2e}$

Work Step by Step

$\int_{0}^{1}x~e^{-x^2}~dx$ Let $u = -x^2$ $\frac{du}{dx} = -2x$ $dx = -\frac{du}{2x}$ When $x = 0$, then $u = 0$ When $x = 1$, then $u = -1$ $\int_{0}^{-1} x~e^u(-\frac{du}{2x})$ $= \int_{-1}^{0} \frac{1}{2}~e^u~du$ $= \frac{1}{2}~(e^u~\vert_{-1}^{0}~)$ $= \frac{1}{2}~(e^0-e^{-1})$ $= \frac{1}{2}~(1-\frac{1}{e})$ $= \frac{1}{2}-\frac{1}{2e}$
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