Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises: 59



Work Step by Step

Let $u=-x^2$ Then $du=-2xdx$, and therefore $\frac{du}{-2x}=dx.$ Substituting this back into our integral we get $\int xe^{u}\frac{du}{-2x} = \int\frac{e^u}{-2}\hspace{0.5mm}du=-\frac{1}{2}\int e^u\hspace{0,5mm}du=-\frac{1}{2}\left(e^u+C\right)$ $=-\frac{1}{2}\left(e^{-x^2}+C\right)$ Plugging in our integration limits and ignoring our integration constant we get $-\frac{1}{2}e^{-x^2}\bigg\vert_0^1=-\frac{1}{2}e^{-1}-\left(-\frac{1}{2}e^0\right)=-\frac{1}{2}e^{-1}+\frac{1}{2}$ =$\frac{1}{2}-\frac{1}{2e}=\frac{e}{2e}-\frac{1}{2e}=\frac{e-1}{2e}$
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