Calculus, 10th Edition (Anton)

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

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.3 Derivatives Of Inverse Functions; Derivatives And Integrals Involving Exponential Functions - Exercises Set 6.3 - Page 433: 66



Work Step by Step

Let $u=-\frac{x}{2}$. Then $du=-\frac{1}{2}dx$, so $dx=-2du$. $\int e^{-x/2}dx$ $=\int e^u*(-2)du$ $=-2e^u+C$ $=-2e^{-\frac{x}{2}}+C$
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