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 - 8.2 Exercises: 12

Answer

$-\frac 5 2 xe^{-2x}-\frac 5 4 e^{-2x}+C$

Work Step by Step

$\int 5xe^{-2x}dx$ $u=5x, v=-\frac 1 2 e^{-2x}, du5, dv=e^{-2x}$ $-\frac 5 2 xe^{-2x}-\int 5xe^{-2x}dx$ $-\frac 5 2 xe^{-2x}-(-\frac 5 2)\int e^{-2x}dx$ $u=-2x, du=-2dx$ $-\frac 5 2 xe^{-2x}-\frac 5 4 e^{-2x}+C$
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