Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises: 31

Answer

$= 2e^{-1} - 6e^{-5}$

Work Step by Step

$\int^{5}_1 \frac{M}{e^{M}} dM$ $u = M$ $u' = 1$ $\frac{du}{dM} = 1$ $du = 1 dM$ $du = dM$ $dv = e^{-M}$ $v = -e^{-M}$ $uv - \int vdu$ $= (M)(-e^{-M}) - \int -e^{-M} dM$ $= (M)(-e^{-M}) + \int e^{-M} dM$ $= (M)(-e^{-M}) - e^{-M} |^{5}_1$ $= [(5)(-e^{-5}) - e^{-5}]-[(1)(-e^{-1}) - e^{-1}]$ $= [-5e^{-5} - e^{-5}] - [-2e^{-1}]$ $= -6e^{-5} + 2e^{-1}$ $= 2e^{-1} - 6e^{-5}$
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