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 - Page 476: 19

Answer

$z^{3}e^{z}-3z^{2}e^{z}+6ze^{z}-6e^{z}+C$

Work Step by Step

$\int$udv = uv - $\int$vdu $\int$$z^{3}e^{z}$ u=$z^{3}$ du=3$z^{2}$dz dv=$e^{z}$dz v=$e^{z}$ $\int$$z^{3}e^{z}$ = $z^{3}$$e^{z}$ - $\int$$e^{z}$3$z^{2}$dz = $z^{3}$$e^{z}$ - 3$\int$$e^{z}$$z^{2}$dz Integration by parts again: u=$z^{2}$ du=$2z$dz dv=$e^{z}$dz v=$e^{z}$ $z^{3}$$e^{z}$ - 3$\int$$e^{z}$$z^{2}$dz = $z^{3}$$e^{z}$ - 3($z^{2}$$e^{z}$ - $\int$$e^{z}$$2z$dz) = $z^{3}$$e^{z}$ - 3($z^{2}$$e^{z}$ - 2$\int$$e^{z}$$z$dz) Integration by parts one last time: u=$z$ du=dz dv=$e^{z}$dz v=$e^{z}$ $z^{3}$$e^{z}$ - 3($z^{2}$$e^{z}$ - 2$\int$$e^{z}$$z$dz) = $z^{3}$$e^{z}$ - 3($z^{2}$$e^{z}$ - 2($z$$e^{z}$-$\int$$e^{z}$dz)) = $z^{3}$$e^{z}$ - 3($z^{2}$$e^{z}$ - 2($z$$e^{z}$-$e^{z}$)) = $z^{3}$$e^{z}$ - 3$z^{2}$$e^{z}$ - 6($z$$e^{z}$-$e^{z}$) = $z^{3}e^{z}-3z^{2}e^{z}+6ze^{z}-6e^{z}+C$
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