Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.6 - Integration Using Tables and Computer Algebra Systems - 7.6 Exercises - Page 513: 26


$24-\displaystyle \frac{65}{e}\approx 0.0878363238562$

Work Step by Step

Table of integrals: $\color{blue}{ 96. \quad \displaystyle \int ue^{au}du=\frac{1}{a^{2}}(au-1)e^{au}+C \\\\ 97. \quad \displaystyle \int u^{n}e^{au}du=\frac{1}{a}u^{n}e^{au}-\frac{n}{a}\int u^{n-1}e^{au}du}$ --- $\displaystyle \int x^{4}e^{-x}dx= \quad$ ... apply (97)$, a=-1,n=4$ $=-x^{4}e^{-x}+4\displaystyle \int x^{3}e^{-x}dx\quad$ ... apply (97)$, a=-1,n=3$ $=-x^{4}e^{-x}+4(-x^{3}e^{-x}+3\displaystyle \int x^{2}e^{-x}dx)\quad$ ... apply (97)$, a=-1,n=2$ $=-(x^{4}+4x^{3})e^{-x}+12(-x^{2}e^{-x}+2\displaystyle \int xe^{-x}dx)\quad$ ... apply (96)$, a=-1$ $=-(x^{4}+4x^{3}+12x^{2})e^{-x}+24[(-x-1)e^{-x}]+C$ $=-(x^{4}+4x^{3}+12x^{2}+24x+24)e^{-x}+C$ $\displaystyle \int_{0}^{1}x^{4}e^{-x}dx=-(x^{4}+4x^{3}+12x^{2}+24x+24)e^{-x}|_{0}^{1}$ $=-(1+4+12+24+24)\displaystyle \frac{1}{e}+(24)$ $=24-\displaystyle \frac{65}{e}\approx 0.0878363238562$
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