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