Answer
Diverges.
Work Step by Step
From the Definition of Improper Integrals with Infinite Integration Limits
Case 2.
If $f$ is continuous on the interval $(-\infty, b]$, then
$\displaystyle \int_{-\infty}^{b}f(x)dx=\lim_{a\rightarrow-\infty}\int_{a}^{b}f(x)dx$.
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$I=\displaystyle \int_{-\infty}^{0}xe^{-4x}dx=\lim_{b\rightarrow-\infty}\int_{b}^{0}xe^{-4x}dx$
...Integration by parts
$\displaystyle \int xe^{-4x}dx=\int udv\quad\left[\begin{array}{ll}
u=x & dv=e^{-4x}\\
du=dx & v=-\frac{1}{4}e^{-4x}
\end{array}\right]=uv-\displaystyle \int vdu$
$=-\displaystyle \frac{xe^{-4x}}{4}+\frac{1}{4}\int e^{-4x}dx$
$=-\displaystyle \frac{xe^{-4x}}{4}+\frac{1}{-16}e^{-4x}+C$
$I=\displaystyle \lim_{b\rightarrow-\infty}[\frac{-xe^{-4x}}{4}-\frac{1}{16}e^{-4x}]_{b}^{0}$
$=\displaystyle \lim_{b\rightarrow-\infty}[-\frac{1}{16}-(-\frac{b}{4}-\frac{1}{16}e^{-4b})]$
$=-\infty$
Diverges.
