Answer
$$\frac{-1}{2}$$
Work Step by Step
\begin{aligned}
\int_{-\infty}^{0} x e^{-x^{2}} d x &=\lim _{R \rightarrow-\infty} \int_{R}^{0} x e^{-x^{2}} d x \\
&=\lim _{R \rightarrow-\infty} -\frac{1}{2} \int_{R}^{0}(-2 x) e^{-x^{2}} d x \\ &=\lim _{R \rightarrow-\infty}-\frac{1}{2} e^{-(x)^{2}} \bigg|_{R}^{0} \\ &=\lim _{R \rightarrow-\infty}\frac{1}{2} e^{-R^{2}}-\frac{1}{2}\\
&= \frac{-1}{2} \end{aligned}
Converges.
