Answer
$$-1$$
Work Step by Step
$$\int_{-\infty}^{0} \theta e^{\theta} d \theta$$
Let
\begin{align*}
u&=\theta \ \ \ \ \ \ dv= e^{\theta }d\theta\\
u&=d\theta \ \ \ \ \ \ v= e^{\theta }
\end{align*}
Then by parts, we get
\begin{align*}
\int_{-\infty}^{0} \theta e^{\theta} d \theta&=\lim _{b \rightarrow-\infty}\left[\theta e^{\theta}-e^{\theta}\right]_{b}^{0}\\
&=\lim _{b \rightarrow-\infty}\left[\left(0 \cdot e^{0}-e^{0}\right)-\left(b e^{b}-e^{b}\right)\right]\\
&=-1-\lim _{b \rightarrow-\infty}\left(\frac{b-1}{e^{-b}}\right)\\
&=-1-\lim _{b \rightarrow-\infty}\left(\frac{1}{-e^{-b}}\right)\\
&=-1
\end{align*}
