Answer
$-1$
Work Step by Step
Here, we have $\lim\limits_{a \to -\infty}\int_a^{0} \theta e^{\theta} d\theta=\lim\limits_{a \to -\infty}[ (\theta e^{\theta})_a^{0} -\int_a^{0} e^{\theta} d\theta$
This implies that
$\lim\limits_{a \to -\infty}[ (\theta e^{\theta})_a^{0} -\int_a^{0} e^{\theta} d\theta=\lim\limits_{a \to -\infty}[ e^{\theta}(\theta-1)]_a^{0}$
or, $\lim\limits_{a \to -\infty}[ e^{\theta}(\theta-1)]_a^{0}=-1-\lim\limits_{a \to -\infty}[ e^{a}(1-a)]$
But $\lim\limits_{a \to -\infty}[ e^{\theta}(\theta-1)]_a^{0}=0(-\infty)$
This shows an indeterminate form of a limit, so we will use L-Hospital's rule.
$\lim\limits_{a \to -\infty}\dfrac{-1}{-e^{-a}}=0$
Thus,
$-1-\lim\limits_{a \to -\infty}[ e^{a}(1-a)]=-1-0=-1$
