\( \begin{array}{l}\text { The integral in this exercise converges. Evaluate the integral without using a table. } \\ \int_{-\infty}^{0} \theta e^{\theta} \mathrm{d} \theta \\ 0\end{array} e^{\theta} \mathrm{d} \theta=\square \) (Type an exact answer.)

To evaluate the integral \( \int_{-\infty}^{0} \theta e^{\theta} \mathrm{d} \theta \), we can use integration by parts. Let's set \( u = \theta \) and \( dv = e^{\theta} d\theta \). Then, we find \( du = d\theta \) and \( v = e^{\theta} \). Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have: \[ \int \theta e^{\theta} d\theta = \theta e^{\theta} - \int e^{\theta} d\theta \] Now we compute the remaining integral: \[ \int e^{\theta} d\theta = e^{\theta} \] Substituting back, we get: \[ \int \theta e^{\theta} d\theta = \theta e^{\theta} - e^{\theta} + C = e^{\theta}(\theta - 1) + C \] To evaluate the definite integral, we evaluate \( e^{\theta}(\theta - 1) \) at the bounds from \(-\infty\) to \(0\): At \( \theta = 0 \): \[ e^{0}(0 - 1) = -1 \] At \( \theta = -\infty \): As \( \theta \to -\infty \), \( e^{\theta} \to 0 \) and \( \theta - 1 \to -\infty \), thus \( e^{\theta}(\theta - 1) \to 0 \). So we find: \[ \int_{-\infty}^{0} \theta e^{\theta} d\theta = \left[ e^{\theta}(\theta - 1) \right]_{-\infty}^{0} = -1 - 0 = -1 \] The value of the integral is: \[ \boxed{-1} \]