Use integration by parts to evaluate the following integral. \[ \int_{-\infty}^{-1} \theta e^{\theta} d \theta \] Use the integration by parts formula to rewrite the given integral. Because the given integral is improper, rewrite it using a limit. A. \( \lim _{a \rightarrow-\infty}-\frac{1}{e}-a e^{a}-\int_{a}^{-1} e^{\theta} d \theta \) B. \( \lim _{a \rightarrow \infty}-\frac{1}{e}-a e^{a}-\int_{-\infty}^{a} e^{\theta} d \theta \) C. \( \lim _{a \rightarrow \infty}-\frac{1}{e}-a e^{a}-\int_{a}^{-1} e^{\theta} d \theta \) D. \( \lim _{a \rightarrow-\infty}-\frac{1}{e}-a e^{a}-\int_{a}^{\infty} e^{\theta} d \theta \) Select the correct choice and, if necessary, fill in the answer box to complete your choice. A. The improper integral converges and \( \int_{-\infty}^{-1} \theta e^{\theta} d \theta=\square \). (Type an exact answer.) B. The improper integral diverges.

Let \[ u = \theta,\quad dv = e^\theta d\theta. \] Then \[ du = d\theta,\quad v = e^\theta. \] Using integration by parts, \[ \int \theta e^\theta d\theta = \theta e^\theta - \int e^\theta d\theta = \theta e^\theta - e^\theta. \] Thus, in antiderivative form, \[ \int \theta e^\theta d\theta = e^\theta (\theta - 1) + C. \] Because the given integral is improper, we introduce a limit: \[ \int_{-\infty}^{-1} \theta e^\theta d\theta = \lim_{a\to -\infty} \int_a^{-1} \theta e^\theta d\theta. \] Applying the integration by parts result to the bounds, \[ \int_a^{-1} \theta e^\theta d\theta = \Bigl[ e^\theta (\theta -1) \Bigr]_a^{-1} = e^{-1}(-1-1) - e^a(a-1). \] This simplifies to \[ e^{-1}(-2) - e^a(a-1)= -\frac{2}{e} - e^a(a-1). \] Alternatively, if we apply integration by parts directly in its formula form, we can set it up as: \[ \int_a^{-1} \theta e^\theta d\theta = \left[\theta e^\theta \right]_a^{-1} - \int_a^{-1} e^\theta d\theta = \left(-\frac{1}{e} - a e^a\right) - \int_a^{-1} e^\theta d\theta, \] which matches option A: \[ \lim_{a \rightarrow -\infty}\left(-\frac{1}{e} - a e^a - \int_a^{-1} e^\thetad\theta\right). \] Now, taking the limit as \( a\to -\infty \): \[ \lim_{a\to-\infty} e^a(a-1)=0, \] because \( e^a \) decays to zero much faster than \( |a| \) grows. Thus, the value of the integral is \[ -\frac{2}{e}. \] So, select option A and fill in the answer box with \[ -\frac{2}{e}. \]