the integral in this exercise converges. evaluate the integral without using a table.\n int_{-infty}^{4}…

the integral in this exercise converges. evaluate the integral without using a table.\n int_{-infty}^{4} \theta e^{\theta} d \theta \n int_{-infty}^{4} \theta e^{\theta} d \theta=square \text { (type an exact answer.) }

the integral in this exercise converges. evaluate the integral without using a table.\n int_{-infty}^{4} \theta e^{\theta} d \theta \n int_{-infty}^{4} \theta e^{\theta} d \theta=square \text { (type an exact answer.) }

Answer

Explanation:

Step1: Use integration by parts

Recall the integration - by - parts formula (\int_{a}^{b}u\mathrm{d}v=uv|{a}^{b}-\int{a}^{b}v\mathrm{d}u). Let (u = \theta) and (\mathrm{d}v=e^{\theta}\mathrm{d}\theta). Then (\mathrm{d}u=\mathrm{d}\theta) and (v = e^{\theta}).

So, (\int\theta e^{\theta}\mathrm{d}\theta=\theta e^{\theta}-\int e^{\theta}\mathrm{d}\theta=\theta e^{\theta}-e^{\theta}+C=e^{\theta}(\theta - 1)+C)

Step2: Evaluate the improper integral

(\int_{-\infty}^{4}\theta e^{\theta}\mathrm{d}\theta=\lim_{a\rightarrow-\infty}\int_{a}^{4}\theta e^{\theta}\mathrm{d}\theta)

[ \begin{align*} \lim_{a\rightarrow-\infty}\int_{a}^{4}\theta e^{\theta}\mathrm{d}\theta&=\lim_{a\rightarrow-\infty}\left[e^{\theta}(\theta - 1)\right]{a}^{4}\ &=\lim{a\rightarrow-\infty}\left(e^{4}(4 - 1)-e^{a}(a - 1)\right) \end{align*} ]

We use L'Hopital's rule for (\lim_{a\rightarrow-\infty}e^{a}(a - 1)=\lim_{a\rightarrow-\infty}\frac{a - 1}{e^{-a}}).

Differentiating the numerator and denominator: (\lim_{a\rightarrow-\infty}\frac{1}{-e^{-a}} = 0)

So, (\lim_{a\rightarrow-\infty}\left(e^{4}(4 - 1)-e^{a}(a - 1)\right)=3e^{4}-0)

Answer:

(3e^{4})