Use integration by parts to evaluate the following integral. $$ \int_{-\infty}^{-6} heta e^{ heta} d heta $$

Question

Use integration by parts to evaluate the following integral. $$ \int_{-\infty}^{-6} 	heta e^{	heta} d 	heta $$

UpStudy AI · Accepted Answer

We begin with the integral

$$
\int_{-\infty}^{-6} \theta e^{\theta} \, d\theta.
$$

We use integration by parts, where

$$
u = \theta \quad \text{and} \quad dv = e^{\theta}\,d\theta.
$$

Then

$$
du = d\theta \quad \text{and} \quad v = e^{\theta}.
$$

By the integration by parts formula

$$
\int u\,dv = uv - \int v\,du,
$$

we have

$$
\int \theta e^{\theta} \, d\theta = \theta e^{\theta} - \int e^{\theta} \, d\theta.
$$

The remaining integral is

$$
\int e^{\theta} \, d\theta = e^{\theta}.
$$

Thus, the antiderivative becomes

$$
\theta e^{\theta} - e^{\theta} = e^{\theta}(\theta - 1) + C.
$$

We now evaluate the definite integral

$$
\int_{-\infty}^{-6} \theta e^{\theta} \, d\theta = \left[ e^{\theta}(\theta - 1) \right]_{-\infty}^{-6}.
$$

This gives

$$
\left. e^{\theta}(\theta - 1) \right|_{\theta=-6} - \lim_{\theta\to -\infty} e^{\theta}(\theta - 1).
$$

First, at $$\theta = -6$$:

$$
e^{-6}(-6 - 1) = -7e^{-6}.
$$

Next, we evaluate the limit as $$\theta \to -\infty$$:

$$
\lim_{\theta \to -\infty} e^{\theta}(\theta - 1) = 0.
$$

This is because $$e^{\theta} \to 0$$ very rapidly as $$\theta \to -\infty$$ compared to the linear growth of $$\theta - 1$$.

Thus, the value of the integral is

$$
-7e^{-6} - 0 = -7e^{-6}.
$$
The integral evaluates to $$ -7e^{-6} $$.

Use integration by parts to evaluate the following integral. \( \int_{-\infty}^{-6} \theta e^{\theta} d \theta \)