Use integration by parts to evaluate the following integral. $$ \int_{-\infty}^{-6} heta e^{ heta} d heta $$ Use the integration by parts formula to rewrite the given integral. Because the given integral is improper, rewrite it using a limit. A. $$ \lim _{a ightarrow \infty}-\frac{6}{e^{6}}-a e^{a}-\int_{-\infty}^{a} e^{ heta} d heta $$ B. $$ \lim _{a ightarrow-\infty}-\frac{6}{e^{6}}-a e^{a}-\int_{a}^{-6} e^{ heta} d heta $$ C. $$ \lim _{a ightarrow-\infty}-\frac{6}{e^{6}}-a e^{a}-\int_{a}^{\infty} e^{ heta} d heta $$ O D. $$ \lim _{a ightarrow \infty}-\frac{6}{e^{5}}-a e^{a}-\int_{a}^{-6} e^{ heta} d heta $$

Question

Use integration by parts to evaluate the following integral. $$ \int_{-\infty}^{-6} 	heta e^{	heta} d 	heta $$ Use the integration by parts formula to rewrite the given integral. Because the given integral is improper, rewrite it using a limit. A. $$ \lim _{a ightarrow \infty}-\frac{6}{e^{6}}-a e^{a}-\int_{-\infty}^{a} e^{	heta} d 	heta $$ B. $$ \lim _{a ightarrow-\infty}-\frac{6}{e^{6}}-a e^{a}-\int_{a}^{-6} e^{	heta} d 	heta $$ C. $$ \lim _{a ightarrow-\infty}-\frac{6}{e^{6}}-a e^{a}-\int_{a}^{\infty} e^{	heta} d 	heta $$ O D. $$ \lim _{a ightarrow \infty}-\frac{6}{e^{5}}-a e^{a}-\int_{a}^{-6} e^{	heta} d 	heta $$

UpStudy AI · Accepted Answer

We begin with the integral

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

Because the lower limit is $$-\infty$$, we rewrite the integral as an improper integral:

$$
\lim_{a\to -\infty} \int_a^{-6} \theta e^{\theta}\, d\theta.
$$

Next, we apply integration by parts. Let

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

Then the derivatives and antiderivatives are

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

The integration by parts formula

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

yields

$$
\int_a^{-6} \theta e^{\theta}\, d\theta = \Bigl[\theta\, e^{\theta}\Bigr]_a^{-6} - \int_a^{-6} e^{\theta}\, d\theta.
$$

Evaluating the boundary term we have

$$
\Bigl[\theta\, e^{\theta}\Bigr]_a^{-6} = (-6)e^{-6} - \bigl(a\, e^{a}\bigr) = -6e^{-6} - a\, e^{a}.
$$

Thus, the integral becomes

$$
\int_a^{-6} \theta e^{\theta}\, d\theta = -6e^{-6} - a\, e^{a} - \int_a^{-6} e^{\theta}\, d\theta.
$$

Finally, reinstating the limit gives

$$
\lim_{a\to -\infty} \left[-6e^{-6} - a\, e^{a} - \int_a^{-6} e^{\theta}\, d\theta\right].
$$

Since $$e^{-6}=\frac{1}{e^6}$$, we can also write the answer as

$$
\lim_{a\to -\infty} \left[-\frac{6}{e^{6}} - a\, e^{a} - \int_a^{-6} e^{\theta}\, d\theta\right].
$$

This matches option B.
The correct answer is option B: $$ \lim_{a \rightarrow-\infty}-\frac{6}{e^{6}}-a e^{a}-\int_{a}^{-6} e^{\theta} d \theta $$.

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