Is the Improper Integral Convergent and What is its Value?

[karush]

206.8.8.11
$\text{determine if the improper integral is convergent} \\
\text{and calculate its value if it is convergent. } $

$$\displaystyle
\int_{0}^{\infty}8e^{-8x} \,dx
=-e^{-8x}+C$$
$$\text{not sure about the coverage thing from the text } $$
View attachment 6032

 

[Greg]

What is $$-\lim_{b\to\infty}e^{-8b}$$ ?

 

[karush]

$\text{ 206.8.8.11}$
$$-\lim_{b\to\infty}e^{-8b}=0$$

$\text{so then the limit exists and it is continuous}$

 

[MarkFL]

We are given to evaluate:

$$I=\int_0^{\infty} 8e^{-8x}\,dx$$

We see that the integrand is differentiable over the reals, therefore it must be continuous over the interval of integration. Therefore, by 1.), we may state:

$$I=\lim_{t\to\infty}\left(\int_0^{t} 8e^{-8x}\,dx\right)$$

$$I=\lim_{t\to\infty}\left(\int_0^{t} e^{-8x}\,(8\,dx)\right)$$

$$I=\lim_{t\to\infty}\left(\int_0^{t} e^{-u}\,du\right)$$

Apply the FTOC:

$$I=\lim_{t\to\infty}\left(-\left[e^{-u}\right]_0^t\right)$$

$$I=\lim_{t\to\infty}\left(1-e^{-t}\right)=1-0=1$$

The limit exists, and so the given definite integral converges to the value of the limit.