evaluate the following integral or state that it diverges. ∫0,∞ 8e^(-4x)dx select the correct choice and, if…

Explanation:

Step1: Find antiderivative

The antiderivative of $8e^{-4x}$ is $8\times\frac{e^{-4x}}{-4}=- 2e^{-4x}$ since the antiderivative of $e^{ax}$ is $\frac{e^{ax}}{a}$ ($a=-4$ here).

Step2: Evaluate improper - integral

$\int_{0}^{\infty}8e^{-4x}dx=\lim_{b\rightarrow\infty}\int_{0}^{b}8e^{-4x}dx=\lim_{b\rightarrow\infty}\left[-2e^{-4x}\right]{0}^{b}$. Substitute the upper and lower limits: $\lim{b\rightarrow\infty}\left(-2e^{-4b}+2e^{0}\right)$. As $b\rightarrow\infty$, $e^{-4b}=\frac{1}{e^{4b}}\rightarrow0$. And $e^{0} = 1$. So $\lim_{b\rightarrow\infty}\left(-2e^{-4b}+2e^{0}\right)=2$.

Answer:

A. The improper integral converges and $\int_{0}^{\infty}8e^{-4x}dx = 2$