consider: $\\int_{a}^{b}h(x)dx$ for $a = 2$, $b=\\infty$, and $h(x)=xe^{-2x}$\n(a) determine whether the…

Explanation:

Step1: Determine if the integral is improper

An improper integral is either of the form (\int_{a}^{\infty}f(x)dx), (\int_{-\infty}^{b}f(x)dx) or (\int_{-\infty}^{\infty}f(x)dx) (infinite - limits) or has a discontinuity in the interval ([a,b]). Here, (a = 2), (b=\infty) and (h(x)=xe^{-2x}) is continuous for all (x\in[2,\infty)). Since one of the limits of integration ((b = \infty)) is infinite, the integral (\int_{2}^{\infty}xe^{-2x}dx) is improper.

Step2: Evaluate the improper integral

We use the formula for an improper integral (\int_{a}^{\infty}f(x)dx=\lim_{t\rightarrow\infty}\int_{a}^{t}f(x)dx). First, find (\int xe^{-2x}dx) using integration by parts. Let (u = x) and (dv=e^{-2x}dx). Then (du = dx) and (v=-\frac{1}{2}e^{-2x}). By the integration - by - parts formula (\int u;dv=uv-\int v;du), we have: (\int xe^{-2x}dx=-\frac{1}{2}xe^{-2x}+\frac{1}{2}\int e^{-2x}dx) (=-\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}+C=-\frac{e^{-2x}}{2}(x + \frac{1}{2})+C)

Now, (\lim_{t\rightarrow\infty}\int_{2}^{t}xe^{-2x}dx=\lim_{t\rightarrow\infty}\left[-\frac{e^{-2x}}{2}(x+\frac{1}{2})\right]{2}^{t}) (=\lim{t\rightarrow\infty}\left(-\frac{e^{-2t}}{2}(t+\frac{1}{2})+\frac{e^{-4}}{2}(2 + \frac{1}{2})\right)) Using L'Hopital's rule for (\lim_{t\rightarrow\infty}\frac{t+\frac{1}{2}}{e^{2t}}), since (\lim_{t\rightarrow\infty}\frac{t+\frac{1}{2}}{e^{2t}}) is of the form (\frac{\infty}{\infty}), and (\lim_{t\rightarrow\infty}\frac{t+\frac{1}{2}}{e^{2t}}=\lim_{t\rightarrow\infty}\frac{1}{2e^{2t}} = 0) (\lim_{t\rightarrow\infty}\left(-\frac{e^{-2t}}{2}(t+\frac{1}{2})+\frac{5e^{-4}}{4}\right)=\frac{5}{4e^{4}})

Answer:

(a) The integral (\int_{2}^{\infty}xe^{-2x}dx) is improper because one of the limits of integration ((b = \infty)) is infinite. (b) The integral converges and its value is (\frac{5}{4e^{4}})