Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, give the answer -1.

Compute the definite integral:
${\int }_4^{\mathrm{\infty }}{e}^{-2x}xdx$
For the integrand $e^{-2x}x$, integrate by parts, $\int fdg=fg-\int gdf$, where
$f=x,dg=e^{-2x}dx,df=dx,g=-\frac{1}{2}{e}^{-2x}$:
$=(-\frac{1}{2}{e}^{-2x}x){|}_4^{\mathrm{\infty }}+\frac{1}{2}{\int }_4^{\mathrm{\infty }}{e}^{-2x}dx$
INTERMEDIATE STEPS:
Possible derivation:
$\frac{d}{dx}\left(x\right)$
Use the power rule, $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$, where n = 1.
$\frac{d}{dx}\left(x\right)=\frac{d}{dx}\left(x^1\right)=x^0$:
Answer: = 1
Evaluate the antiderivative at the limits and subtract.
$(-\frac{1}{2}{e}^{-2x}x){|}_4^{\mathrm{\infty }}=(\underset{b\to \mathrm{\infty }}{lim}-\frac{1}{2}{e}^{-2b}b)-(-\frac{1}{2}{e}^{-24}4)=(\underset{b\to \mathrm{\infty }}{lim}-\frac{1}{2}{e}^{-2b}b)-(-\frac{2}{e^8})$:
$=(\underset{b\to \mathrm{\infty }}{lim}-\frac{1}{2}{e}^{-2b}b)+\frac{2}{e^8}+\frac{1}{2}{\int }_4^{\mathrm{\infty }}{e}^{-2x}dx$
$\underset{b\to \mathrm{\infty }}{lim}-\frac{1}{2}{e}^{-2b}b=0$:
$=\frac{2}{e^8}+\frac{1}{2}{\int }_4^{\mathrm{\infty }}{e}^{-2x}dx$
INTERMEDIATE STEPS:
Find the following limit:
$\underset{x\to \mathrm{\infty }}{lim}-\frac{1}{2}{e}^{-2x}x$
Hint: Factor a constant multiple out of the limit.
$\underset{x\to \mathrm{\infty }}{lim}-\frac{1}{2}x{e}^{-2x}=-\frac{1}{2}\underset{x\to \mathrm{\infty }}{lim}x{e}^{-2x}$:
$-\frac{\underset{x\to \mathrm{\infty }}{lim}{e}^{-2x}x}{2}$
Hint: | Linear functions grow asymptotically slower than exponential functions.
Since the polynomial x grows asymptotically slower than $e^{2x}$ as x approaches ∞, <