Start with integration by parts.
$$ \int_0^\infty kxe^{-2x} dx $$
$$ u = kx \hskip{2em} dv = e^{-2x} dx $$
$$ du=k dx \hskip{2em} v=-\frac{1}{2}e^{-2x} $$
$$ \int u dv = uv - \int v dv $$
$$ \int_0^\infty kxe^{-2x} dx = -\frac{kx}{2}e^{-2x} - \int -\frac{k}{2}e^{-2x}$$
$$ = -\frac{kx}{2}e^{-2x} -\frac{k}{4}e^{-2x} $$
$$ = \frac{-\frac{1}{2}kx-\frac{k}{4}}{e^{2x}} $$
$$ = \frac{-2kx-k}{4e^{2x}} \Big]_0^\infty $$
We obtain the improper integral above, which we can rewrite as:
$$ = \lim\limits_{b \to \infty}\frac{-2kx-k}{4e^{2x}} \Big]_0^b$$
$$ = \lim\limits_{b\to\infty}\frac{-2kb-k}{4e^{2b}} - \frac{-2k(0)-k}{4e^{2(0)}} $$
We can use inspection or L'Hopital's rule to evaluate the first fraction to zero, resulting in:
$$ = \frac{-(-k)}{4} = \frac{k}{4} $$
We need the initial integral to evaluate to 1.
$$ \frac{k}{4} = 1 $$
$$ k= \boxed{4} $$