= Solution
Repeated <integration by parts> gives, for every nonnegative integer $j$,
$$
\int_0^\infty x^j e^{-\lambda x}dx=\frac{j!}{\lambda^{j+1}}.
$$
With $A=\lambda^2$, the first two <moments> are $\mathbb E[X]=\lambda^2(2!/\lambda^3)=2/\lambda$ and $\mathbb E[X^2]=\lambda^2(3!/\lambda^4)=6/\lambda^2$. Subtracting the square of the <mean> gives
$$
\boxed{\mu=\frac2\lambda,\qquad\sigma^2=\frac2{\lambda^2},\qquad\sigma=\frac{\sqrt2}{\lambda}.}
$$
If $X$ carries units, $\lambda$ has inverse units of $X$, so the <mean> and <standard deviation> have the units of $X$ and the <variance> has their square.
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