= Solution
Assume $\lambda>0$, as required for normalization of the <exponential distribution>. Its <cumulative distribution function> is the integral of its <probability density function> from $-\infty$ to $x$:
$$
\boxed{F_X(x)=\begin{cases}0,&x<0,\\1-e^{-\lambda x},&x\geq0.\end{cases}}
$$
For the <expected value>, <integration by parts> gives
$$
\begin{aligned}
\mathbb E[X]
&=\int_0^\infty x\lambda e^{-\lambda x}\,dx\\
&=\left[-xe^{-\lambda x}\right]_0^\infty+\int_0^\infty e^{-\lambda x}\,dx
=\frac1\lambda.
\end{aligned}
$$
Similarly,
$$
\mathbb E[X^2]=\left[-x^2e^{-\lambda x}\right]_0^\infty
+2\int_0^\infty xe^{-\lambda x}\,dx=\frac2{\lambda^2}.
$$
The endpoint terms vanish because exponential decay dominates each fixed power of $x$. Therefore
$$
\boxed{\mathbb E[X]=\frac1\lambda,\qquad
\operatorname{Var}(X)=\mathbb E[X^2]-\mathbb E[X]^2=\frac1{\lambda^2}.}
$$
The mean has the same units as $X$, while the <variance> has their square; $\lambda$ has reciprocal units.
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