Let $x: \Omega \to \R $ a random variable and
$p: \Omega \to \R $ a distribution.
The mean square of $x$
is $\E (x^2)$.
Define $\mu = \E (x)$.
We can express $\E (x^2) = \E (x)^2 + \cov(x)$
since
\[
\begin{aligned}
\cov(x) = \E ((x -\mu )^2)
&= \E (x^2 - 2\mu x + \mu ^2) \\
&= \E (x^2) -2\mu \E (x) + \mu ^2 \\
&= \E (x^2) - \mu ^2.
\end{aligned}
\]
The n’th moment of $x$ is $\E (x^n)$. The mean is the first moment. The covariance is the second moment minus the square of the first moment.