How do we relate RMS and standard deviation for continuous signals? Because the discrete formula for RMS, $\displaystyle X_{RMS}=\sqrt{{1 \over N}(x[1]^2+x[2]^2+...+x[N]^2)}$, is almost the same as the formula for standard deviation (assuming mean zero), except for a factor of $\sqrt{\frac{N}{N-1}}$ the RMS and standard deviation are often said to be approximately the same.
See, for example, https://www.allaboutcircuits.com/technical-articles/how-standard-deviations-relates-rms-values/
But if we use the continuous formula for RMS,
$\displaystyle X_{RMS}=\sqrt{{1 \over {T_2-T_1}}\int_{T_1}^{T_2}x(t)^2 dt}$
, it wouldn't match the continuous formula for standard deviation of a continuous distribution, because it would have included the pdf $f_X(x)$ inside the integral. I.e. $\sigma_X=\sqrt{\int x^2 f_X(x)dx}$.
Is there a way to draw this comparison in the the continuous regime? Thank you.

The reason for the lack of the $f_X(x)$ in the formula, is because that term would appear only when integrating over all possible values of X. This is not the case here. We are integrating over samples of X in the given interval $[T_1,T_2]$.
We have that
$Var[X]=E[(X-\mu)^2]=E[X^2]$
To estimate the variance of the signal $x(t)$ over the interval $[T_1,T_2]$ we must calculate the average of $x^2(t)$ in that interval:
$\displaystyle {1 \over {T_2-T_1}}\int_{T_1}^{T_2}x(t)^2 dt$
Then apply the square root to get the standard deviation.
