Q:

Why the RMS of a PSD curve is the root of the area below

I will try to explain what is my level of understanding of this problem, please correct me if I'm wrong:

RMS is the Root Mean Square, it represent the mean value of the input signal.
PSD is the measurement of the responses that shows me at which frequencies most of the energy is concentrated.
The area below a curve is the integration of that function.

My situation is that several random vibration tests are performed. These tests are called random tests because of the input signal. In contrast to a sine test where the structure is excited with a sinusoidal input, only one frequency is excited at a time, here 'all' frequencies are excited at the same time.
In this case PSD is measured in ${{g^2}/{Hz}}$ and RMS in ${g_{RMS}}$. Armed with that it easy to see that if you multiply PSD per the frequency range and you take the root of the result you will get something in ${g}$'s, but I don't know how exactly derive the famous relation:
\begin{equation}
{g_{RMS}=\sqrt{\int_{f_1}^{f_2}PSD(f)df}}
\end{equation}
The understanding that I have is very basic and it would be great if someone give me a clear idea of the relations among the RMS, PSD and the real signal. Thank you very much.
In the figure I have plotted a standard random signal.
 

A:

I think this is simply an aspect of Parseval's Theorem (e.g. click me)
It simply says: sum of squares in the time time domain equals sum of squares in the frequency domain. Substitute "sum" for "integral" if using the analog domain. In other words: total energy in the time domain equals total energy in the frequency domain. 
This can easily reproduce your formula, $g_{RMS}$ represents the  time domain energy and the integral on the right represents the energy in the frequency domain. The exact scaling depends on the details like length of the signal, periodicity, sample rate, etc. 

