Q:

RMS value of any periodic voltage or current

How using calculus can be derived the RMS value of any periodic voltage or current?

A:

If you know the formula for the periodic waveform then you square the formula then take the average over the waveform's period.
Taking the average firstly requires integrating the formula and plugging in the limits of time for the start of the period and the end of the period. Secondly, you divide through by the period converting area of the waveform to a mean amplitude.
Then take the square root. 
EDIT  An alterative way of looking at RMS - imagine the periodic waveform were applied to a resistor and ask yourself what the power would be - it would be \$ V^2/R \$
. To get the "effective" power you need to average the power waveform (integration then dividing by time period).
The mean/average power you have now calculated could have been created by a DC voltage and to find this dc voltage, equate the average power to \$ V(dc)^2/R \$
. This conveniently gets rid of the resistor, R and by taking the square root you return an equivalent DC voltage that would achieve the same power into any abitrary resistor. This is RMS; an equivalent DC voltage that would create the same power as the waveform.
Squaring is not done to make the waveform always positive (rather like the ABS function) so that the average isn't zero. Squaring is absolutely fundamental to the process of converting to a power and then finding the equivalent DC voltage that would generate that same power. If you did ABS(sinewave) then averaged it, then took the square root you would not get true RMS.

