a simple question about average power of uniformly distributed random variable

[MamadJun]

can anyone explain and say the reason of this sentence?

Any uniformly distributed random variable in the range (-a/2 to a/2) has an average power(variance) given by a^2/12.

is it ok?

 

[Mityan]

Yes.
Variable - v.
Mean-square value = integral (or sum) of v^2 divided by a (i.e. in interval from -a/2 to a/2).
It means (v^3)/3/a. From -a/2 to a/2.
After placing in this formula the integration limits instead of v, you get a^2/12.