Q:

Why signal interval is -T/2?

While going over some notes for the past year, I never really understood why many of the integrals for signals and systems, integrate over \$ [- \frac{T}{2},\frac{T}{2}]\$ rather than just \$[0,T]\$. (The same is can be true for \$ [-T,T]\$ vs \$[0,2T]\$ )
Why is that a common interval ?
Here is an example of what I mean.

Edit:
I corrected the internal from \$ [0, \frac{T}{2}]\$ to \$[0,T] \$. So you may see some answers talk about a range that wasn't in the question. 

A:

One simple reason : \$[-\frac{T}{2} , \frac{T}{2}]\$ cover one entier period of the signal while \$[0 , \frac{T}{2}]\$ cover only half a period. 
But, I think your true question is : Why \$[-\frac{T}{2} , \frac{T}{2}]\$ rather than \$[0 , T]\$? 
In both case, an entier period is cover. So? What is the difference? 
It doesn't really matter since you will have the same result if you do the calculation.

Using the \$[-\frac{T}{2} , \frac{T}{2}]\$ interval makes calculation easier. As you may notice, signal use to be periodic, that mean, the expression use to be like "\$cos(2πxT)\$". When "\$cos(2πxT)\$" is integrated between \$[-\frac{T}{2} , \frac{T}{2}]\$, "/2" get rid of "2*". This makes things a little easier. 
When a periodic signal is represented, most of the time (at least on exercice book), only one period is drawn with \$x = 0\$ as middle point. It is then logic to integrate from \$-\frac{T}{2}\$ to \$\frac{T}{2}\$.

