Assume f is periodic with T; that is f(x + T) - f(x) for all x...

Assume f is periodic with T; that is f(x + T) - f(x) for all x.show that the integral from 0 to T of f(x)dx = the integral of a to a+T of f(x)dx for any real number a. (Hint: First, work with 0<=a<=T.

It's not really necessary to consider separately whether a is between 0 and T, a < 0 or a > T. Assuming f is everywhere integrable,

T+a . . . . . a . . . . . . T+a
∫ f(x) dx = ∫ f(x) dx + ∫ f(x) dx,
0 . . . . . . 0 . . . . . . . a

and similarly

T+a . . . . .T . . . . . . T+a
∫ f(x) dx = ∫ f(x) dx + ∫ f(x) dx.
0 . . . . . . 0 . . . . . . . T

That is, you can break up the integral from 0 to T+a at either a or at T. (Note: I am not making any assumptions whatsoever about the relationship between a and T. In fact T + a may very well be negative. It doesn't matter.)

It follows that

T+a . . . . . T . . . . . . T+a . . . . a
∫ f(x) dx = ∫ f(x) dx + ∫ f(x) dx - ∫ f(x) dx. <--- call this equation (*)
a . . . . . . 0 . . . . . . . T . . . . . .0

To get the desired conclusion, we have to show that the last two integrals on the right cancel each other.

Consider the change of variables u = x + T, du = dx, u(0) = T and u(a) = T + a

a . . . . . . . T+a
∫ f(x) dx = ∫ f(u-T) du
0 . . . . . . . T

But f(u - T) = f(u - T + T) = f(u) for all u. So

a . . . . . . . T+a
∫ f(x) dx = ∫ f(x) dx
0 . . . . . . . T

So that (*) reduces to

T+a . . . . . T
∫ f(x) dx = ∫ f(x) dx.
a . . . . . . 0