Several questions about power of periodic signal I am currently learning Fourier series (periodic signals) and there are few things I am not sure about. 

(1)

$$
P = \frac{1}{T}\int_{-T/2}^{T/2}|s(t)|^2dt = \frac{1}{T}\int_{0}^{T}|s(t)|^2dt
$$

Let's say, that period is $T = 2$ and my signal $s(t) = t$:

$$
P = \frac{1}{T}\int_{-T/2}^{T/2}|s(t)|^2dt = \frac{1}{2}\int_{-1}^{1}t^2dt = \frac{1}{3}
$$

but 

$$
P = \frac{1}{2}\int_{0}^{T}|s(t)|^2dt = \frac{1}{2}\int_{0}^{2}t^2dt = \frac{4}{3}
$$

What am I missing here? Are these equations wrong? I've checked several sources and all of them state it this way. What kind of power do I calculate this way? Is it total power of periodic signal?

(2) I also have a problem, I don't understand how to calculate.
I am given:

\begin{align}
x(t) &amp;= t\\T &amp;= \frac{1}{3}
\end{align}
what is the power on interval 
$$
\left(\frac{1}{3};\frac{2}{3}\right)
$$
How do I calculate this? 

(3) Am I right if I say, that I calculate power of first $N$ harmonics with Parseval's theorem? 

$$
P_n = a_0^2 + \frac{1}{2}(a_n^2 + b_n^2)
$$

I am really trying to understand it, but the more I read, the more confused I am. So I really need to point me in the right direction.
You are missing the effect of the integration interval on the integrand function...

If you choose a period; T = 2 , and a periodic signal whose base period is $s_0(t) = t$  in the interval [0,2]... Then the integrand in the shifted interval [-1,1] will be different as given by 

$$
s_1(t) = \begin{cases}{ t + 2  ~~~, -1 &lt; t &lt; 0 \\ t   ~~~~~~~~~~ , ~   0&lt; t &lt; 1}\end{cases} 
$$  

Then you should write the integrals as follows: :
$$
P = \frac{1}{2}\int_{-1}^{1}|s_1(t)|^2dt = \frac{1}{2}\int_{0}^{2}|s_0(t)|^2dt
$$

You should get the same result now...