Several questions about power of periodic signal
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...