Raghav Gupta said:
Thanks very much Aditya. I don't know that time period property. Can you give a link or reference for the proof of it?
Suppose that ##f## is a ##T## periodic function. Then for any real number ##\alpha##, we can write:
$$\int_{\alpha}^{\alpha + T} f(x) \space dx = \int_0^T f(x) \space dx$$
To prove this, write:
$$\int_{\alpha}^{\alpha + T} f(x) \space dx = \int_{\alpha}^0 f(x) \space dx + \int_{0}^{\alpha + T} f(x) \space dx$$
Now suppose ##u = x + T \Rightarrow x = u - T \Rightarrow dx = du## for the first integral on the right, then:
$$ \int_{\alpha}^0 f(x) \space dx + \int_{0}^{\alpha + T} f(x) \space dx = \int_{\alpha + T}^{T} f(u - T) \space du + \int_{0}^{\alpha + T} f(x) \space dx$$
Now due to the ##T## periodicity of ##f##, we may write:
$$ \int_{\alpha + T}^{T} f(u - T) \space du + \int_{0}^{\alpha + T} f(x) \space dx = \int_{\alpha + T}^{T} f(u) \space du + \int_{0}^{\alpha + T} f(x) \space dx$$
Simply replace the dummy variable ##u## by ##x## now to obtain:
$$\int_{\alpha + T}^{T} f(u) \space du + \int_{0}^{\alpha + T} f(x) \space dx = \int_{\alpha + T}^{T} f(x) \space dx + \int_{0}^{\alpha + T} f(x) \space dx$$
The integrals represent the area from ##[0, T]## because we integrate from ##0## to ##\alpha + T## and then from ##\alpha + T## to ##T##. Therefore:
$$\int_{\alpha + T}^{T} f(x) \space dx + \int_{0}^{\alpha + T} f(x) \space dx = \int_0^T f(x) \space dx$$
So really, the integral of a ##T## periodic function over any interval of length ##T## is the same.