Integrate

[Guest]

[tex]\int_{B+T}^{A+T } f(x) dx = \int_{B}^{A } f(x) dx[/tex]
proof please !
T-period

[shyamjayakannan]

Let [tex]u=x-T\Rightarrow du=dx[/tex], and the integral becomes [tex]\int\limits_{B}^{A}f(u+T)\ du[/tex]. Since [tex]T[/tex] is the period, [tex]f(u+T)=f(u)[/tex].

So, [tex]\int\limits_{B}^{A}f(u+T)\ du=\int\limits_{B}^{A}f(u)\ du=\int\limits_{B}^{A}f(x)\ dx[/tex] because the variable does not matter. Hence, proved