integral over a period interval

Theorem.  If the real function $f$ is periodic and integrable (http://planetmath.org/RiemannIntegrable) over a period (http://planetmath.org/Periodic) interval, the value of integral over a period interval is always the same, i.e.

${\displaystyle {\int }_a^{a+p}}f(x)𝑑x={\displaystyle {\int }_0^p}f(x)𝑑x\mathit{\hspace{1em}}\forall a\in ℝ$(1)

where $p$ is the period of $f$.

Proof.  The right hand side of the equation (1) is manipulated, with one substitution (http://planetmath.org/ChangeOfVariableInDefiniteIntegral)  $x=t+p$:

${\displaystyle {\int }_0^p}f(x)𝑑x$$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_0^a}f(x)𝑑x+{\displaystyle {\int }_a^p}f(x)𝑑x$

$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_0^a}f(x)𝑑x+{\displaystyle {\int }_a^{a+p}}f(x)𝑑x-{\displaystyle {\int }_p^{a+p}}f(x)𝑑x$

$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_0^a}f(x)𝑑x+{\displaystyle {\int }_a^{a+p}}f(x)𝑑x-{\displaystyle {\int }_0^a}f(t+p)𝑑t$

$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_0^a}f(x)𝑑x+{\displaystyle {\int }_a^{a+p}}f(x)𝑑x-{\displaystyle {\int }_0^a}f(t)𝑑t$

$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_a^{a+p}}f(x)𝑑x$