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If $f(x)$ is a continuous periodic function with period $T$, then the integral $I=\int_a^{a+\tau} f(x) d x$ is
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Independent of a
Consider the function $g(a)=\int_a^{a+T} f(x) d x$
$=\int_a^0 f(x) d x+\int_0^T f(x) d x+\int_T^{a+T} f(x) d x$
Putting $x-T=y$ in last integral, we get $\int_T^{a+T} f(x) d x=\int_0^a f(y+T) d y=\int_0^a f(y) d y$
$\Rightarrow g(a)=\int_a^0 f(x) d x+\int_0^T f(x) d x+\int_T^{a+T} f(x) d x=\int_0^T f(x) d x$
Hence $g(a)$ is independent of $a$.
$=\int_a^0 f(x) d x+\int_0^T f(x) d x+\int_T^{a+T} f(x) d x$
Putting $x-T=y$ in last integral, we get $\int_T^{a+T} f(x) d x=\int_0^a f(y+T) d y=\int_0^a f(y) d y$
$\Rightarrow g(a)=\int_a^0 f(x) d x+\int_0^T f(x) d x+\int_T^{a+T} f(x) d x=\int_0^T f(x) d x$
Hence $g(a)$ is independent of $a$.
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