Prove that $\int_a^{a+T}f(x)dx$ is independent of $a$ if $f$ is continuous and periodic with period $T$
I indeed don't how to treat to this problem.
Prove that $\int_a^{a+T}f(x)dx$ is independent of $a$ if $f$ is continuous and periodic with period $T$
I indeed don't how to treat to this problem.
Let $nT$ denote the unique multiple of $T$ in $[a,a+T)$. Then $I=J+K$ where $ J=\int_a^{nT}f(x)\mathrm dx,\qquad K=\int_{nT}^{a+T}f(x)\mathrm dx. $ The change of variables $x=t-T+nT$ in $J$ yields $ J=\int_{a-nT+T}^Tf(t)\mathrm dt. $ The change of variables $x=s+nT$ in $K$ yields $ K=\int_0^{a-nT+T}f(s)\mathrm ds. $ Thus $I=\displaystyle\int_0^Tf(u)\mathrm du$ is independent of $a$.
So, clearly, $I$ is a function of $a$. Let us find $I'(a)$. Since $f(x)$ is continuous on the closed interval $[a, a + k]$, the First Fundamental Theorem of Calculus applies to give $I'(a) = f(a + T) - f(a).$ Since $f$ is periodic of period $T$, $f(a + T) = f(a)$, so $I'(a) = 0$. Therefore, $I(a)$ is the constant function. In other words, as $a$ changes, $I(a)$ does not, so $I$ does not depend on $a$.