Question

If $f\left(x\right)$ is a continuous periodic function with period T, then the integral $I={\int }_a^{a+t}f\left(x\right)dx$ is

• A. Equal to $2a$
• B. Equal to $3a$
• C. Independent of $a$
• D. None of these

Answer 1

The correct option is C Independent of $a$

Consider the function $g\left(a\right)={\int }_a^{a+T}f\left(x\right)dx={\int }_a^{0}f\left(x\right)dx+{\int }_0^{T}f\left(x\right)dx+{\int }_T^{a+T}f\left(x\right)dx$ Putting $x-T=y$ in last integral, and $dx=dy$
Also at $x=T,y=0$ and $x=a+T,y=a$
$\Rightarrow {\int }_T^{a+T}f\left(x\right)dx={\int }_0^{a}f(y+T)dy={\int }_0^{a}f\left(y\right)dy={\int }_0^{a}f\left(x\right)dxg\left(a\right)={\int }_a^{0}f\left(x\right)dx+{\int }_0^{T}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dxg\left(a\right)={\int }_a^{0}f\left(x\right)dx+{\int }_0^{T}f\left(x\right)dx-{\int }_a^{0}f\left(x\right)dxg\left(a\right)={\int }_0^{T}f\left(x\right)dx$ which is independent of $a$.