Let f be a real function satisfying $f\left(x\right)=f(x+3)-f(x+6)+f(x+9)$. Then the function $g\left(x\right)=\underset{x}{\overset{x+12}{\int }}f\left(t\right)dt$ is

$f\left(x\right)=f(x+3)-f(x+6)+f(x+9)$

Replacing x with x + 3, $f(x+3)=f(x+6)-f(x+9)+f(x+12)$

Adding we get $f\left(x\right)=f(x+12)$

Now $g\left(x\right)=\underset{x}{\overset{x+12}{\int }}f\left(t\right)dt\Rightarrow g^{|}\left(x\right)=f(x+12)-f\left(x\right)=0\forall x$

$\therefore g\left(x\right)$ is a constant function