If $f\left(x\right)$  be a real valued function, $f\left(x\right)+f\left(x+4\right)=f\left(x+2\right)+f\left(x+6\right),g\left(x\right)={\displaystyle {\int }_x^{x+8}f\left(t\right)dt}$ . Then $g^{\text{'}}\left(x\right)$  is equal to

$\because f\left(x\right)+f\left(x+4\right)=f\left(x+2\right)+f\left(x+6\right)$  ….(i)

Replacing $x$ by $x+2$ in Eq. (i), then

$f\left(x+2\right)+f\left(x+6\right)=f\left(x+4\right)+f\left(x+8\right),$ ….(ii)

From eqs. (i) and (ii) we get

$f\left(x+8\right)=f\left(x\right)$

$\Rightarrow f\left(x+8\right)-f\left(x\right)=0$   …….(iii)

Now, $g\left(x\right)={\displaystyle {\int }_x^{x+8}f\left(t\right)dt}$

$\therefore g^{\text{'}}\left(x\right)=f(x+8)-f\left(x\right)$   [from eq.(iii)]

$=0$