Let f(x)=

$f(x)=\int_{0}^{x} g(g) d t$............(i)

$\because g$ is a non-zero even function.

$\therefore g(-x)=g(x)$,...........(ii)

Given, $f(x+5)=g(x)$...........(iii)

From (i) $f^{\prime}(x)=g(x)$

Let, $I=\int_{0}^{x} f(t) d t$,

Put $t=\lambda-5 \Rightarrow I=\int_{5}^{x+5} f(\lambda-5) d \lambda$

$\because f(x+5)=\mathrm{g}(x)$

$\Rightarrow f(-x+5)=g(-x)=g(x)$ ...(iv)

$I=\int_{5}^{x+5} f(\lambda-5) d \lambda$

$\because f(0)=0, g(x)$ is even $\Rightarrow f(x)$ is odd

$\therefore I=\int_{5}^{x+5}-f(5-\lambda) d \lambda$

$\Rightarrow I=\int_{5}^{x+5} g(\lambda) d \lambda=\int_{x+5}^{5} g(t) d t$ (from (iv))