Definite integral related to even function property

Question

If $\displaystyle f\left( x \right)=\int _{ 0 }^{ x }{ g\left( t \right) dt }$,where g is an even function and $f(x+5)=g(x)$ then how to prove the following $$g(0)-g(x)=\int _{ 0 }^{ x }{ f\left( t \right) dt }$$

Answer 1

We know that $g$ is even and $f$ odd. Then for all $x$:

$$f(x+5)=g(x)=g(-x)=f(-x+5)=-f(x-5)$$ Hence we get $$f(x+10)+f(x)=0$$ This is $$g(x+5)+f(x)=0$$ As $f^{\prime}(x)=g(x)$ (from the first relation) we have $$f^{\prime}(x+5)+f(x)=0$$

But as $g(x)=f(x+5)$ this is $$g^{\prime}(x)+f(x)=0$$ And now it is easy to finish.