Let $f(x)=\int_0^xg(t)dt,$ where g is a non-zero even function. If $f (x+5)=g(x)$, then $\int_0^xf(t)dt$ is equal to:
i)$\int_5^{x+5}g(t)dt$
ii)$\int_{x+5}^5g(t)dt$
Follwing is my approach:
since g(x) is even $\implies$ g(x)=g(-x)
From the given question $f(x)=\int_0^xg(t)dt$-------(1) ; on replacing 'x' with '-x' we get
$$f(-x)=\int_0^{-x}g(t)dt$$
Let $p=-t \implies dp=-dt$. Substituting this we get
$$f(-x)=\int_0^{x}g(-p)(-dp)$$
$$\implies f(-x)=-\int_0^{x}g(p)dp\enspace ----(2)$$
As the integral of a function does not depend on the variable when the limits are same, on adding eqation (1) and (2) we get
$$f(x)+f(-x)=0\enspace i.e.\enspace an \enspace odd \enspace function$$
It is given in the question that $f(x+5)=g(x)$, and on replacing $'x'$ with $'x+5'$ we get $f(x+10)=g(x+5)-----(3)$
Cosider option i)
$$\int_5^{x+5}g(t)dt$$
On shifting the limits we would get
$$\int_{5-5}^{x+5-5}g(t+5)dt=\int_{0}^{x}g(t+5)dt$$
$$\implies \int_{0}^{x}f(t+10)dt$$
Here I am stuck. I am unable to further simplify it to get the desired option. Here I used option i) for simplification.
The correct answer is option ii)