I want to differentiate the following function with respect to $x$: $$G(x) = \int_{0}^{x} (x-y)^nf(y)dy$$ I know how to solve the integral using the Leibniz rule, it is in the exact form which the formula requires. But the thing is that the course I am doing is fairly rigorous and hence it is not allowed to use results without proof. I feel that this can be calculated by using the fundamental theorem of calculus, but I haven't been able to find proof. Any help is appreciated. The original question is :
Let $f$ be a continuous real-valued function on $[0,\infty)$, and define the sequence of functions on $[0,\infty)$ \begin{split} F_0(x)&=\int_0^x f(y) dy,\\ F_1(x)&=\int_0^x (x-y )f(y) dy,\\ \dots \dots \\ F_n(x)&=\frac{1}{n!} \int_0^x (x-y)^nf(y) dy\\ \dots \dots \end{split}
Verify that $F'_n=F_{n-1}$, $F_{n}^{(n+1)}=f$, and the first $n$ derivatives of $F_n$ vanish at $0$.
Since $$ (x-y)^n=\sum_{k=0}^{n}{n \choose k}x^k(-y)^{n-k}, $$ we have $$ G(x)=\int_0^x(x-y)^nf(y)dy=\sum_{k=0}^n{n \choose k}x^k\int_0^x (-y)^{n-k}f(y)dy. $$ Computing the derivative of $G$, we get \begin{eqnarray} G'(x) &=&\sum_{k=1}^n{n \choose k}kx^{k-1}\int_0^x (-y)^{n-k}f(y)dy\cr &&+\sum_{k=0}^n{n \choose k}x^k(-x)^{n-k}f(x)\cr &=&\sum_{k=1}^n k\cdot \frac{n!}{k!(n-k)!}x^{k-1}\int_0^x (-y)^{n-k}f(y)dy\cr &&+f(x)\sum_{k=0}^n{n \choose k}x^k(-x)^{n-k}\cr &=& \sum_{k=1}^n \frac{n!}{(k-1)!(n-k)!}x^{k-1}\int_0^x (-y)^{n-k}f(y)dy\cr &&+f(x)\cdot(x-x)^n \cr &=&\sum_{p=0}^{n-1} \frac{n!}{p!(n-1-p)!}x^{p}\int_0^x (-y)^{n-p-1}f(y)dy+f(x)\cdot0\cr &=&n\sum_{p=0}^{n-1} \frac{(n-1)!}{p!(n-1-p)!}x^{p}\int_0^x (-y)^{n-1-p}f(y)dy\cr &=&n\sum_{p=0}^{n-1} {n-1 \choose p}x^{p}\int_0^x (-y)^{n-1-p}f(y)dy\cr &=&n\int_0^x\sum_{p=0}^{n-1} {n-1 \choose p}x^{p} (-y)^{n-1-p}f(y)dy\cr &=&n\int_0^x(x-y)^{n-1}f(y)dy \end{eqnarray}