Suppose $f:\Bbb{R}\rightarrow\Bbb{R}$ is continuous. Define $G(x)=\int_0^x(x-t)f(t)dt$ for all $x\in\Bbb{R}$. Prove $G''(x)=f(x)$ This is a tricky example of the Fundamental Theorem of Calculus 1. I am looking for confirmation of my proof or answer to see if I am doing it correctly because this is different than the basic applications. 
$\underline{Proof}$
$G(x)=\int_0^x(x-t)f(t)dt\rightarrow G(x)=\int_0^xxf(t)dt-\int_0^xtf(t)dt$.
So, let $Q(x)=\int_0^xf(t)dt$ and let $P(x)=\int_0^xtf(t)dt$ then we have, 
$G(x)=xQ(x)-P(x)$
$G'(x)=xQ'(x)+Q(x)-P'(x)=xf(x)+Q(x)-xf(x)=Q(x)$
hence,
$G''(x)=Q'(x)=f(x)$
∎

Your proof is correct. An alternative is to use integration by parts. With
$$
 f_1(x) = \int_0^x f(t) \, dt \, , \quad f_2(x) = \int_0^x f_1(t) \, dt
$$
we have
$$
G(x) =  \bigl\lbrack (x-t)f_1(t) \bigr\rbrack_{t=0}^{t=x} +
\int_0^x f_1(t) \, dt = f_2(x)
$$
and therefore $G'' = f_2'' = f_1' = f$.
This generalizes to higher anti-derivatives:
$$
\int_0^x \frac{(x-t)^{n-1}}{(n-1)!} f(t) \, dt
= \int_0^x \int_0^{t_1} \cdots \int_0^{t_{n-1}} f(t_n) \, dt_n \cdots dt_1
$$
for $n \ge 1$.
