I am trying to prove the following identity: $$\int_0^x(f*g)(y)dy = (\int_0^xf(y)dy)*g(x) = f(x)*(\int_0^xg(y)dy)$$, where $(p*q)(t) = \int_0^tp(x)q(t-x)dx$.
I thought that since I already know that $(f*g)' = f'*g=f*g'$, setting $F(x) = \int_0^xf(y)dy$ and $G(x) = \int_0^xg(y)dy$, it should be sufficient to prove that $(F*G)'(x) = \int_0^x(f*g)(y)dy$, which I can't do... I also tried Fubini's theorem to manipulate the double integrals, with no success...
You haven't stated the conditions on $f$ and $g$ that make the formula true, but through formal manipulation, we find
$$\int_0^x (f * g)(y)\, dy = \int_0^x \int_0^y f(t)g(y - t)\, dt\, dy = \int_0^x \int_t^x f(t)g(y-t)\, dy\, dt,$$
and
$$\int_0^x \int_t^x f(t)g(y-t)\, dy\, dt = \int_0^x \int_0^{x-t} f(t)g(y)\, dy\, dt = \int_0^x f(t)G(x - t)\, dt = (f * G)(x),$$
hence
$$\int_0^x (f * g)(y)\, dy = (f * G)(x).$$