Applying the Fundamental Theorem of Calculus with derivatives

Question

Following the next exercise.
Given a function $g$ continuous for all $x$ which $g(1)=5$ and $\int_{0}^{1}g(t)dt=2$. Let $f(x)=\frac{1}{2} \int_{0}^{x}(x-t)^2g(t)dt$. Prove that: $$f'(x) = x \int_{0}^{x}g(t)dt - \int_{0}^{x}g(t)dt$$
Proof (attempt): First of all, we have that $f(x) = [...]$ so, by the product rule for derivatives we've got: $$f' (x)= (0)(f(x)) + {\frac{1}{2}}{\int_{0}^{x}(x-t)^2g(t)dt}'$$ Then, I was trying to apply integration by parts to compute the integral or at least leave it in terms of g, but that led me to nothing. So, I don't know what to do. If anyone can help me a little bit, I'd be very grateful.

Answer 1

$$ f(x) = \frac12\int_0^x(x - t)^2 g(t)\,dt = \frac{x^2}2\int_0^x g(t)\,dt - x\int_0^x tg(t)\,dt + \frac12\int_0^x t^2g(t)\,dt, $$ $$f'(x) = x\int_0^x g(t)\,dt - \int_0^x tg(t)\,dt.$$