I'm having trouble with this problem that envolves derivate of a integral. Take a look please.
$g(x)$ is a definite function such that $g(1)=6$. Consider the function: $$f(x) = \frac{1}{2}\int_{0}^{x}(x-t)^2g(t) dt$$ Find $f'''(1)$.
I've done this so far:
$$f'(x)=\frac{1}{2}[0.g(x) -x^2.g(0)]$$
$$f''(x)= -x.g(0)$$
$$f'''(x)=-g(0)$$
I'm in a dead end. I supose that I'm wrong because the exercise gave me $g(1)$ that I'm suppose to use.
$$\frac{1}{2}\int_{0}^{x}(x-t)^2g(t) dt = \frac{1}{2}\int_{0}^{x}(x^2-2xt+t^2)g(t) dt =$$
$$\frac{1}{2}x^2\int_{0}^{x} g(t)dt-x\int_{0}^{x} tg(t)dt+\frac{1}{2}\int_{0}^{x} t^2g(t)dt$$
Take derivative using product rule and Fundamental Theorem of calculus.