$$\dfrac{1}{2}\int_{0}^{x}t^2g(t)dt$$
By theorem fundamental of calculus, $\int_{0}^xt^2g(t)=h(x)-h(0)$, $\dfrac{d(h(t))}{dt}=t^2g(t)$, but i dont know what h(x) satisfaes that.
Help pls
Edit: The exercise is
Let be $f(x)=\dfrac{1}{2}\int_{0}^{x}(x-t)^2g(t)dt$
Describe $\dfrac{d(f(x))}{dx}$ in terms of $\int_{0}^{x}g(t)dt$ and $\int_{0}^{x}tg(t)dt$.
What i did was separate the integral:
$f(x)=\dfrac{1}{2}(x^2\int_{0}^{x}g(t)dt-2x\int_{0}^{x}tg(t)dt+\int_{0}^{x}t^2g(t)$)
but an integral is $\dfrac{1}{2}\int_{0}^{x}t^2g(t)dt$
im correct?
The context matters quite a bit. In many applications and contexts, the answer $\frac{1}{2}\int_0^x t^2g(t) dt$ is perfectly satisfactory. If you want to change the form of the integral, consider using integration by parts to get a new expression.