Be $g(x)=x^3-3x+1$ I have to draw the graph of $f(x)=g(4\int_0^x e^{-y^2}dy)$
Because materially calculate the integral is impossible (this is what I understand), I thought I could simply replace the integral in the first expression of g(x) and then derive with respect to x this new expression of g until order 3. However I don't know if derivation is correct, because when I derived I considered the integral like his derivative was the expression that appears in the integral.
At the third order of derivation I obtain $$g'''\left (4\int_0^x e^{-y^2}dy\right )=f'''(x)=6$$
And then I integrate f'''(x) three times in order to get f(x)
Something seems to be wrong, probably the entire procedure. Thanks for your time!