Ratio of areas under the curve of polynomials in an interval

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Let $g(x)=\dfrac{\int_0^xs^2ds}{\int_0^ns^2ds}$, and let $f(x)=\dfrac{\int_x^ns^2ds}{\int_0^ns^2ds}$. Note that $g(x)+f(x)=1$. Let $v=\{x:f(x)=g(x)\}$, which is just going to be a number and only depends on the numerator since $f(x),g(x)$ have exactly the same denominator. Then I am interested in the quantity $\alpha(n)=\dfrac{v}{n}$.

One can explicitly calculate $v$ (I use $x$ as variable) in terms of $n$. That yields the equation:

$$2x^3-6nx^2+6n^2x=n^3$$

which after plotting it here, we can see it's just a line, namely $v=\beta n$, then $\alpha(n)=\beta$.

That can be confirmed here in this animation. Furthermore, after playing with the animation, if instead of $\int x^2dx$ I consider $\int x^3dx$, the result still holds, namely $\alpha(n)=K$ is just some constant.

This defies my intuition, because I first I assumed the bigger the $n$, the bigger the $\alpha$, in other words, I thought $v$ grew than faster than $\beta n$, but apparently it doesn't.

My questions are:

  1. I think my analysis is correct, but I'd appreciate if someone can point out any mistake.
  2. Is there a more general result that deals with this? some theorem that says this is true for all polynomials or something like that?
  3. My intuition failed me, but is there another way where one can see this without going through all the analysis? in other words, a perspective that makes this fact more straightforward to see?

Edit: The equation I had was wrong. The actual equation is $2x^3=n^3$, as pointed out by @Andrei.