take the set of polynomials of the form $(a+b)^n$ and generalize them: let $P_f(a,b;n)$ be a sequence of polynomials where $f:(-c,c)\to \mathbb{R}^+$ is a function with $\int_{-c}^c f=1$ and $c$ can be infinity. further $P_f(a,b;1)=a+b$ and the polynomial always has degree $n$, for all $f$. the surface under $f$ describes the coefficients for the resulting polynomials. for example $f(x):={1\over \sqrt{2\pi}}e^{-x^2\over 2}$ and $c$ infinite, gives above $P_f(a,b;n)=(a+b)^n$.
now set $f$ to be the closed form expression of an upside-down cycloid shifted horizontally by half of its period-length and shifted up by $2r$, normalized to surface 1, and let $c$ be half the cycloid's period-length so that the cusps are excluded from the function. what is $P_f(a,b;n)$? what if the cycloid was further shifted up, so it doesn't touch the x axis? can further insights be gained when looking at it from a projective perspective (whith homogenous polynomials in 3 variables, where each term has equal degree)?
wikipedia says a cycloid is $x(t)=r(t-\sin t)$ and $y(t)=r(1-\cos t)$ and that the area under it is $3\pi r^2$. because of the cusps having infinite derivative, there is no closed form expression...
