Can anyone please explain what is geometric center of a graph $f(x)$?

Question

Can anyone please explain what is geometric center of a graph $f(x)$ ?

What will be the geometric center of $f(x) = x^2$ for all $x \in [0 , 3]$?

Answer 1

Think of this the curve as made out of a metallic wire, with uniform length density $\lambda$. Then a small piece, of length $dl$ has a mass $dm=\lambda dl$. The total mass of the wire is then $$M=\int dm=\lambda\int dl$$ Since we are given the curve in terms of $x$, we rewrite the above equation as $$M=\lambda\int_0^3\sqrt{1+[f'(x)]^2}dx$$ Then the center of mass in the $x$ direction is $$\bar x=\frac 1M\int x dm=\frac{\int_0^3x\sqrt{1+[f'(x)]^2}dx}{\int_0^3\sqrt{1+[f'(x)]^2}dx}$$ Similarly, $$\bar y=\frac 1M=\int ydm=\frac{\int_0^3x^2\sqrt{1+[f'(x)]^2}dx}{\int_0^3\sqrt{1+[f'(x)]^2}dx}$$ So calculate $f'(x)$, then the three integrals, and you get the center. Notice that the center is independent of the mass density, so it's a geometric property of the curve.