We have a parabola
y=f(x)=4-x^2.
The zeroes of f(x) are located at x=-2, and x=+2 (see attached graph).
The centroid is defined as
[tex]y0=\frac{\int_{-2}^2{f(x)*(y/2)}dx}{\int_{-2}^2{f(x)}dx}[/tex]
[tex]=\frac{\int_{-2}^2{(4-x^2)^2/2}dx}{\int_{-2}^2{4-x^2}dx}[/tex]
[tex]=\frac{\int_{-2}^2{(4+x^4/2-4x^2)}dx}{\int_{-2}^2{4-x^2}dx}[/tex]
[tex]=\frac{[(8x+x^5/10-4x^3/3)]_{-2}^2}{[4x-x^3/3]_{-2}^2}[/tex]
[tex]=\frac{[(32+32/5-64/3)]}{[16-16/3]}[/tex]
[tex]=\frac{[(256/15)]}{[32/3]}[/tex]
[tex]=\frac{8}{5}[/tex]
=1.6
Or more precisely, the centroid is at C(0,1.6)
