Question

Consider the functions

f(x) = xn and g(x) = xm
on the interval [0, 1], where m and n are positive integers and n > m. Find the centroid of the region bounded by f and g.

Answer 1

Assuming the region is of uniform density, compute the mass of the region and its moments.

If n > m, then in the interval [0, 1] we have f(x) ≤ g(x). This means the mass is

`M=\displaystyle\int_0^1(g(x)-f(x))\,\mathrm dx=\left((x^(m+1))/(m+1)-(x^(n+1))/(n+1)\right)\bigg|_0^1=\frac1{m+1}-\frac1{n+1}`

The moments are

`M_x=\displaystyle\int_0^1\frac{g(x)^2-f(x)^2}2\,\mathrm dx=\frac12\left((x^(2m+1))/(2m+1)-(x^(2n+1))/(2n+1)\right)\bigg|_0^1=\frac12\left(\frac1{2m+1}-\frac1{2n+1}\right)`

`M_y=\displaystyle\int_0^1x(g(x)-f(x))\,\mathrm dx=\int_0^1(x^(m+1)-x^(n+1))\,\mathrm dx=\left((x^(m+2))/(m+2)-(x^(n+2))/(n+2)\right)\bigg|_0^1=\frac1{m+2}-\frac1{n+2}`

Then the centroid is

`(\overline x,\overline y)=\left(\frac{M_y}M,\frac{M_x}M\right)=\boxed{\left(((m+1)(n+1))/((m+2)(n+2)),((m+1)(n+1))/((2m+1)(2n+1))\right)}`