# Is the integral of a linear function over a triangle always the average of the vertices?

Say I have a triangle with vertices$\ r_0$,$\ r_1$and$\ r_2$. At each vertex the value of some function$\ f(r)$ is known. I know that I can use barycenter coordinates to describe any point$\ t$ in the triangle with $$\ t(u,v) = (1 - u - v)r_0 + ur_1 + vr_2$$

If the function$\ f(r)$ can be assumed linear than any value of$\ f(r)$ can be approximated by $$\ f(r)=(1-u-v)f(r_0)+uf(r_1)+vf(r_2)$$ If I integrate$\ f(r)$ over a triangle using $$I=\int f(r)dr = 2A\int_{0}^1\int_{0}^{1-u}((1-u-v)f(r_0)+uf(r_1)+vf(r_2))dvdu$$ I get that $$\ I=\frac{A(f(r_0)+f(r_1)+f(r_2))}{3}$$ This result seems overly simple. I was looking at a simple pressure distribution across a triangle and wanted the total pressure on the surface triangle. I know the pressure at the corners and the area. Is this the correct approach and result (assuming the linear assumption is valid)?

• I'm not familiar with barycentric coordinates myself, but as $f$ is linear, I don't find your result that surprising – Shuri2060 Jul 12 '17 at 19:00

If $f$ is linear, then it is completely determined by its value on the vertices. The barycentric expression that you have for the value of $f$ on the triangle isn’t just an approximation—it’s exact.
As for the integral of $f$, think of the function as giving the water level in a triangular basin in which the water is sloshed over toward some of the vertices. The integral of $f$ over this region is just the total water volume. The same volume is given by the average water level times the area of the basin, but that average water level is what you’d get when the water was perfectly still. Linearity ensures that this average water level will be a simple average of the “sloshed” levels at the vertices.