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A barycentric coordinate system assigns to a point in the affine span of the vertices of a nondegenerate simplex its unique vertex weights ${\lambda }_0,\dots ,{\lambda }_d$, constrained to sum to one. Uniqueness follows because $v_1-v_0,\dots ,v_d-v_0$ form a basis of the translation space: expand $x-v_0$ in that basis and set ${\lambda }_0=1-{\sum }_{i=1}^d{\lambda }_i$. A point lies in the closed simplex exactly when every weight is nonnegative. An affine function satisfies $f(x)={\sum }_i{\lambda }_{i}f(v_i)$, which makes these coordinates useful for interpolation and the integral of an affine function over a simplex.

For a nondegenerate $d$-dimensional simplex $S$ with vertices $v_0,\dots ,v_d$ and an affine function $f$, the integral equals the volume times the value at the centroid $c=(v_0+\cdots +v_d)/(d+1)$. To prove this, write $x={\sum }_i{\lambda }_i(x)v_i$ in a barycentric coordinate system. On the standard simplex, symmetry permuting the vertices gives equal integrals of all ${\lambda }_i$; since ${\sum }_i{\lambda }_i=1$, each is $\mathrm{vol}(S)/(d+1)$. An invertible affine map transfers this result to any simplex. Affineness gives $f(x)={\sum }_i{\lambda }_i(x)f(v_i)$, so linearity of integration proves the formula. For a tetrahedron the factor is one quarter. The rule is exact for affine integrands; a nonlinear integrand generally needs additional moments.