Can we use $3$ arbitary nodes for numerical integration? A function $f(x)$ is given at $3$ arbitary nodes lying in the interval $[a,b]$.

Can we find an approximation of $\int_a^b f(x) dx$ using a formula containing all the $3$ nodes ?
Or do we have to use a formula with $2$ nodes twice ?

The simpson-rule requires equidistant nodes and methods like Gauss-Legendre or Gauss-Tchebycheff have fixed nodes, so the only method I can think of is to construct the parabola containing the $3$ given points and integrating it.
 A: You can approximate the function with a parabola, $y=ax^2+bx+c$, through any three distinct nodes, and integrate the quadratic polynomial.  The nodes do not have to be equally spaced to make the approximation.  Simpson's Rule is just a special case where the nodes are uniformly spaced.
You do lose some accuracy by having your three nodes unequally spaced, compared with having them equally spaced with Simpson's Rule.  That is why mathematicians prefer the equal spacing when that option is available.
Say your function is $y=x^3$ and you integrate from $x=0$ to $x=3$, but you are allowed only to use the nodes $(0,0)$, $(1,1)$ and $(3,27)$.  The parabola through these nodes is $y=4x^2-3x$.  You integrate the latter function form $x=0$ to $x=3$ and you get $22.5$.  Two applications of the trapezoid rule with the given nodes would give you $28.5$ whereas the exact value with the $y=x^3$ function is $20.25$.  The parabola is the more accurate choice between these two.  However, if you could have used $(1.5, 3.375)$ and applied Simpson's Rule with the equal node spacing, you would have hit $20.25$ exactly (Simpson's Rule is exact for cubic polynomials).  The unequal-node parabola is generally between the trapezoidal and Simpson rules in accuracy.
