Theorem. If f(x) is a polynomial of degree 3 or less, then

Proof. We first prove the theorem for the case of an interval from x = -h to x = h, the mid-point being at x = 0. Letting

            f(x) = Ax³ + Bx² + Cx + D

we find, after going through the algebra, that

We will now evaluate the expression

and find that it evaluates to

which will constitute the proof.

Adding these and multiplying by , which in this case is equal to 2h/6 or h/3, we obtain the result:

This proves the theorem for the case of an interval that is bisected by the origin. The proof for the general case follows immediately because, by letting x = x' + ½ (a + b), we can translate the axes so that the new origin is at the mid-point of the interval. The equation y = Ax³ + Bx² + Cx + D takes the form y = Ax³ + B'x² + C'x + D', and the new interval is of the type for which we have just proved the theorem.