[by Chris N. and Andrew K., 11/14/2002]
[with minor editing by Jimmy L. and Mr. Hansen]

Prove (2M + T)/3 = Simpson’s Rule:

Let y = f(x) be continuous on a closed interval [a, b].
Let n = no. of trapezoids = no. of rectangles, with n Î N, n ³ 2. (Problem is uninteresting for smaller n.)
Let M = the midpoint Riemann sum using midpoints x₁, x₃, x₅, . . . , x_2n–1
Let T = the sum of the Trapezoidal Rule using mesh points x₀, x₂, x₄, . . . , x_2n

Note: Here, width (horizontal step size) of trapezoids and midpoint rectangles is 2Dx. For Simpson’s Rule we use the more conventional mesh size of Dx.

M = 2Dx(y₁ + y₃ + y₅ + . . . + y_2n–1)

T = 2Dx(½ y₀ + y₂ + y₄ + . . . + ½ y_2n) = Dx(y₀ + 2y₂ + 2y₄ + . . . + y_2n)

Simpson’s Rule  = 1/3 · Dx(y₀ + 4y₁ + 2y₂ + 4y₃ + 2y₄ + 4y₅ + . . . + y_2n)
                          = 1/3 (Dx(y₀ + 2y₂ + 2y₄ + . . . + y_2n) + 4Dx(y₁ + y₃ + y₅ + . . . + y_2n–1))
                          = 1/3 (T + 2M)
                          = (2M + T)/3, Q.E.D.