DerivationIntegrationadvanced

Derivation: Simpson 1/3

From three equally spaced nodes to Simpson's 1, 4, 1 weights by integrating the quadratic Lagrange polynomial.

Weights of the interpolating parabola

  1. Take x0=ax_0=a, x1=a+b2x_1=\frac{a+b}{2} and x2=bx_2=b. Write h=b−a2h=\frac{b-a}{2} and use t=x−aht=\frac{x-a}{h}, so tt runs over [0,2][0,2].

    dx=h dtdx=h\,dt

  2. The quadratic bases in the variable t are:

    ℓ0(t)=(t−1)(t−2)2,ℓ1(t)=−t(t−2),ℓ2(t)=t(t−1)2\ell_0(t)=\frac{(t-1)(t-2)}{2},\quad \ell_1(t)=-t(t-2),\quad \ell_2(t)=\frac{t(t-1)}{2}

  3. Integrate each basis on [0,2] and multiply by h:

    h∫02ℓ0(t)dt=h3,h∫02ℓ1(t)dt=4h3,h∫02ℓ2(t)dt=h3h\int_0^2\ell_0(t)dt=\frac{h}{3},\quad h\int_0^2\ell_1(t)dt=\frac{4h}{3},\quad h\int_0^2\ell_2(t)dt=\frac{h}{3}

  4. Collecting the terms gives Simpson 1/3:

    ∫abf(x) dx≈h3[f(a)+4f ⁣(a+b2)+f(b)]\int_a^b f(x)\,dx\approx\frac{h}{3}\left[f(a)+4f\!\left(\frac{a+b}{2}\right)+f(b)\right]