Numerical Integration Methods: Learn It 3

Simpson’s Rule

We have seen how the midpoint rule approximates curves with rectangles and the trapezoidal rule uses straight lines. Simpson’s rule takes this progression further by approximating curves with parabolas (quadratic functions).

This method uses piecewise quadratic functions to follow the curve more accurately than either rectangles or straight-line segments.

How Simpson’s Rule Works

The setup requires partitioning the interval into an even number of subintervals, each with equal width. This even number requirement is essential because we work with pairs of subintervals.

For each pair of subintervals: Over the first pair of subintervals, we approximate [latex]\int_{x_0}^{x_2} f(x)dx[/latex] with [latex]\int_{x_0}^{x_2} p(x)dx[/latex], where [latex]p(x) = Ax^2 + Bx + C[/latex] is the unique quadratic function passing through the three points [latex](x_0, f(x_0))[/latex], [latex](x_1, f(x_1))[/latex], and [latex](x_2, f(x_2))[/latex].

Continuing the pattern: Over the next pair of subintervals, we approximate [latex]\int_{x_2}^{x_4} f(x)dx[/latex] with the integral of another quadratic function passing through [latex](x_2, f(x_2))[/latex], [latex](x_3, f(x_3))[/latex], and [latex](x_4, f(x_4))[/latex]. This process continues with each successive pair of subintervals.

This figure has two graphs, both of the same non-negative function in the first quadrant. The function increases and decreases. The quadrant is divided into a grid. The first graph, beginning on the x-axis at the point labeled x sub 0, there are trapezoids shaded whose heights are represented by the function p(x), which is a curve following an approximate path of the original graph. The x-axis is scaled by increments of x sub 0, x sub 1, x sub 2. The second graph has on the x-axis at the point labeled x sub 0. There are shaded regions under the curve, divided by x sub 0, x sub 1, x sub 2, x sub 3, and x sub 4. The curve is sectioned into two different parts above the shaded areas. These two parts are labeled p sub 1(x) and p sub 2(x).
Figure 4. With Simpson’s rule, we approximate a definite integral by integrating a piecewise quadratic function.
Why Quadratics? Three points determine a unique parabola, just like two points determine a unique line. Using quadratic approximations captures the curvature of functions much better than straight-line approximations.

Deriving Simpson’s Rule Formula

To understand the formula for Simpson’s rule, we need to derive the approximation over the first two subintervals. This derivation requires keeping track of several key relationships.

The quadratic function [latex]p(x) = Ax^2 + Bx + C[/latex] must pass through our three points:

[latex]\begin{array}{c}f(x_0) = p(x_0) = Ax_0^2 + Bx_0 + C\\ f(x_1) = p(x_1) = Ax_1^2 + Bx_1 + C\\ f(x_2) = p(x_2) = Ax_2^2 + Bx_2 + C\end{array}[/latex]

We also need these geometric facts:

  • [latex]x_2 - x_0 = 2\Delta x[/latex], where [latex]\Delta x[/latex] is the length of a subinterval
  • [latex]x_2 + x_0 = 2x_1[/latex], since [latex]x_1 = \frac{(x_2 + x_0)}{2}[/latex]

We can now integrate the quadratic approximation:

[latex]\begin{array}{ccccc}\hfill {\displaystyle\int _{{x}_{0}}^{{x}_{2}}f(x)dx}& \approx {\displaystyle\int _{{x}_{0}}^{{x}_{2}}p(x)dx}\hfill & & & \\ & ={\displaystyle\int _{{x}_{0}}^{{x}_{2}}(A{x}^{2}+Bx+C)dx}\hfill & & & \\ & =\frac{A}{3}{x}^{3}+\frac{B}{2}{x}^{2}+Cx|\begin{array}{c}{}^{{x}_{2}} \\ {}_{{x}_{0}}\end{array}\hfill & & & \text{Find the antiderivative.}\hfill \\ & =\frac{A}{3}({x}_{2}{}^{3}-{x}_{0}{}^{3})+\frac{B}{2}({x}_{2}{}^{2}-{x}_{0}{}^{2})+C({x}_{2}-{x}_{0})\hfill & & & \text{Evaluate the antiderivative.}\hfill\end{array}[/latex]

After factoring and using algebraic identities:

[latex]\begin{array}{ccccc} & =\frac{A}{3}({x}_{2}-{x}_{0})({x}_{2}{}^{2}+{x}_{2}{x}_{0}+{x}_{0}{}^{2})  +\frac{B}{2}({x}_{2}-{x}_{0})({x}_{2}+{x}_{0})+C({x}_{2}-{x}_{0})\hfill & & & \\ & =\frac{{x}_{2}-{x}_{0}}{6}(2A({x}_{2}{}^{2}+{x}_{2}{x}_{0}+{x}_{0}{}^{2})+3B({x}_{2}+{x}_{0})+6C)\hfill & & & \text{Factor out}\frac{{x}_{2}-{x}_{0}}{6}.\hfill\end{array}[/latex]

Through careful algebraic manipulation and substitution of our relationships, this eventually simplifies to:

[latex]\int_{x_0}^{x_2} f(x)dx = \frac{\Delta x}{3}(f(x_0) + 4f(x_1) + f(x_2))[/latex]

When we apply the same method to approximate [latex]\int_{x_2}^{x_4} f(x)dx[/latex], we get:

[latex]\int_{x_2}^{x_4} f(x)dx \approx \frac{\Delta x}{3}(f(x_2) + 4f(x_3) + f(x_4))[/latex]

Combining these approximations over the interval [latex][x_0, x_4][/latex]:

[latex]\int_{x_0}^{x_4} f(x)dx = \frac{\Delta x}{3}(f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + f(x_4))[/latex]

Notice the Pattern: The coefficients follow a specific pattern: 1, 4, 2, 4, 1. Interior points where parabolas meet get coefficient 2, while other interior points get coefficient 4

This pattern continues as we add pairs of subintervals, leading us to the general Simpson’s rule formula.

Simpson’s Rule

Assume that [latex]f\left(x\right)[/latex] is continuous over [latex]\left[a,b\right][/latex]. Let n be a positive even integer and [latex]\Delta x=\frac{b-a}{n}[/latex]. Let [latex]\left[a,b\right][/latex] be divided into [latex]n[/latex] subintervals, each of length [latex]\Delta x[/latex], with endpoints at [latex]P=\left\{{x}_{0},{x}_{1},{x}_{2},\ldots ,{x}_{n}\right\}[/latex].

Set

[latex]{S}_{n}=\frac{\Delta x}{3}\left(f\left({x}_{0}\right)+4f\left({x}_{1}\right)+2f\left({x}_{2}\right)+4f\left({x}_{3}\right)+2f\left({x}_{4}\right)+\cdots +2f\left({x}_{n - 2}\right)+4f\left({x}_{n - 1}\right)+f\left({x}_{n}\right)\right)[/latex].

Then,

[latex]\underset{n\to \text{+}\infty }{\text{lim}}{S}_{n}={\displaystyle\int }_{a}^{b}f\left(x\right)dx[/latex].