Simpson's Rule

Where a straight top falls short

The trapezium rule tops each strip with a straight chord. On a curve that bends, the chord cuts across the bend and leaves a sliver on every strip.

Take $y=2.2\sqrt{x}$y = 2.2√x from x = 0 to x = 4 as one strip. The chord from (0, 0) to (4, 4.4) is the line y = 1.1x, and the triangle under it has area $\left(\frac{1}{2}\right)\times 4\times 4.4=8.8$(1/2) × 4 × 4.4 = 8.8. The exact area is $2.2\times \left(\frac{2}{3}\right)\times 4^{\frac{3}{2}}=2.2\times \frac{16}{3}\approx 11.733$2.2 × (2/3) × 4^(3/2) = 2.2 × 16/3 ≈ 11.733. The curve bulges above the chord, so the chord misses almost 3 square units.

The curve $y=2.2\sqrt{x}$y = 2.2√x and the straight line y = 1.1x, which meet at (0, 0) and (4, 4.4). Between them, the curve bulges above the line, and that gap, about 2.93 square units, is what a single straight top leaves out.

Three points fix a parabola

A parabola $y=A{x}^2+Bx+C$y = Ax² + Bx + C has three coefficients, so three points fix it. Simpson’s rule takes the strips two at a time and tops each pair with the parabola through the curve at the pair’s left edge, middle and right edge.

Its area has a short form. Slide the pair so that its middle is at x = 0 and its edges are at −h and h, with heights $y_0$y₀, $y_1$y₁ and $y_2$y₂. The area under the parabola is the integral of $A{x}^2+Bx+C$Ax² + Bx + C from −h to h, which is $\left(\frac{2}{3}\right)A{h}^3+2Ch$(2/3)A h³ + 2C h, since the Bx part cancels. At the middle, $y_1=C$y₁ = C, and at the edges, $y_0+y_2=2A{h}^2+2C$y₀ + y₂ = 2A h² + 2C. So $\left(\frac{h}{3}\right)\left(y_0+4y_1+y_2\right)=\left(\frac{h}{3}\right)\left(2A{h}^2+6C\right)=\left(\frac{2}{3}\right)A{h}^3+2Ch$(h/3)(y₀ + 4y₁ + y₂) = (h/3)(2A h² + 6C) = (2/3)A h³ + 2C h, the same area.

One pair on $x^2$x² from 1 to 3 has h = 1 and heights 1, 4 and 9, so it gives $\left(\frac{1}{3}\right)\left(1+16+9\right)=\frac{26}{3}$(1/3)(1 + 16 + 9) = 26/3. The exact area is $\frac{27-1}{3}=\frac{26}{3}$(27 − 1)/3 = 26/3. The curve is itself a parabola, so the fitted top is the curve, and the rule is exact.

The gold curve $y=x^2$y = x², with the three points (1, 1), (2, 4) and (3, 9) that fix the parabola over one pair of strips from x = 1 to x = 3. The shaded area is $\left(\frac{1}{3}\right)\left(1+4\times 4+9\right)=\frac{26}{3}$(1/3)(1 + 4 × 4 + 9) = 26/3, exactly the integral.

The weights 1, 4, 2, 4, 1

With four strips there are two pairs and five heights, $y_0$y₀ to $y_4$y₄. The first pair gives $\left(\frac{h}{3}\right)\left(y_0+4y_1+y_2\right)$(h/3)(y₀ + 4y₁ + y₂) and the second gives $\left(\frac{h}{3}\right)\left(y_2+4y_3+y_4\right)$(h/3)(y₂ + 4y₃ + y₄). The height $y_2$y₂ is the right edge of one pair and the left edge of the next, so it is counted in both: $S=\left(\frac{h}{3}\right)\left(y_0+4y_1+2y_2+4y_3+y_4\right)$S = (h/3)(y₀ + 4y₁ + 2y₂ + 4y₃ + y₄).

With more pairs the same thing happens at every shared edge. The weights run 1, 4, 2, 4, 2, …, 2, 4, 1: the two ends 1, the middle of every pair 4, and every shared edge 2. All of it is multiplied by $\frac{h}{3}$h/3.

The strips must pair up, so their number must be even. With 7 strips one is left over, and no parabola can be fitted to it on its own.

Same strips, far smaller error

Estimate the integral of $\frac{1}{x}$1/x from 1 to 2, whose exact value is $\ln 2\approx 0.693147$ln 2 ≈ 0.693147. With 2 strips, h = 0.5 and the heights are 1, 0.6667 and 0.5. The trapezium rule gives 0.708333, too much by 0.0152. Simpson’s rule gives $\left(\frac{0.5}{3}\right)\left(1+4\times 0.6667+0.5\right)=0.694444$(0.5/3)(1 + 4 × 0.6667 + 0.5) = 0.694444, too much by 0.0013.

With 4 strips the trapezium rule gives 0.697024, off by 0.0039, and Simpson’s rule gives 0.693254, off by 0.00011. With 8 strips the errors are 0.00097 and 0.000007. Halving the strips cuts the trapezium error to a quarter each time. It cuts Simpson’s error by a factor of about 12 and then about 15, closing on 16 as the strips get thinner.

Simpson’s rule is exact for every cubic, not only for parabolas. On $x^3$x³ from 0 to 2, one pair with h = 1 gives $\left(\frac{1}{3}\right)\left(0+4\times 1+8\right)=4$(1/3)(0 + 4 × 1 + 8) = 4, and the exact area is $\frac{2^4}{4}=4$2⁴/4 = 4. On $x^4$x⁴ from 0 to 2 it gives $\left(\frac{1}{3}\right)\left(0+4+16\right)=\frac{20}{3}\approx 6.667$(1/3)(0 + 4 + 16) = 20/3 ≈ 6.667 against the exact $\frac{32}{5}=6.4$32/5 = 6.4.

The trapezium rule with 4 strips on $y=e^{-x^2}$y = e^(−x²) from 0 to 1.5: each chord cuts across the bend, an error of 0.0037. Switch to Simpson’s rule on the same five heights: two parabolas, each over a pair of strips, and the estimate is 0.8561 against the exact 0.8562, an error of 0.0001.

The usual mistakes

Using an odd number of strips. Simpson’s rule needs the strips in pairs, so 7 strips will not do; take another reading, or treat the odd strip with the trapezium rule.

Giving the end heights a weight of 4 or 2. They belong to one pair each and take weight 1.

Multiplying by $\frac{h}{2}$h/2. That is the trapezium rule’s multiplier; the area of a parabola over a pair gives $\frac{h}{3}$h/3.

Taking h as the width of a pair. h is the width of one strip, the gap between neighboring heights.

A river gauged from soundings

In the application below, nine depth soundings across a river make eight strips, an even number, so Simpson’s rule applies. Set beside it, the trapezium rule comes out smaller, because its straight chords cut inside a bed that dips away beneath them.