Numerical Integration

An area with no formula

The curve $y=e^{-x^2}$y = e^(−x²) is the bell shape behind the normal distribution. No combination of powers, exponentials, logarithms and trigonometric functions differentiates to $e^{-x^2}$e^(−x²), so its integral cannot be worked by finding an antiderivative.

The area under it from x = −1.5 to x = 1.5 is still a definite number, about 1.7124. It can be estimated as closely as needed by cutting the region into strips and adding their areas.

The gold curve $y=e^{-x^2}$y = e^(−x²), shaded from x = −1.5 to x = 1.5. The curve is 1 high at x = 0 and about 0.105 high at each end of the shading. The shaded area is about 1.7124.

Strips with slanted tops

Split the interval from a to b into n strips of equal width $h=\frac{b-a}{n}$h = (b − a)/n. Over each strip, join the two points of the curve at its edges with a straight line. The strip becomes a trapezium, whose area is its width times the average of its two parallel sides: h × (left height + right height) ÷ 2.

For the bell from −1.5 to 1.5 with n = 3, h = 1 and the heights are taken at x = −1.5, −0.5, 0.5 and 1.5. They are $e^{-2.25}\approx 0.1054$e^(−2.25) ≈ 0.1054, $e^{-0.25}\approx 0.7788$e^(−0.25) ≈ 0.7788, 0.7788 and 0.1054. The three trapezia have areas 0.4421, 0.7788 and 0.4421, which add to 1.6630.

Three strips, each 1 wide, under the gold curve $y=e^{-x^2}$y = e^(−x²). Each top is the straight line between the curve’s heights at the strip’s two edges. Together they cover 1.6630, short of the true 1.7124.

The formula

Number the heights $y_0$y₀, $y_1$y₁, …, $y_n$yₙ from left to right. Adding the strips gives $\left(\frac{h}{2}\right)\left(y_0+y_1\right)+\left(\frac{h}{2}\right)\left(y_1+y_2\right)+\cdots +\left(\frac{h}{2}\right)\left(y_{n-1}+y_n\right)$(h/2)(y₀ + y₁) + (h/2)(y₁ + y₂) + … + (h/2)(yₙ₋₁ + yₙ). Every inner height is the right side of one strip and the left side of the next, so it appears twice, while $y_0$y₀ and $y_n$yₙ appear once each.

That is the trapezium rule: $T=\left(\frac{h}{2}\right)\left(y_0+2y_1+2y_2+\cdots +2y_{n-1}+y_n\right)$T = (h/2)(y₀ + 2y₁ + 2y₂ + … + 2yₙ₋₁ + yₙ). For the bell with n = 3 it gives $\left(\frac{1}{2}\right)\left(0.1054+2\times 0.7788+2\times 0.7788+0.1054\right)=\left(\frac{1}{2}\right)\left(3.3260\right)=1.6630$(1/2)(0.1054 + 2 × 0.7788 + 2 × 0.7788 + 0.1054) = (1/2)(3.3260) = 1.6630, as before.

n strips need n + 1 heights. With heights 0, 1 and 4 at h = 1, the rule gives $\left(\frac{1}{2}\right)\left(0+2\times 1+4\right)=3$(1/2)(0 + 2 × 1 + 4) = 3.

Thinner strips

More strips follow the curve more closely. For the bell from −1.5 to 1.5: with 3 strips the estimate is 1.6630, 0.0494 short; with 6 strips it is 1.6994, 0.0130 short; with 8 it is 1.7050; with 12 it is 1.7091, 0.0033 short; with 24 it is 1.7116, 0.0008 short.

Each time the strips are halved in width the error falls to about a quarter: 0.0494 to 0.0130, then 0.0130 to 0.0033, then 0.0033 to 0.0008. The gap between a chord and a curve shrinks with the square of the strip width.

Eight strips, each 0.375 wide, under the gold curve $y=e^{-x^2}$y = e^(−x²). Their tops lie almost on the curve, and together they cover 1.7050, short of 1.7124 by 0.0074.

Too much or too little

Which way the rule errs depends on the bend. Where a curve bends upward, each chord lies above the curve, so each trapezium holds a sliver too much and the rule overestimates. For $x^2$x² from 0 to 2 with h = 1, the heights are 0, 1 and 4, the rule gives 3, and the exact area is $\frac{8}{3}$8/3, so the error is $\frac{1}{3}$1/3 too much. With h = 0.5 the heights are 0, 0.25, 1, 2.25 and 4, the rule gives 2.75, and the error is $\frac{1}{12}$1/12, a quarter of $\frac{1}{3}$1/3.

Where a curve bends downward, each chord lies below it and the rule underestimates. For $\sqrt{x}$√x from 0 to 4 with h = 1 the heights are 0, 1, 1.4142, 1.7321 and 2, and the rule gives 5.1463, short of the exact $\frac{16}{3}\approx 5.3333$16/3 ≈ 5.3333.

The bell bends downward in its middle, between x = −0.707 and x = 0.707, and upward outside that. The middle strips carry most of the area, so their shortfall outweighs the excess in the outer strips, and every estimate above came out low.

Two strips, each 1 wide, under the gold curve $y=x^2$y = x² from 0 to 2. The curve bends upward, so each straight top runs above it. The strips cover 3, which is $\frac{1}{3}$1/3 more than the exact $\frac{8}{3}$8/3.

The trapezium rule with 4 strips on $y=e^{-x^2}$y = e^(−x²) from 0 to 1.5. The estimate is 0.8525 against the exact 0.8562, an error of 0.0037, tinted between the chords and the curve. Drag n to 8: the estimate becomes 0.8553 and the error 0.0009, about a quarter.

The usual mistakes

Adding the heights as they are. With heights 0, 1 and 4 at h = 1 that gives 5; the inner height counts twice and the bracket is then multiplied by $\frac{h}{2}$h/2, which gives 3.

Halving without doubling. $\left(\frac{1}{2}\right)\left(0+1+4\right)=2.5$(1/2)(0 + 1 + 4) = 2.5 forgets that the inner height belongs to two strips.

Doubling the end heights as well. $y_0$y₀ and $y_n$yₙ each belong to one strip only.

Taking h as the width divided by the number of heights. Four strips from 0 to 2 have h = 0.5 and five heights.

Five readings across a roof

In the application below, the trapezium rule runs on five readings of a curved roof taken 4 meters apart. It halves the two readings at the walls, which a plain average does not, and so comes much nearer the true mean height.