Mid-ordinate and Simpson's rulesAQA A-Level Further Maths: Revision notes

Section 1

Why numerical integration

Many integrands, such as $\frac{1}{1+x}$ in a measured data set or $\sqrt{1+x^3}$ , have no simple antiderivative, or only discrete data are known. Numerical integration approximates $\int_a^bf(x)\,dx$ by splitting $[a,b]$ into $n$ strips of equal width $h=\frac{b-a}{n}$ and adding approximate strip areas. The ordinates are $y_r=f(a+rh)$ for $r=0,1,\ldots,n$ .

Key termsstripordinatestrip width

Section 2

The mid-ordinate rule

The mid-ordinate rule replaces each strip by a rectangle whose height is the function value at the midpoint of the strip: $\int_a^bf(x)\,dx\approx h\left[y_{1/2}+y_{3/2}+\cdots+y_{n-1/2}\right],$ where $y_{r+1/2}=f\left(a+(r+\tfrac12)h\right)$ . Example. $\int_0^1\frac{1}{1+x}dx$ with $n=4$ , $h=0.25$ : the midpoints are $0.125,0.375,0.625,0.875$ and the estimate is $0.25(0.8889+0.7273+0.6154+0.5333)=0.6912$ . The exact value is $\ln2=0.6931$ , so this is an underestimate.

Key termsmid-ordinate rule

Common mistake

Using the strip boundaries instead of the midpoints. Mid-ordinates are at $a+\tfrac12h$ , $a+\tfrac32h$ , and so on.

Section 3

Simpson's rule

Simpson's rule fits a parabola through each pair of strips (three ordinates). For an even number $n$ of strips: $\int_a^bf(x)\,dx\approx\frac{h}{3}\Big[y_0+y_n+4(y_1+y_3+\cdots+y_{n-1})+2(y_2+y_4+\cdots+y_{n-2})\Big].$ The two end ordinates have coefficient 1, the odd-numbered ordinates 4, and the even-numbered interior ordinates 2. Example. $\int_0^1\frac{1}{1+x}dx$ with $n=4$ : $\frac{0.25}{3}\left[1+0.5+4(0.8+0.5714)+2(0.6667)\right]=0.6933$ , much closer to $0.6931$ than the mid-ordinate value.

Key termsSimpson's rule

Common mistake

Using an odd number of strips. Simpson's rule needs $n$ even, which means an odd number of ordinates.

Section 4

Working from data and checking

If only data at equal spacing are given (a river's depth, a velocity at 1 s intervals), apply the rule to the values directly. Simpson's rule uses the data at the strip boundaries. For the mid-ordinate rule, use wider strips so that the data points lie at the midpoints: values at $x=1,3,5$ give three strips of width 2. Give a unit with the area, and keep full calculator values until the final rounding.

Exam tip

Write the ordinates in a list first, then substitute into the formula. Check by counting that the coefficients are $1,4,2,4,\ldots,4,1$ .

Section 5

Accuracy and error

Accuracy improves as $h$ decreases (more strips). Simpson's rule is exact for cubics (or any polynomial of degree up to 3). For example $\int_0^2x^3dx=4$ is given exactly by Simpson's rule with two strips. The mid-ordinate rule underestimates when the function is convex ( $f''>0$ ) and overestimates when it is concave ( $f''<0$ ). Quote an error as a percentage: $\frac{|\text{estimate}-\text{exact}|}{\text{exact}}\times100$ .

Key termsconvexconcave

Exam tip

Justify over- or under-estimation using the sign of $f''$ , not just by comparing with the exact answer.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Mid-ordinate and Simpson's rules

The integral $I=\int_0^1\frac{1}{1+x}\,dx$ is to be estimated using four strips of equal width. Let $y_r=\frac{1}{1+0.25r}$ .
Use the mid-ordinate rule with four strips to estimate $I$ , to 4 decimal places.2 marks
The depth of a river is measured at 1 m intervals across its 6 m width. The depths, in metres, from one bank to the other are $0,\ 1.2,\ 2.0,\ 2.4,\ 1.8,\ 1.0,\ 0$ . The cross-sectional area is the area under the depth profile.
Use the mid-ordinate rule with three strips of width 2 m to estimate the cross-sectional area.2 marks
Consider $I=\int_0^2x^3\,dx$ .
Use Simpson's rule with two strips to estimate $I$ , and show that it equals the exact value of $I$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).