A LevelFurther MathsEdexcelFurther Pure 1Revision NotesFurther Numerical MethodsFurther Numerical MethodsSimpson's Rule

Simpson's Rule (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Exam code: 9FM0

Written by: Mark Curtis

Updated on 23 January 2026

Simpson's rule

What is Simpson's rule?

Simpson's rule is a method of numerical integration given by
- $\int_{a}^{b} f (x) d x \approx \frac{h}{3} [f (x_{0}) + 4 f (x_{1}) + 2 f (x_{2}) + 4 f (x_{3}) + . . . + 2 f (x_{n - 2}) + 4 f (x_{n - 1}) + f (x_{n})]$
- where
  - $n$ is the number of intervals
  - $n$ must be even
  - $x_{0} = a$ and $x_{n} = b$
  - $h$ is the width of an interval, $\frac{b - a}{n}$
The $n$ intervals mean there are $(n + 1)$ points
- $x_{0}$ to $x_{n}$

Increasing the number of intervals, $n$
- gives a better estimate of the area under the curve

Examiner Tips and Tricks

Simpson's rule is not given in the formula booklet, so it helps to remember it as

$\int_{a}^{b} f (x) d x \approx \frac{h}{3} [(ends) + 4 \times (odds) + 2 \times (evens)]$
- Just don't include $x_{0}$ and $x_{n}$ in the 'evens' (as they are the 'ends')!

How is Simpson's rule different to the trapezium rule?

The trapezium rule splits the area under a curve into vertical strips
- then connects the points on the curve with straight lines
  - to form trapezia
- but these can overestimate or underestimate the area of the strip
Simpson's rule connects the points on the curve with quadratics
- which match the shape of the curve more closely
  - allowing the strips to better estimate the area
In particular, quadratics are fitted across pairs of intervals
- The first quadratic goes through points 0, 1 and 2
  - $(x_{0}, f (x_{0}))$ , $(x_{1}, f (x_{1}))$ and $(x_{2}, f (x_{2}))$
- The second quadratic goes through points 2, 3 and 4, etc
  - $(x_{2}, f (x_{2}))$ , $(x_{3}, f (x_{3}))$ and $(x_{4}, f (x_{4}))$
- This is why the number of intervals, $n$ , must be even

The curve y=f(x) split into intervals from x_0 at x=a to x_n at x=b. The points with coordinates (x_0, f(x_0)), (x_1, f(x_1)) and (x_2, f(x_2)) are shown with a parabola joining them labelled 1st quadratic. The points x_3 and x_4 are shown with points 2, 3 and 4 being joined by another parabola, labelled 2nd quadratic.

The equation of the first quadratic works out to be
- $y = f (x_{1}) + \frac{f (x_{2}) - f (x_{0})}{2 h} (x - x_{1}) + \frac{f (x_{2}) - 2 f (x_{1}) + f (x_{0})}{2 h^{2}} {(x - x_{1})}^{2}$
- and the area under the first quadratic, $\int_{x_{0}}^{x_{2}} y d x$ , works out to be
  - $\frac{h}{3} (f (x_{0}) + 4 f (x_{1}) + f (x_{2}))$
- Summing all these quadratic areas gives Simpson's rule
  - $\frac{h}{3} (f (x_{0}) + 4 f (x_{1}) + f (x_{2})) + \frac{h}{3} (f (x_{2}) + 4 f (x_{3}) + f (x_{4})) + . . .$
  - $\frac{h}{3} [f (x_{0}) + 4 f (x_{1}) + 2 f (x_{2}) + 4 f (x_{3}) + . . .]$

Examiner Tips and Tricks

You do not need to learn the proof of Simpson's rule given here, you just need to be able to use the rule.

How do I use Simpson's rule?

The following worked example shows how to use Simpson's rule

Examiner Tips and Tricks

Correct final answers with no working score no marks! Examiners want to see:

Values presented in a table
Calculations showing the correct form of Simpson's rule

Worked Example

Use Simpson’s rule with 6 intervals to find an estimate for $\int_{0}^{1.2} e^{- x^{2}} d x$ .

Give your answer to 3 significant figures.

Answer:

Find $h$ , the interval width, using $\frac{b - a}{n}$

$h = \frac{1.2 - 0}{6} = 0.2$

Create a table of values

It helps to include a row for $n = 0$ to $n = 6$ at the top to identify 'odd' and 'even' points

$n$	0	1	2	3	4	5	6
$x$	0	0.2	0.4	0.6	0.8	1.0	1.2
$f (x)$	$e^{- 0^{2}}$	$e^{- 0 . 2^{2}}$	$e^{- 0 . 4^{2}}$	$e^{- 0 . 6^{2}}$	$e^{- 0 . 8^{2}}$	$e^{- 1 . 0^{2}}$	$e^{- 1 . 2^{2}}$
	1	0.96078...	0.85214...	0.69767...	0.52729...	0.36787...	0.23692...

Substitute these values into Simpson's rule, $\frac{h}{3} [(ends) + 4 \times (odds) + 2 \times (evens)]$

$\int_{0}^{1.2} e^{- x^{2}} d x \approx \frac{0.2}{3} [(e^{- 0^{2}} + e^{- 1 . 2^{2}}) + 4 \times (e^{- 0 . 2^{2}} + e^{- 0 . 6^{2}} + e^{- 1 . 0^{2}}) + 2 \times (e^{- 0 . 4^{2}} + e^{- 0 . 8^{2}})]$

Work out this value

$\begin{array}{rcl} \int_{0}^{1.2} e^{- x^{2}} d x & \approx & \frac{0.2}{3} \times 12.10117 . . . \\ = & 0.80674 . . . \end{array}$

Give your final answer to 3 significant figures

$\begin{array}{rcl} \int_{0}^{1.2} e^{- x^{2}} d x & \approx & 0.807 \end{array}$ to 3 s.f.

What else could I be asked about Simpson's rule?

You need to know that Simpson's rule is generally more accurate than the trapezium rule
You need to know how to find the percentage error of your estimate if given the actual value
- using the formula
  - $percentage error = \frac{| estimate - actual |}{actual} \times 100 %$
- You will either be given the actual answer
  - or you need to calculate the definite integral yourself
- You may need to state if the estimate is
  - an underestimate
  - or an overestimate
You need to be able to comment on the accuracy
- by comparing the estimate's value to the actual value
  - e.g. calculate the percentage error to deduce if it's accurate or not
  - e.g. find how many decimal places the estimate is accurate to

Examiner Tips and Tricks

If the actual answer is not given to you in the question, you will need to find it using integration.

Worked Example

Simpson's rule with 4 intervals gives an estimate of $\int_{1}^{5} \frac{1}{x} d x$ as $\frac{73}{45}$ .

(a) Calculate the percentage error, to 2 significant figures.

(b) Comment on the accuracy of the estimate.

Answer:

(a)

The formula is $percentage error = \frac{| estimate - actual |}{actual} \times 100 %$

Find the actual value of the integral by integration

$\begin{array}{rcl} \int_{1}^{5} \frac{1}{x} d x & = & {[\ln | x |]}_{1}^{5} \\ = & \ln 5 - \ln 1 \\ = & \ln 5 \end{array}$

Substitute the estimate and the actual answer into the formula

$\frac{|\frac{73}{45} - \ln 5|}{\ln 5} \times 100$

Work out this value

$0.7943338 . . .$

Round to 2 significant figures

The percentage error is 0.79% to 2 s.f.

(b)

It helps to write out $\frac{73}{45}$ and $\ln 5$ as decimals

$\begin{array}{rcl} \frac{73}{45} & = & 1.6222222 . . . \\ \ln 5 & = & 1.6094379 . . . \end{array}$

Now write a comment about accuracy, giving a reason

Any one out of:

The percentage error is very small (0.79%), so the estimate is very accurate
The estimate is accurate to 1 decimal place (2 significant figures)

Examiner Tips and Tricks

When commenting on accuracy, mark schemes do not like the phrase "is very close to". You must refer to numbers instead, such as the percentage error or number of significant figures.

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Simpson's Rule (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Simpson's rule

What is Simpson's rule?

How is Simpson's rule different to the trapezium rule?

How do I use Simpson's rule?

What else could I be asked about Simpson's rule?

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Mark Curtis