Simpson's rule for numerical integration in QCE Specialist Mathematics Unit 4

Syllabus dot point

“Use Simpson's rule to approximate the value of a definite integral or an area, applying the rule with an even number of subintervals and recognising when a numerical method is required”

QCESpecialist MathematicsUnit 4: Further calculus, and…8 min readUpdated 2 Jun 2026

Quick answer

Simpson's rule approximates $\int_a^b f(x)\, dx \approx \frac{h}{3}[f(x_0) + 4f(x_1) + 2f(x_2) + \cdots + 4f(x_{n-1}) + f(x_n)]$ , where $h = \frac{b-a}{n}$ and $n$ must be even. The coefficient pattern is $1, 4, 2, 4, \dots, 4, 1$ : ends weight $1$ , odd interior points weight $4$ , even interior points weight $2$ . It fits parabolas, is exact for cubics, and is used when an exact antiderivative is unavailable or the integrand is given as data.

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What this dot point is asking
The answer
Exam-style questions

What this dot point is asking

QCAA wants you to approximate a definite integral numerically when an exact antiderivative is unavailable or impractical, using Simpson's rule. The assessable skill is applying the rule correctly: setting up an even number of subintervals, computing the function values, and applying the alternating $4$ and $2$ weighting pattern accurately.

The answer

When a numerical method is needed

Some integrands have no elementary antiderivative, for example $e^{-x^2}$ , and some are given only as tabulated data. In these cases you approximate the definite integral numerically. Simpson's rule fits parabolic arcs through groups of three points, giving a more accurate estimate than the trapezoidal rule for smooth functions.

The strip width

Divide $[a, b]$ into $n$ equal subintervals, where $n$ must be even. The strip width is

h = \frac{b - a}{n},

and the sample points are $x_0 = a$ , $x_1 = a + h$ , up to $x_n = b$ .

Simpson's rule

The approximation is

\int_a^b f(x)\, dx \approx \frac{h}{3}\Big[ f(x_0) + 4 f(x_1) + 2 f(x_2) + 4 f(x_3) + \cdots + 4 f(x_{n-1}) + f(x_n) \Big].

The end values $f(x_0)$ and $f(x_n)$ have weight $1$ . Interior values alternate: odd-indexed points get weight $4$ , even-indexed interior points get weight $2$ . The whole bracket is multiplied by $\frac{h}{3}$ .

The weighting pattern

The pattern of coefficients is $1, 4, 2, 4, 2, \dots, 4, 1$ . The first and last are $1$ , every odd-position interior value is $4$ , every even-position interior value is $2$ , and the values at odd positions always outnumber those at even positions. This pattern is why $n$ must be even: the points group into pairs of strips, each capped by a parabola.

Why it is accurate

Because Simpson's rule fits parabolas rather than straight lines, it is exact for polynomials up to degree three and very accurate for smooth curves. Halving $h$ reduces the error sharply, so a modest number of strips usually suffices.

Reading off function values

Tabulate the points $x_0, x_1, \dots, x_n$ and the corresponding $f(x_i)$ before applying the rule. Organising the values in a table with their weights prevents the most common arithmetic slips.

Working from tabulated data

In many applied questions the function is never given as a formula at all; instead you are handed a table of measured values, such as river depths or terrain heights at equal spacings. Simpson's rule applies directly: the spacing between readings is $h$ , the number of strips is one fewer than the number of readings, and you must check that this strip count is even before proceeding. If a data set has an odd number of strips, Simpson's rule cannot be applied to the whole set, and you either drop to a different method or combine Simpson's rule on most strips with a trapezoidal strip at the end.

Areas, volumes and symmetry

Because a definite integral can represent an area or, via $\pi\int y^2\,dx$ , a volume of revolution, Simpson's rule estimates these too when no formula integration is possible. Symmetry is a useful shortcut: if a region is symmetric about an axis, apply the rule to one half and double, which halves the arithmetic. The dam-surface question above uses exactly this idea, integrating the upper half of an ellipse and doubling for the lower half.

Accuracy and error behaviour

Simpson's rule is exact for any polynomial of degree three or less, because a cubic is captured perfectly by the parabolic fitting over paired strips. For smoother functions the error decreases rapidly as the strip width shrinks; roughly, halving $h$ cuts the error by a factor of about sixteen. This is far better than the trapezoidal rule, whose error only falls by a factor of four under the same refinement, which is why Simpson's rule is the preferred numerical method when the data spacing permits an even strip count.

Strip count rule

The number of strips is one less than the number of function values, and it must be even for Simpson's rule. Seven readings give six strips; five readings give four strips.

Worked example

Use Simpson's rule with $n = 4$ to approximate $\displaystyle\int_0^1 e^{-x^2}\, dx$ .

The integrand $e^{-x^2}$ has no elementary antiderivative, so numerical integration is required. Four strips gives $n = 4$ (even), and we apply the $1, 4, 2, 4, 1$ weighting pattern.

Step 1: Find the strip width and the sample points

Divide the interval $[0, 1]$ into $n = 4$ equal strips:

h = \frac{1 - 0}{4} = 0.25.

The five sample points are $x_0 = 0$ , $x_1 = 0.25$ , $x_2 = 0.5$ , $x_3 = 0.75$ , $x_4 = 1$ .

Step 2: Evaluate the function at each point

Compute $f(x) = e^{-x^2}$ at each sample point:

$f(x_0) = e^0 = 1$ ; $\quad f(x_1) = e^{-0.0625} \approx 0.939413$ ; $\quad f(x_2) = e^{-0.25} \approx 0.778801$ ; $\quad f(x_3) = e^{-0.5625} \approx 0.569783$ ; $\quad f(x_4) = e^{-1} \approx 0.367879$ .

Step 3: Apply Simpson's rule with weights $1, 4, 2, 4, 1$

The end values carry weight $1$ , the odd-indexed interior values carry weight $4$ , and the even-indexed interior value carries weight $2$ :

\int_0^1 e^{-x^2}\, dx \approx \frac{0.25}{3}\Big[ 1 + 4(0.939413) + 2(0.778801) + 4(0.569783) + 0.367879 \Big].

Step 4: Compute the weighted sum

Apply the weights to each value and add:

$4(0.939413) = 3.757652$ ; $\quad 2(0.778801) = 1.557602$ ; $\quad 4(0.569783) = 2.279132$ .

Sum: $1 + 3.757652 + 1.557602 + 2.279132 + 0.367879 = 8.962265$ .

Step 5: Multiply by $h/3$

\frac{0.25}{3} \times 8.962265 \approx 0.083333 \times 8.962265 \approx 0.74686.

Step 6: Check against the accepted value

The accepted value is about $0.74682$ , so the estimate is accurate to four decimal places with only four strips, illustrating why Simpson's rule is preferred over the trapezoidal rule.

Final answer: $\displaystyle\int_0^1 e^{-x^2}\, dx \approx 0.74686$ .

Common trap

The most penalised error is muddling the coefficient pattern, for example weighting an interior point $2$ when it should be $4$ or vice versa. Remember: ends are $1$ , odd-indexed interior points are $4$ , even-indexed interior points are $2$ . A second error is using an odd number of subintervals; Simpson's rule requires $n$ to be even so the strips pair up under parabolas. Always tabulate values with their weights before summing.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

Paper 2 (complex familiar). A river is

12

m wide. Depths (m) at

2

m intervals across the width are

0.52, 2.15, 3.70, 4.27, 3.32, 1.28, 0.59

. A biologist estimates the cross-sectional area as

15

^2

. Use Simpson's rule to evaluate the reasonableness of this estimate, justifying your decision.

Show worked answer →

Seven readings give $6$ strips of width $h = 2$ (even, so Simpson's rule applies). Weights $1, 4, 2, 4, 2, 4, 1$ :

Area $= \dfrac{h}{3}\big[f_0 + 4(f_1 + f_3 + f_5) + 2(f_2 + f_4) + f_6\big]$ $= \dfrac{2}{3}\big[0.52 + 4(2.15 + 4.27 + 1.28) + 2(3.70 + 3.32) + 0.59\big]$ $= \dfrac{2}{3}(45.95) \approx 30.6$ m $^2$ .

The estimate $30.6$ m $^2$ is about double the biologist's $15$ m $^2$ , so $15$ m $^2$ is not reasonable; it substantially underestimates the area.

Markers reward the correct strip count, the weighting pattern, the evaluated area, and a justified decision.

Practice4 marks

Paper 2 (complex familiar). The edge of a dam is modelled by the ellipse

\dfrac{(x - 2)^2}{16} + \dfrac{y^2}{9} = 1

for

0 \leq x \leq 6

, symmetric about the

x

-axis. Use Simpson's rule with four strips to approximate the surface area.

Show worked answer →

Upper half: $y = 3\sqrt{1 - \dfrac{(x - 2)^2}{16}}$ over $0 \leq x \leq 6$ ; double for symmetry. Four strips gives $h = \dfrac{6}{4} = 1.5$ .

Values: $y(0) \approx 2.598$ , $y(1.5) \approx 2.976$ , $y(3) \approx 2.905$ , $y(4.5) \approx 2.342$ , $y(6) = 0$ .

Upper area $= \dfrac{1.5}{3}\big[2.598 + 4(2.976 + 2.342) + 2(2.905) + 0\big] = 0.5(29.68) \approx 14.84$ km $^2$ . Total $= 2 \times 14.84 \approx 29.68$ km $^2$ .

Markers reward the half-curve setup, the strip width, the function values, and doubling for symmetry.

Practise this

Sources & how we know this

Specialist Mathematics General senior syllabus 2025 — QCAA

What this dot point is asking

The answer

When a numerical method is needed

The strip width

Simpson's rule

The weighting pattern

Why it is accurate

Reading off function values

Working from tabulated data

Areas, volumes and symmetry

Accuracy and error behaviour

Use Simpson's rule with n=4n = 4n=4 to approximate ∫01e−x2 dx\displaystyle\int_0^1 e^{-x^2}\, dx∫01​e−x2dx.

Step 1: Find the strip width and the sample points

Step 2: Evaluate the function at each point

Step 3: Apply Simpson's rule with weights 1,4,2,4,11, 4, 2, 4, 11,4,2,4,1

Step 4: Compute the weighted sum

Step 5: Multiply by h/3h/3h/3

Step 6: Check against the accepted value

Exam-style questions

Practise this

Sources & how we know this

Use Simpson's rule with $n = 4$ to approximate $\displaystyle\int_0^1 e^{-x^2}\, dx$ .

Step 3: Apply Simpson's rule with weights $1, 4, 2, 4, 1$

Step 5: Multiply by $h/3$