What are strip width and points?

h = 1 - 04 = 0.25, so x_0 = 0, x_1 = 0.25, x_2 = 0.5, x_3 = 0.75, x_4 = 1.

What is compute the bracket?

4(0.939413) = 3.757652; \;2(0.778801) = 1.557602; \;4(0.569783) = 2.279132. Sum: 1 + 3.757652 + 1.557602 + 2.279132 + 0.367879 = 8.962265.

0.253 × 8.962265 ≈ 0.083333 × 8.962265 ≈ 0.74686.

The accepted value is about 0.74682, so the estimate is accurate to four decimal places with only four strips, as expected of Simpson's rule.

QLDSpecialist MathematicsSyllabus dot point

Topic 1: Integration and applications of integration

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

A focused answer to the QCE Specialist Mathematics Unit 4 dot point on Simpson's rule. Covers the rule and its weighting pattern, the requirement for an even number of subintervals, choosing the strip width, and when numerical integration is needed, with a verified worked example and the coefficient-pattern trap.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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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.

Worked example

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

Strip width and points. $h = \dfrac{1 - 0}{4} = 0.25$ , so $x_0 = 0$ , $x_1 = 0.25$ , $x_2 = 0.5$ , $x_3 = 0.75$ , $x_4 = 1$ .

Function values.

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

Apply the rule with weights $1, 4, 2, 4, 1$ :

\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].

Compute the bracket: $4(0.939413) = 3.757652$ ; $\;2(0.778801) = 1.557602$ ; $\;4(0.569783) = 2.279132$ . Sum: $1 + 3.757652 + 1.557602 + 2.279132 + 0.367879 = 8.962265$ .
Multiply: $\dfrac{0.25}{3} \times 8.962265 \approx 0.083333 \times 8.962265 \approx 0.74686.$
Check: The accepted value is about $0.74682$ , so the estimate is accurate to four decimal places with only four strips, as expected of Simpson's rule.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 QCAA4 marksAt a certain location, a biologist measures the width of a river to be 12 m. She also records the depth of the river at regular 2 m interval widths as shown. Width (m): 0, 2, 4, 6, 8, 10, 12. Depth (m): 0.52, 2.15, 3.70, 4.27, 3.32, 1.28, 0.59. The biologist estimates the cross-sectional area of the river at this location to be 15 m^2. Use Simpson's rule to evaluate the reasonableness of this estimation. Justify your area calculation and decision regarding reasonableness using mathematical reasoning.

Show worked answer →

There are 7 depth readings, giving 6 strips of width h = 2 (an even number of strips, so Simpson's rule applies).

Apply the pattern 1, 4, 2, 4, 2, 4, 1:
Area = (h/3)[f0 + 4(f1 + f3 + f5) + 2(f2 + f4) + f6]
= (2/3)[0.52 + 4(2.15 + 4.27 + 1.28) + 2(3.70 + 3.32) + 0.59]
= (2/3)[0.52 + 4(7.70) + 2(7.02) + 0.59]
= (2/3)(45.95)
= 30.6 m^2 (to 1 d.p.). [3 marks for correct set-up and evaluation]

Decision [1 mark]: the Simpson's rule estimate is about 30.6 m^2, which is roughly double the biologist's estimate of 15 m^2. The estimate of 15 m^2 is therefore not reasonable; it substantially underestimates the cross-sectional area.

2022 QCAA4 marksAn aerial view of the surface of a dam, 6 km in length, is symmetrically positioned on a Cartesian plane. A dam wall is located along the y-axis. The surrounding edge of the dam can be modelled by the ellipse (x - 2)^2 / 16 + y^2 / 9 = 1, for 0 less than or equal to x less than or equal to 6. Use Simpson's rule with four strips to determine an approximate area of the surface of the dam.

Show worked answer →

Use the upper half of the ellipse, y = 3 * sqrt(1 - (x - 2)^2 / 16), over 0 <= x <= 6, then double for symmetry about the x-axis. Four strips means h = (6 - 0)/4 = 1.5. [1 mark for h]

Evaluate y at x = 0, 1.5, 3, 4.5, 6 [1 mark]:
y(0) = 2.598, y(1.5) = 2.976, y(3) = 2.905, y(4.5) = 2.342, y(6) = 0.

Simpson's rule for the upper half [1 mark]:
A_top = (1.5/3)[2.598 + 4(2.976 + 2.342) + 2(2.905) + 0]
= 0.5[2.598 + 21.272 + 5.810] = 0.5(29.68) = 14.84 km^2.

Total surface area = 2 * 14.84 = 29.68 km^2 (to 2 d.p.). [1 mark]