The Trapezium Rule

This lesson covers the trapezium rule — a numerical method for approximating definite integrals — as required by the Edexcel A-Level Mathematics specification (9MA0). You need to know the formula, apply it with a given number of strips, understand whether the approximation is an overestimate or underestimate, and know how to improve accuracy.

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Why Use Numerical Integration?

Some functions cannot be integrated analytically — that is, there is no formula for their antiderivative in terms of standard functions. For example:

• ∫ e^(-x²) dx
• ∫ √(1 + x³) dx
• ∫ sin(x²) dx

In these cases, we use numerical methods to approximate the value of the definite integral. The trapezium rule is the most common method at A-Level.

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The Idea Behind the Trapezium Rule

The area under a curve y = f(x) between x = a and x = b is approximated by dividing the region into n equal strips (intervals) and treating each strip as a trapezium rather than following the exact curve.

The area of a trapezium is:

Area = (1/2)(a + b) × h

where a and b are the parallel sides and h is the distance between them.

For each strip under the curve, the two parallel sides are the y-values at the left and right edges, and h is the width of the strip.

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The Formula

With n strips (n + 1 ordinates) from x = a to x = b:

h = (b - a) / n

where h is the width of each strip.

The approximation is:

∫ from a to b of y dx ≈ (h/2)[y₀ + y_n + 2(y₁ + y₂ + ... + y_{n-1})]

In words: (h/2) × [first y + last y + 2 × (sum of all middle y-values)]

Where:

• y₀ = f(a), y₁ = f(a + h), y₂ = f(a + 2h), ..., y_n = f(b)
• h = (b - a)/n

Exam Tip: This formula is given in the formula booklet, but you should be very comfortable using it quickly and accurately.

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Step-by-Step Method

1. Calculate h = (b - a) / n.
2. List the x-values: x₀ = a, x₁ = a + h, x₂ = a + 2h, ..., x_n = b.
3. Calculate the y-values: y₀ = f(x₀), y₁ = f(x₁), ..., y_n = f(x_n).
4. Apply the formula: (h/2)[y₀ + y_n + 2(y₁ + y₂ + ... + y_{n-1})].

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Worked Examples

Example 1: Simple Application

Use the trapezium rule with 4 strips to estimate ∫ from 0 to 2 of √(4 + x²) dx.

Step 1: h = (2 - 0)/4 = 0.5

Step 2: x-values: 0, 0.5, 1, 1.5, 2

Step 3: y-values:

x	y = √(4 + x²)
0	√4 = 2
0.5	√4.25 = 2.0616
1	√5 = 2.2361
1.5	√6.25 = 2.5
2	√8 = 2.8284

Step 4: Apply the formula: (0.5/2)[2 + 2.8284 + 2(2.0616 + 2.2361 + 2.5)] = 0.25[4.8284 + 2(6.7977)] = 0.25[4.8284 + 13.5954] = 0.25[18.4238] = 4.6060 (4 d.p.)

Example 2: Using a Table of Values

The table below shows values of y = e^(x²) for selected values of x.

x	0	0.25	0.5	0.75	1
y	1	1.0645	1.2840	1.7551	2.7183

Use the trapezium rule with all the values in the table to estimate ∫ from 0 to 1 of e^(x²) dx.

h = 0.25, n = 4

Estimate = (0.25/2)[1 + 2.7183 + 2(1.0645 + 1.2840 + 1.7551)] = 0.125[3.7183 + 2(4.1036)] = 0.125[3.7183 + 8.2072] = 0.125[11.9255] = 1.4907 (4 d.p.)

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Overestimates and Underestimates

Whether the trapezium rule gives an overestimate or underestimate depends on the shape of the curve:

Curve Shape	Effect
Concave up (curves upward, like y = x²)	Overestimate — the trapezium tops lie above the curve
Concave down (curves downward, like y = √x)	Underestimate — the trapezium tops lie below the curve

How to Determine Concavity

A curve y = f(x) is:

• Concave up when d²y/dx² > 0 (the second derivative is positive)
• Concave down when d²y/dx² < 0 (the second derivative is negative)

Example: For y = x², d²y/dx² = 2 > 0, so the curve is concave up. The trapezium rule will give an overestimate.

Example: For y = √x = x^(1/2), d²y/dx² = -1/(4x^(3/2)) < 0, so the curve is concave down. The trapezium rule will give an underestimate.

Exam Tip: A quick sketch of the curve often makes it obvious whether the trapezia lie above or below the curve. If the curve bends towards the tops of the trapezia, it is an underestimate; if it bends away, it is an overestimate.

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Improving Accuracy

The accuracy of the trapezium rule can be improved by:

1. Increasing the number of strips (n) — More strips means each strip is narrower, so the straight-line top of each trapezium more closely follows the curve.

2. Using more strips in regions where the curve changes rapidly — If the curve is nearly straight, even a few strips give a good approximation.

How Accuracy Scales

Doubling the number of strips roughly quarters the error (the error is proportional to h², where h is the strip width). So:

• 4 strips → error E
• 8 strips → error approximately E/4
• 16 strips → error approximately E/16

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Trapezium Rule with Trigonometric and Exponential Functions

Example: Estimate ∫ from 0 to π/2 of √(sin x) dx using 4 strips

h = (π/2 - 0)/4 = π/8 ≈ 0.3927

x	0	π/8	π/4	3π/8	π/2
sin x	0	0.3827	0.7071	0.9239	1
√(sin x)	0	0.6186	0.8409	0.9612	1

Estimate = (0.3927/2)[0 + 1 + 2(0.6186 + 0.8409 + 0.9612)] = 0.19635[1 + 2(2.4207)] = 0.19635[1 + 4.8414] = 0.19635[5.8414] = 1.1469 (4 d.p.)

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When the Trapezium Rule Is Tested in Exams

Typical exam questions will:

• Give you a function and ask you to use the trapezium rule with a specified number of strips
• Provide a table of values and ask you to apply the rule
• Ask whether the estimate is an overestimate or underestimate, with justification
• Ask you to explain how the estimate could be improved
• Combine the trapezium rule with other topics (e.g., compare with an exact answer found by integration)

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Common Mistakes

1. Using the wrong value of h — Double-check h = (b - a)/n.
2. Including y₀ or y_n in the "middle" sum — Only y₁ to y_{n-1} are doubled.
3. Arithmetic errors — Use a calculator and check your working.
4. Forgetting the h/2 factor — The formula is (h/2)[...], not h[...].
5. Confusing n (number of strips) with n + 1 (number of ordinates/y-values).

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Summary

• The trapezium rule approximates ∫ from a to b of y dx using n trapezia.
• Formula: (h/2)[y₀ + y_n + 2(y₁ + y₂ + ... + y_{n-1})] where h = (b - a)/n.
• Concave up curves → overestimate. Concave down curves → underestimate.
• Accuracy is improved by increasing the number of strips.
• The error is proportional to h², so doubling n quarters the error.
• The method is essential for functions that cannot be integrated analytically.

A-Level Deep Dive: The Trapezium Rule

Spec mapping

Edexcel 9MA0 specification section 10 — Integration, sub-strand 10.5 (numerical integration) covers the trapezium rule with equally spaced ordinates to estimate the area under a curve; relate the accuracy of the estimate to the concavity of the integrand and to the number of strips (refer to the official specification document for exact wording). The rule is examined on 9MA0-02 (Paper 2 — Pure Mathematics) and frequently appears in section 8 (Differentiation) synoptic contexts where the second derivative is used to justify whether the estimate is an over- or under-approximation. The trapezium rule formula $\int_a^b f(x)\,dx \approx \tfrac{h}{2}[y_0 + 2(y_1 + y_2 + \ldots + y_{n-1}) + y_n]$∫ab​f(x)dx≈2h​[y0​+2(y1​+y2​+…+yn−1​)+yn​] with $h = (b-a)/n$h=(b−a)/n is printed in the Edexcel formula booklet — but the over/under reasoning is not, and that is where most of the AO2 marks live.

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Worked example with full mark scheme

Question (8 marks): Use the trapezium rule with 5 strips to estimate $\int_0^1 \sqrt{1 + x^2}\,dx$∫01​1+x2​dx, giving your answer to 4 decimal places. State, with justification, whether the estimate is an over- or under-estimate of the true value. (8)

Solution with mark scheme:

Step 1 — set up the strip width and ordinates.

With $n = 5$n=5 strips on $[0, 1]$[0,1], $h = (1 - 0)/5 = 0.2$h=(1−0)/5=0.2. The ordinates are at $x = 0, 0.2, 0.4, 0.6, 0.8, 1.0$x=0,0.2,0.4,0.6,0.8,1.0.

M1 — correct $h$h and 6 ordinate values listed.

Step 2 — compute the $y$y-values.

Let $f(x) = \sqrt{1 + x^2}$f(x)=1+x2​.

$x$x	$y = \sqrt{1+x^2}$y=1+x2​
0.0	1.0000
0.2	1.0198
0.4	1.0770
0.6	1.1662
0.8	1.2806
1.0	1.4142

M1 — at least 4 of 6 ordinates correct to 4 d.p. A1 — all six correct.

Step 3 — apply the rule.

$\int_0^1 \sqrt{1+x^2}\,dx \approx \dfrac{0.2}{2}\bigl[1.0000 + 1.4142 + 2(1.0198 + 1.0770 + 1.1662 + 1.2806)\bigr]$∫01​1+x2​dx≈20.2​[1.0000+1.4142+2(1.0198+1.0770+1.1662+1.2806)]

The interior sum is $1.0198 + 1.0770 + 1.1662 + 1.2806 = 4.5436$1.0198+1.0770+1.1662+1.2806=4.5436, doubled to $9.0872$9.0872. The end pair contributes $1.0000 + 1.4142 = 2.4142$1.0000+1.4142=2.4142. Total inside the bracket: $11.5014$11.5014.

M1 — correct structure (end ordinates once, interior ordinates doubled). A1 — correct bracket value.

$\approx 0.1 \times 11.5014 = 1.1501$≈0.1×11.5014=1.1501

A1 — final answer $1.1501$1.1501 (4 d.p.).

Step 4 — over/under justification.

Compute $f''(x)$f′′(x). With $f(x) = (1 + x^2)^{1/2}$f(x)=(1+x2)1/2:

$f'(x) = x(1+x^2)^{-1/2}, \qquad f''(x) = (1+x^2)^{-3/2}$f′(x)=x(1+x2)−1/2,f′′(x)=(1+x2)−3/2

For all $x \in [0, 1]$x∈[0,1], $f''(x) > 0$f′′(x)>0, so the curve is concave up throughout the interval. The trapezium chords therefore lie above the curve, and the trapezium estimate is an over-estimate of the true integral.

M1 — computing or stating sign of $f''$f′′ on the interval. A1 — correct conclusion (over-estimate) with concavity reason.

Total: 8 marks (M4 A4).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): The function $f$f is defined by $f(x) = \ln(2 + x^2)$f(x)=ln(2+x2) for $x \geq 0$x≥0.

(a) Use the trapezium rule with 4 strips to estimate $\int_0^2 \ln(2 + x^2)\,dx$∫02​ln(2+x2)dx, giving your answer to 3 decimal places. (4)

(b) By considering the second derivative, determine whether your answer to (a) is an over- or under-estimate. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — $h = 0.5$h=0.5, ordinates at $x = 0, 0.5, 1.0, 1.5, 2.0$x=0,0.5,1.0,1.5,2.0, with $y$y-values $\ln 2, \ln 2.25, \ln 3, \ln 4.25, \ln 6$ln2,ln2.25,ln3,ln4.25,ln6.
• M1 (AO1.1b) — substitution into $\tfrac{h}{2}[y_0 + y_n + 2(y_1 + y_2 + y_3)]$2h​[y0​+yn​+2(y1​+y2​+y3​)].
• A1 (AO1.1b) — bracket evaluated correctly.
• A1 (AO2.5) — answer $\approx 1.971$≈1.971 (3 d.p.) with method visible.

(b)

• M1 (AO3.2a) — computing $f''(x)$f′′(x) and analysing its sign on $[0, 2]$[0,2]. With $f(x) = \ln(2+x^2)$f(x)=ln(2+x2), $f'(x) = \dfrac{2x}{2+x^2}$f′(x)=2+x22x​ and $f''(x) = \dfrac{2(2 - x^2)}{(2+x^2)^2}$f′′(x)=(2+x2)22(2−x2)​. This changes sign at $x = \sqrt{2} \approx 1.414$x=2​≈1.414.
• A1 (AO3.5b) — interpretation: $f''$f′′ is positive on $[0, \sqrt{2})$[0,2​) (concave up — over-estimate) and negative on $(\sqrt{2}, 2]$(2​,2] (concave down — under-estimate), so no clean over/under conclusion holds globally; a fully reasoned answer notes the sign change and concludes that the trapezium estimate's bias is indeterminate without strip-by-strip analysis.

Total: 6 marks split AO1 = 3, AO2 = 1, AO3 = 2. This is exactly the kind of question Edexcel uses to separate B from A* candidates: the rote calculation is AO1, but recognising that concavity changes inside the interval is AO3 reasoning.

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Synoptic links

Connects to:

1. Section 8 — Definite integration (exact methods): when the integrand has an elementary antiderivative, the trapezium rule's error can be benchmarked against the true value. $\int_0^1 (1 + x^2)\,dx = 4/3$∫01​(1+x2)dx=4/3 exactly; comparing against the trapezium estimate with $n = 5$n=5 shows roughly a 0.4% over-estimate, validating the rule.

2. Section 9 — Concavity from the second derivative: $f''(x) > 0$f′′(x)>0 on $[a, b]$[a,b] means the curve is concave up, so chord segments lie above the curve and the trapezium rule over-estimates; $f''(x) < 0$f′′(x)<0 means concave down, chords lie below, so the rule under-estimates. This is the single most-tested AO2 reasoning step on trapezium-rule questions.

3. Simpson's rule (Further Maths 9FM0): Simpson's rule uses parabolic arcs through three consecutive ordinates and is exact for any cubic. Its error scales as $O(h^4)$O(h4) versus the trapezium rule's $O(h^2)$O(h2) — a fundamental jump in convergence rate that motivates higher-order quadrature.

4. Numerical methods (section 10 sub-strand): trapezium rule is one member of a family that includes the midpoint rule (one-point quadrature) and rectangle rule (left/right Riemann sums). Each represents a different polynomial fit to the integrand; trapezium uses linear interpolation between consecutive ordinates.

5. Experimental physics and statistics: when an integrand is known only as a table of measured data points (e.g. velocity vs time from a sensor), no antiderivative exists and numerical integration is the only option. The trapezium rule is the workhorse for computing displacement, work done, or expected values from tabulated data.

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Mark-scheme literacy

Trapezium-rule questions on 9MA0-02 split AO marks as follows: