The Trapezium Rule

This lesson covers the trapezium rule, a method for numerical integration — approximating the value of a definite integral when exact integration is difficult or impossible. This is a required topic on the AQA A-Level Mathematics specification and is frequently examined.

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Why Numerical Integration?

Not all functions can be integrated analytically. For example, there is no elementary antiderivative of e^(x²), 1/ln x, or √(1 + x³). In such cases, we approximate the integral numerically.

The trapezium rule approximates the area under a curve by dividing it into trapeziums (trapezoids) rather than rectangles, giving a better approximation.

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The Formula

For n equally spaced strips on the interval [a, b], the strip width is:

h = (b − a)/n

The x-values are x₀ = a, x₁ = a + h, x₂ = a + 2h, ..., xₙ = b.

The corresponding y-values are y₀ = f(x₀), y₁ = f(x₁), ..., yₙ = f(xₙ).

The trapezium rule formula is:

∫(from a to b) f(x) dx ≈ h/2 × [y₀ + yₙ + 2(y₁ + y₂ + ... + yₙ₋₁)]

Or equivalently:

∫(from a to b) f(x) dx ≈ h/2 × [(first + last) + 2 × (sum of middle y-values)]

Exam Tip: The formula is given in the AQA formula booklet, but you must know how to apply it. Set up a clear table of x and y values — this makes the arithmetic much easier to follow and the examiner can award method marks.

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Example 1: Basic Application

Use the trapezium rule with 4 strips to estimate ∫₀² x² dx.

Step 1: Calculate h:

h = (2 − 0)/4 = 0.5

Step 2: Set up the table:

x	0	0.5	1.0	1.5	2.0
y = x²	0	0.25	1	2.25	4

Step 3: Apply the formula:

Area ≈ 0.5/2 × [0 + 4 + 2(0.25 + 1 + 2.25)]
= 0.25 × [4 + 2(3.5)]
= 0.25 × [4 + 7]
= 0.25 × 11
= 2.75

Compare with exact value: ∫₀² x² dx = [x³/3]₀² = 8/3 ≈ 2.667.

The estimate (2.75) is slightly too high — we will see why shortly.

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Example 2: Function Without Elementary Integral

Use the trapezium rule with 5 strips to estimate ∫₁² (1/ln x) dx, giving your answer to 3 decimal places. Use x-values 1.0, 1.2, 1.4, 1.6, 1.8, 2.0.

Step 1:

h = (2 − 1)/5 = 0.2

Step 2: Calculate y-values (note: at x = 1, ln 1 = 0, so 1/ln 1 is undefined — this means the integrand has a singularity at x = 1. In practice the question would either avoid this point or use a different interval. Let us assume the question uses the interval [1.2, 2.0] with 4 strips instead.)

Revised: Use 4 strips on [1.2, 2.0]:

h = (2.0 − 1.2)/4 = 0.2

x	1.2	1.4	1.6	1.8	2.0
ln x	0.1823	0.3365	0.4700	0.5878	0.6931
1/ln x	5.4854	2.9726	2.1277	1.7013	1.4427

Area ≈ 0.2/2 × [5.4854 + 1.4427 + 2(2.9726 + 2.1277 + 1.7013)]
= 0.1 × [6.9281 + 2(6.8016)]
= 0.1 × [6.9281 + 13.6032]
= 0.1 × 20.5313
≈ 2.053

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Example 3: With Given y-values

The table shows values of y = √(4 − x²):

x	0	0.5	1.0	1.5	2.0
y	2	1.9365	1.7321	1.3229	0

Use the trapezium rule to estimate ∫₀² √(4 − x²) dx.

h = 0.5
Area ≈ 0.5/2 × [2 + 0 + 2(1.9365 + 1.7321 + 1.3229)]
= 0.25 × [2 + 2(4.9915)]
= 0.25 × [2 + 9.983]
= 0.25 × 11.983
= 2.996

(The exact value is π, since this is a quarter-circle of radius 2.)

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Overestimates and Underestimates

The accuracy of the trapezium rule depends on the shape of the curve:

Convex Curves (Concave Up)

A curve where the gradient is increasing (d²y/dx² > 0) is called convex (or concave up). The trapeziums lie above the curve, so the trapezium rule gives an overestimate.

Example: y = x² on [0, 2]. The curve is convex (d²y/dx² = 2 > 0). Our estimate of 2.75 was indeed greater than the exact value of 8/3 ≈ 2.667.

Concave Curves (Concave Down)

A curve where the gradient is decreasing (d²y/dx² < 0) is called concave (or concave down). The trapeziums lie below the curve, so the trapezium rule gives an underestimate.

Example: y = √x on [0, 4]. Since d²y/dx² = −1/(4x^(3/2)) < 0, the curve is concave, and the trapezium rule underestimates the area.

Mixed Concavity

If the curve has a point of inflection within [a, b], the overestimate and underestimate may partially cancel.

Exam Tip: A very common exam question asks "State, with a reason, whether the trapezium rule gives an overestimate or underestimate." Your answer should reference the concavity of the curve: "The curve is convex (d²y/dx² > 0) on this interval, so the trapeziums lie above the curve, giving an overestimate."

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Improving Accuracy

The trapezium rule becomes more accurate as the number of strips increases (i.e. as h decreases). Doubling the number of strips typically reduces the error by a factor of about 4 (since the error is proportional to h²).

Example 4

Compare the trapezium rule estimates for ∫₀² x² dx using 2 strips and 4 strips.

2 strips (h = 1):

x	0	1	2
y	0	1	4

Area ≈ 1/2 × [0 + 4 + 2(1)] = 0.5 × 6 = 3

4 strips (h = 0.5): From Example 1, Area ≈ 2.75.

Exact: 8/3 ≈ 2.667.

The error with 2 strips is 3 − 2.667 = 0.333. The error with 4 strips is 2.75 − 2.667 = 0.083. The ratio is approximately 4, confirming that doubling the strips reduces the error by a factor of 4.

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When to Use the Trapezium Rule

• When the function cannot be integrated analytically (e.g. e^(−x²), sin(x²)).
• When you are given tabulated data rather than an algebraic formula.
• When the question specifically asks for a numerical approximation.

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Common Mistakes

1. Miscounting strips vs ordinates: n strips require n + 1 x-values (ordinates). A common error is confusing the two.
2. Forgetting to halve h: The formula uses h/2, not h.
3. Including first/last values in the "middle" sum: Only y₁ to yₙ₋₁ go in the bracket multiplied by 2. y₀ and yₙ are the "first and last" values.
4. Rounding too early: Keep intermediate values to several decimal places and only round the final answer.

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Summary

• The trapezium rule approximates ∫ₐᵇ f(x) dx using n trapeziums.
• Formula: Area ≈ h/2 × [y₀ + yₙ + 2(y₁ + y₂ + ... + yₙ₋₁)] where h = (b − a)/n.
• The rule overestimates for convex curves and underestimates for concave curves.
• Increasing the number of strips improves accuracy.
• Use a clear table of x and y values.

Exam Tip: When answering trapezium rule questions: (1) state h explicitly; (2) draw a clear table of x and y values; (3) substitute into the formula showing the structure clearly; (4) give your answer to the degree of accuracy specified. If asked whether the approximation is an over- or underestimate, consider the concavity (second derivative) of the curve.

A-Level Deep Dive: The Trapezium Rule

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section H: Integration, sub-strand H6 covers the trapezium rule to estimate the area under a curve and determine whether it gives an over-estimate or under-estimate (refer to the official specification document for exact wording). The formula appears in the AQA formula booklet as $\int_a^b y\,dx \approx \tfrac{h}{2}\left[y_0 + 2(y_1 + y_2 + \cdots + y_{n-1}) + y_n\right]$∫ab​ydx≈2h​[y0​+2(y1​+y2​+⋯+yn−1​)+yn​] where $h = (b - a)/n$h=(b−a)/n. Although housed in section H, trapezium-rule problems routinely synopt with section G (definite integration: comparing the estimate against an exact integral where one exists), section J (differentiation: using the second derivative to determine concavity and hence over/under-estimate) and section L (numerical methods, in Year 2: locating roots and iteration as further numerical techniques). Further Mathematics 7367 extends the strand to Simpson's rule.

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Worked example with full mark scheme

Question (8 marks):

(a) Use the trapezium rule with five strips to estimate $\int_0^1 \sqrt{1 + x^2}\,dx$∫01​1+x2​dx, giving your answer to four decimal places. (5)

(b) State, with reasons referring to concavity, whether your answer is an over-estimate or an under-estimate of the true value. (3)

Solution with mark scheme:

(a) Step 1 — set up the strip width.

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.

B1 — correct strip width $h = 0.2$h=0.2 stated or implied by correct $x$x-values.

Step 2 — evaluate the ordinates. Let $f(x) = \sqrt{1 + x^2}$f(x)=1+x2​.

$x$x	$x^2$x2	$1 + x^2$1+x2	$y = \sqrt{1 + x^2}$y=1+x2​
0.0	0.00	1.00	1.000000
0.2	0.04	1.04	1.019804
0.4	0.16	1.16	1.077033
0.6	0.36	1.36	1.166190
0.8	0.64	1.64	1.280625
1.0	1.00	2.00	1.414214

M1 — at least four ordinates evaluated correctly (allow rounding to 4 d.p. or better at this stage).

A1 — all six ordinates correct to 4 d.p.

Step 3 — apply the trapezium rule.

$\int_0^1 \sqrt{1 + x^2}\,dx \approx \dfrac{0.2}{2}\left[1.000000 + 2(1.019804 + 1.077033 + 1.166190 + 1.280625) + 1.414214\right]$∫01​1+x2​dx≈20.2​[1.000000+2(1.019804+1.077033+1.166190+1.280625)+1.414214]

The interior sum: $1.019804 + 1.077033 + 1.166190 + 1.280625 = 4.543652$1.019804+1.077033+1.166190+1.280625=4.543652. Doubled: $9.087304$9.087304.

Bracket total: $1.000000 + 9.087304 + 1.414214 = 11.501518$1.000000+9.087304+1.414214=11.501518.

Multiply by $h/2 = 0.1$h/2=0.1: $0.1 \times 11.501518 = 1.1501518$0.1×11.501518=1.1501518.

M1 — correct structure of the trapezium rule (first and last ordinates with weight 1, interiors weight 2, multiplied by $h/2$h/2).

A1 — answer $1.1502$1.1502 to 4 d.p.

(b) Step 1 — analyse concavity.

Let $f(x) = \sqrt{1 + x^2} = (1 + x^2)^{1/2}$f(x)=1+x2​=(1+x2)1/2. Then $f'(x) = x(1 + x^2)^{-1/2}$f′(x)=x(1+x2)−1/2, and using the quotient/product rule:

$f''(x) = (1 + x^2)^{-1/2} + x \cdot \left(-\dfrac{1}{2}\right)(1 + x^2)^{-3/2} \cdot 2x = \dfrac{1}{(1 + x^2)^{1/2}} - \dfrac{x^2}{(1 + x^2)^{3/2}} = \dfrac{1}{(1 + x^2)^{3/2}}$f′′(x)=(1+x2)−1/2+x⋅(−21​)(1+x2)−3/2⋅2x=(1+x2)1/21​−(1+x2)3/2x2​=(1+x2)3/21​

M1 — attempt at $f''(x)$f′′(x) via the chain rule applied twice (or equivalent).

Step 2 — sign of $f''$f′′ and conclusion.

Since $(1 + x^2)^{3/2} > 0$(1+x2)3/2>0 for all real $x$x, we have $f''(x) > 0$f′′(x)>0 on $[0, 1]$[0,1]. The curve is therefore concave up (convex) on this interval.

A1 — correct deduction that $f''(x) > 0$f′′(x)>0 on $[0, 1]$[0,1], so the curve is concave up.

Step 3 — link concavity to over/under-estimate.

For a concave-up curve, each chord lies above the curve, so the area of each trapezium exceeds the area under the curve over that strip. Summing strips, the trapezium-rule estimate is therefore an over-estimate of the true integral.

A1 — correct conclusion (over-estimate) with justification referencing chord-above-curve for concave-up.

Total: 8 marks (B1 M2 A2 in (a); M1 A2 in (b)).

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Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): The diagram (not shown) depicts $y = \ln(1 + x)$y=ln(1+x) for $0 \leq x \leq 4$0≤x≤4.

(a) Use the trapezium rule with four strips to estimate $\int_0^4 \ln(1 + x)\,dx$∫04​ln(1+x)dx. (4)

(b) Without further calculation, state, with a reason, whether the estimate in (a) is an over-estimate or an under-estimate. (2)

Mark scheme decomposition by AO:

(a)

• B1 (AO1.1a) — $h = 1$h=1 stated or implied; ordinates at $x = 0, 1, 2, 3, 4$x=0,1,2,3,4 giving $y = 0,\,\ln 2,\,\ln 3,\,\ln 4,\,\ln 5$y=0,ln2,ln3,ln4,ln5.
• M1 (AO1.1b) — correct trapezium-rule structure: $\tfrac{1}{2}\left[0 + 2(\ln 2 + \ln 3 + \ln 4) + \ln 5\right]$21​[0+2(ln2+ln3+ln4)+ln5].
• M1 (AO1.1b) — correct evaluation of the interior sum (using $\ln 2 + \ln 3 + \ln 4 = \ln 24 \approx 3.1781$ln2+ln3+ln4=ln24≈3.1781).
• A1 (AO1.1b) — final answer $\approx 4.0830$≈4.0830 to 4 d.p.

(b)

• M1 (AO2.1) — observation that $\ln(1 + x)$ln(1+x) has $f''(x) = -1/(1 + x)^2 < 0$f′′(x)=−1/(1+x)2<0, so the curve is concave down on $[0, 4]$[0,4].
• A1 (AO2.4) — conclusion: under-estimate, because for a concave-down curve each chord lies below the curve, so trapezium areas are smaller than the true area under each strip.

Total: 6 marks split AO1 = 4, AO2 = 2. AQA reliably reserves the AO2 marks for the over/under reasoning — purely numerical answers without the concavity justification cap candidates at AO1 marks only.

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Synoptic links

Connects to:

1. Section G — Definite integration (exact value): when the integrand has an elementary antiderivative, you can compute the exact integral and compare against the trapezium estimate to verify the over/under direction. For instance, $\int_0^1 x^2\,dx = 1/3$∫01​x2dx=1/3 exactly; a trapezium estimate with two strips gives $\tfrac{1}{4}[0 + 2(0.25) + 1] = 0.375$41​[0+2(0.25)+1]=0.375, an over-estimate as predicted by $f''(x) = 2 > 0$f′′(x)=2>0.

2. Section J — Concavity and the second derivative: the sign of $f''$f′′ on $[a, b]$[a,b] is the tool that decides whether the trapezium rule over- or under-estimates. $f'' > 0$f′′>0 throughout means concave up, so chords lie above the curve and the rule over-estimates; $f'' < 0$f′′<0 means concave down, chords below curve, rule under-estimates. If $f''$f′′ changes sign on $[a, b]$[a,b] no global conclusion is possible without splitting the interval.

3. Further Mathematics 7367 — Simpson's rule: Simpson's rule fits parabolas through successive triples of points and is exact for cubics, with error $O(h^4)$O(h4) versus the trapezium rule's $O(h^2)$O(h2). For a smooth integrand, doubling the number of strips reduces trapezium error by a factor of 4 but Simpson error by a factor of 16.

4. Numerical methods (Year 2, sub-strand N): trapezium rule is the simplest quadrature method. Locating roots numerically (Newton-Raphson, fixed-point iteration) and approximating integrals belong to the same family — replacing exact closed-form work with controlled approximations when no elementary form exists.

5. A-Level Physics — experimental data: when only a discrete set of measurements is available (e.g. force-vs-displacement to compute work done, or velocity-vs-time to compute displacement), the trapezium rule is the natural estimator because no functional form is assumed. AQA Physics paper 1 explicitly accepts trapezium-rule arguments in "estimate the area under the curve" questions.

6. Section H — Integration by substitution and parts: trapezium rule is the fallback when these analytical techniques don't apply — for example, $\int_0^1 e^{-x^2}\,dx$∫01​e−x2dx has no elementary antiderivative and must be approximated numerically.

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Mark-scheme literacy

Trapezium-rule questions on AQA 7357 split AO marks fairly evenly across AO1 and AO2: