# Trapezium rule

> Mathematics · CIE A-Level 9709
> Source: https://www.owlsprep.com/study/cie-9709-u2-trapezium-rule/

This module teaches the trapezium rule, a numerical method to approximate definite integrals of functions with no elementary antiderivatives, including formula application, error analysis, and exam strategies. In CIE 9709 the trapezium rule is assessed on Paper 2 only — it does not appear on Paper 1 (Pure 1) or Paper 3 (Pure 3).

**Prerequisites:** [Basic definite integral evaluation](https://www.owlsprep.com/study/cie-9709-u1-definite-integrals/); [Second derivatives and curve concavity](https://www.owlsprep.com/study/cie-9709-u1-second-derivatives/)

## Learning objectives

- Recall the full trapezium rule formula for approximate definite integrals with equal interval widths
- Apply the trapezium rule to calculate area estimates for non-integrable elementary functions
- Compute absolute and percentage error between trapezium estimates and exact integral values
- Identify when the trapezium rule produces an overestimate or underestimate of the true area

## Core Trapezium Rule Formula

**Trapezium Rule for n equal strips** — For an integral over interval [a, b] split into n equal width strips, the total approximate area is the sum of the areas of n individual trapezia, each with two parallel vertical sides of length $y_i$ and $y_{i+1}$

*Notation:* $\int_a^b f(x) dx \approx T_n$

$$T_n = \frac{h}{2} \left( y_0 + 2y_1 + 2y_2 + ... + 2y_{n-1} + y_n \right) \quad \text{where } h = \frac{b-a}{n}$$

**Worked example:** Write out the full trapezium rule expression for $\int_1^5 x^2 dx$ using 4 equal strips

1. First calculate interval width h:
2. $$h = \frac{5-1}{4} = 1$$
3. List all y values at x=1, 2, 3, 4, 5:
4. $$y_0 = 1^2 = 1, y_1 = 4, y_2 =9, y_3=16, y_4=25$$
5. Substitute into the formula, remembering to double all middle terms:
6. $$T_4 = \frac{1}{2} \left( 1 + 2(4 +9 +16) +25 \right) = 42$$

> **tip**
>
> Always confirm you have n+1 total y-values for n strips, as missing a term is the most common arithmetic error.

## Step-by-Step Application Workflow

1. Extract the lower bound a, upper bound b, and number of strips n from the question
2. Calculate interval width h = (b - a)/n and confirm all x points fit evenly between a and b
3. Evaluate f(x) at every x from a to b in steps of h, rounding to the required number of decimal places
4. Multiply all intermediate y-values (not first or last) by 2, sum all terms
5. Multiply the total sum by h/2 to get the final approximate integral

**Worked example:** Approximate $\int_0^2 e^{x^2} dx$ using 4 strips, giving your answer to 3 significant figures

1. Calculate h = (2-0)/4 = 0.5
2. Evaluate y values: $y_0 = e^0 = 1$, $y_1 = e^{0.25} \approx 1.2840$, $y_2 = e^{1} \approx 2.7183$, $y_3 = e^{2.25} \approx 9.4877$, $y_4 = e^{4} \approx 54.5982$
3. Sum adjusted terms: $1 + 2(1.2840 + 2.7183 +9.4877) + 54.5982 = 1 + 26.98 + 54.5982 = 82.5782$
4. Multiply by h/2 = 0.25: $0.25 \times 82.5782 \approx 20.6$ (3 s.f.)

**Check your understanding**

Test your understanding of the workflow:

1. How many y-values do you need for a trapezium rule calculation with 8 strips?

   - 7
   - 8
   - 9
   - 10

   *Why:* You always need n+1 points for n strips, so 8+1=9 total y-values.

## Error Behaviour and Estimate Direction

**Derivation:** Show which way the trapezium rule errs by comparing each strip's chord with the curve

*Starting from:* The top of each strip is a straight chord joining two adjacent points on the curve, so whether that chord sits above or below the curve fixes the error direction

1. Sketch a single strip: its slanted top edge is the chord joining the two adjacent points where the curve is sampled
2. If the curve is concave down ($f''(x) < 0$) it arches above the chord, so the chord lies below the curve; the strip misses the sliver between chord and curve and its area is less than the true area under that segment
3. If the curve is concave up ($f''(x) > 0$) it sags below the chord, so the chord lies above the curve; the strip includes extra area and overshoots the true area under that segment
4. Summing all strips keeps the same direction: chords below the curve give a total below the exact integral, chords above give a total above it

*Conclusion:* For $f''(x) > 0$ (concave up) across [a,b] the trapezium rule overestimates; for $f''(x) < 0$ (concave down) it underestimates; for a straight line the chords lie on the curve and the rule is exact

**Worked example:** State if the trapezium rule estimate for $\int_0^\pi \sin x dx$ with 2 strips is an over or under estimate, and verify with exact value

1. Sketch $y = \sin x$ on $(0, \pi)$: the arch bulges above the straight chords, so every chord lies below the curve and the estimate should be an underestimate. (Supporting check: $f'(x) = \cos x$, $f''(x) = -\sin x < 0$ on $0 < x < \pi$, i.e. concave down.)
2. Trapezium estimate: $h = \pi/2$, $T_2 = \frac{\pi}{4}(0 + 2(1) + 0) = \pi/2 \approx 1.571$
3. Exact integral: $\int_0^\pi \sin x dx = 2$. Since $1.571 < 2$, the estimate is an underestimate, matching the concave-down sketch.

> **warning**
>
> Do not assume the estimate direction based on the function being increasing or decreasing: only concavity (second derivative) determines if the trapezium rule over or under estimates.

## Exam Question Variations

**Exam command terms**

CIE 9709 uses standard command terms for trapezium rule questions, with specific marking requirements:

- **Use the trapezium rule with 3 intervals** — You must use n=3 strips, no alternative method will gain marks *(Commonly the opening part of a numerical integration question)*

- **Hence find the percentage error** — You must use your earlier trapezium estimate and the exact integral value to calculate the relative error

- **Explain with a reason whether the estimate is greater or less than the exact value** — You must give a reason from the shape of the curve — whether the chords lie above it (overestimate) or below it (underestimate) — or equivalently the sign of the second derivative, not just compare numbers

| Question Type | Mark Allocation | Common Rounding Requirement |
| --- | --- | --- |
| Basic 4-strip estimate | 4-5 marks | 3 significant figures |
| Percentage error calculation | 2-3 marks | 1 decimal place |
| Over/under estimate justification | 2 marks | State the sketch or concavity reason |

## Common pitfalls

- **Wrong:** Forgetting to multiply all intermediate y-values by 2 in the formula
  - Why it fails: This mistake produces an estimate roughly half the correct size, and is penalised heavily in marking
  - Correct: Write the full formula $\frac{h}{2}(y_0 + 2y_1 + ... + y_n)$ on your working page before substituting values
- **Wrong:** Using n as the number of points instead of number of strips to calculate h
  - Why it fails: This gives an incorrect interval width, invalidating the entire estimate
  - Correct: Explicitly write n = number of strips, then confirm h = (b-a)/n before calculating any y values
- **Wrong:** Justifying over/under estimate by saying 'the curve is increasing' instead of referencing concavity
  - Why it fails: CIE examiners do not award marks for this reasoning, as increasing functions can produce both over and under estimates
  - Correct: Sketch the curve with its chords and state whether the chords lie above it (overestimate) or below it (underestimate); you may support this with the sign of $f''(x)$ across the interval
- **Wrong:** Rounding individual y-values too early, leading to large final estimate errors
  - Why it fails: Cumulative rounding of intermediate terms can shift your final answer outside the acceptable range for exam marks
  - Correct: Keep all y-values to at least 2 more decimal places than the required final answer before summing
- **Wrong:** Calculating percentage error relative to the trapezium estimate instead of the exact value
  - Why it fails: This reverses the error ratio and gives a wrong final percentage
  - Correct: Memorise the formula $\left|\frac{Estimate - True}{True}\right| \times 100\%$ to avoid swapping terms

## Cheatsheet

| Parameter | Definition | Formula |
| --- | --- | --- |
| Number of strips n | Equal width segments between integration bounds | $n = \frac{b-a}{h}$ |
| Trapezium rule estimate | Approximate integral of f(x) from a to b | $\frac{h}{2}\left(y_0 + 2y_1 + 2y_2 + ... + 2y_{n-1} + y_n\right)$ |
| Percentage error | Relative difference from exact integral value | $\left\|\frac{Estimate - True}{True}\right\| \times 100\%$ |
| Overestimate condition | Trapezium value higher than true integral | $f''(x) > 0$ (concave up): chords lie above the curve |
| Underestimate condition | Trapezium value lower than true integral | $f''(x) < 0$ (concave down): chords lie below the curve |

## What's next

Mastering the trapezium rule is a key stepping stone to the other numerical method examined in Pure 2 (Paper 2): iterative root-finding. The rule is frequently combined with transcendental functions such as exponential and trigonometric curves that have no elementary antiderivatives, so you will see it reappear in later integration questions. You should now practise applying the rule to functions with non-integer interval widths, and cross-check your estimates against exact integral values to build intuition for typical error sizes. Remember that in 9709 the trapezium rule is assessed on Paper 2 only — it does not appear on Paper 1 or Paper 3 — so this foundation makes the rest of the Paper 2 numerical work far easier to master.

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