IB Maths AI & A-Level
Some curves cannot be integrated. Others can, but not by you, not yet. The trapezium rule is how you find the area underneath one anyway — by covering it with straight-sided shapes you already know how to measure.
This page explains where the formula comes from, how to use it without slipping on the usual traps, and how to answer the question examiners nearly always add on the end: is your answer too big or too small?
Check your course first
The trapezium rule is on IB Maths AI (syllabus point SL 5.8, so both SL and HL) and on A-level maths for AQA, Edexcel and OCR. It is not on IB Maths AA at either level — it is not in the AA formula booklet and AA papers do not ask for it. If you are on AA, you can safely skip this.
You want the area between a curve and the x-axis. The shape has a curved top, so there is no area formula for it. But a trapezium — a four-sided shape with two parallel sides — does have one, and you learnt it years ago.
So we cheat. We slice the region into thin vertical strips and pretend the top of each strip is a straight line rather than a curve. Each strip becomes a trapezium, we work out all their areas, and we add them up.
Scroll the diagram sideways.
The whole method in one line
Replace the curved top of each strip with a straight line, add up the trapezia, and accept that the answer is close rather than exact.
Two words you need, because every exam question uses them:
Here it is as it appears in the IB Maths AI formula booklet and in A-level textbooks.
∫ab y dx ≈ ½h [ (y0 + yn) + 2(y1 + y2 + … + yn−1) ]
where h = (b − a) ÷ n
It looks worse than it is. There are only three ideas in it.
| Part | What it means |
|---|---|
| h | The width of one strip. Take the whole interval from a to b and divide it by the number of strips. |
| y0 + yn | The two end heights, counted once each. They are the outside edges, so they belong to only one trapezium. |
| 2(y1 + …) | All the middle heights, counted twice. Each inside line is shared by the trapezium on its left and the one on its right, so it gets used twice. |
| ½h | Half the strip width, from the ordinary area of a trapezium: ½ × (a + b) × height. |
Why the middle ones are doubled
Write the table out, even when you are sure
Nearly every mark lost on this topic comes from a mis-typed y-value, not from misunderstanding the method. A table makes the slip visible, and in an exam it earns method marks even if one number is wrong.
Use the trapezium rule with four strips to estimate ∫04 x2 dx.
Step 1. Find h
h = (4 − 0) / 4 = 1
Step 2. Five y-values, stepping by 1
x: 0, 1, 2, 3, 4
y: 0, 1, 4, 9, 16
Step 3. Ends once, middles doubled
(y₀ + y₄) = 0 + 16 = 16
2(y₁ + y₂ + y₃) = 2(1 + 4 + 9) = 28
Step 4. Multiply by ½h
½ × 1 × (16 + 28)
estimate = 22
This one we can check, because x2 is easy to integrate properly: the exact answer is 64/3 = 21.33…. Our estimate of 22 is too big by about 0.67. Hold that thought — the next section explains why it had to come out too big.
Use the trapezium rule with four strips to estimate ∫13 (1/x) dx, giving your answer to three decimal places.
Step 1. Find h
h = (3 − 1) / 4 = 0.5
Step 2. Five y-values from x = 1 to x = 3
x: 1, 1.5, 2, 2.5, 3
y: 1, 0.6667, 0.5, 0.4, 0.3333
Step 3
(y₀ + y₄) = 1 + 0.3333 = 1.3333
2(0.6667 + 0.5 + 0.4) = 3.1333
Step 4
½ × 0.5 × (1.3333 + 3.1333)
estimate = 1.117 (3 d.p.)
The true value is ln 3 = 1.0986…, so again we are slightly over. Four strips is not many; with more we would get closer.
Use four strips to estimate ∫04 √x dx.
Step 1 and 2. h = 1, and the heights are square roots
x: 0, 1, 2, 3, 4
y: 0, 1, 1.4142, 1.7321, 2
Step 3
(0 + 2) = 2
2(1 + 1.4142 + 1.7321) = 8.2926
Step 4
½ × 1 × (2 + 8.2926)
estimate = 5.146 (3 d.p.)
The exact answer is 16/3 = 5.333…, so this time the estimate is too small. Same method, opposite outcome. The next section is about why.
After the calculation there is nearly always a second part: is this an overestimate or an underestimate, and how do you know? It is worth easy marks and most students guess.
You do not have to guess. It depends on one thing only: which way the curve bends.
Scroll the diagrams sideways.
| If the curve… | The chord sits… | Your answer is… |
|---|---|---|
| bends upwards (convex) | above the curve | an OVERestimate — you counted area that is not there |
| bends downwards (concave) | below the curve | an UNDERestimate — you missed a sliver under the curve |
| is a straight line | on the curve | exact. There is no error at all. |
The quick test, if you are allowed calculus
If the curve changes direction of bend
Some curves bend upwards in one place and downwards in another — sin x over a full cycle, for instance. Then the two errors partly cancel and you cannot say which way the total goes without more work. Exam questions avoid this, so if your interval contains a point of inflection, re-read the question: you have probably misread the limits.
How to say it in an exam
Do not just write “overestimate”. Write why: “The curve is convex on this interval, so each chord lies above the curve and each trapezium includes a little extra area. The estimate is therefore too large.” That sentence is the mark.
Thinner strips follow the curve more closely, so the answer gets better. Here is the same area — ∫04 x2 dx — with more and more strips.
Scroll the diagrams sideways.
| Strips | Estimate | Error |
|---|---|---|
| 1 | 32 | +10.667 |
| 2 | 24 | +2.667 |
| 4 | 22 | +0.667 |
| 8 | 21.5 | +0.167 |
| 16 | 21.375 | +0.042 |
| 32 | 21.344 | +0.010 |
Look at the error column
10.667, then 2.667, then 0.667, then 0.167. Each one is a quarter of the one above. Doubling the number of strips divides the error by four, every time.
That is not a coincidence of this example. The error in the trapezium rule is proportional to h2, and halving the strip width halves h, so the error falls by a factor of 22 = 4. It is worth knowing because it tells you what you get for your effort: going from 4 strips to 8 is worth far more than going from 20 to 24.
Three ways to make the estimate better
On IB Maths AI especially, the commonest version of this question gives you no equation. Just measurements. You cannot integrate a table of numbers, which is exactly why the trapezium rule exists.
A river is 12 m wide. Its depth is measured every 2 m across the river, giving the readings below. Estimate the area of the cross-section.
| Distance (m) | 0 | 2 | 4 | 6 | 8 | 10 | 12 |
|---|---|---|---|---|---|---|---|
| Depth (m) | 0 | 1.2 | 2.1 | 2.6 | 2.3 | 1.4 | 0 |
Count first. Seven readings means six strips
h = 2 (given by the measurements)
Ends once
0 + 0 = 0
Middles doubled
2(1.2 + 2.1 + 2.6 + 2.3 + 1.4)
= 2 × 9.6 = 19.2
Multiply by ½h
½ × 2 × (0 + 19.2)
area ≈ 19.2 m²
Notice there was nothing to differentiate, nothing to integrate and no way to check the answer. With real data, the trapezium rule is not an approximation to something better — it is the whole method.
What else gets asked with measured data
This single mistake costs more marks than everything else on this topic combined, so it gets its own worked comparison. A question gives you six y-values between x = 2 and x = 7. What is h?
Wrong
“Six values, so six strips.”
h = (7 − 2) ÷ 6 = 0.833…
Every y-value is now at the wrong x, and the whole answer is wrong.
Right
“Six values, so five strips.”
h = (7 − 2) ÷ 5 = 1
The readings fall at 2, 3, 4, 5, 6, 7. They line up.
The rule, in four words: ordinates are strips plus one. Think of a fence — five panels need six posts.
| The mistake | What goes wrong | Do this instead |
|---|---|---|
| Dividing by the number of ordinates | h is wrong, so everything after it is wrong. | h = (b − a) ÷ number of strips. |
| Doubling the first and last y | The outer edges belong to one trapezium each, not two. | Ends once, middles twice. |
| Forgetting the ½ | Your answer is exactly double the right one. | Multiply by ½h at the end, not h. |
| Calculator in degrees | Every y-value in a trig question is wrong from the first line. | Radians, unless the question says otherwise. |
| Rounding inside the table | Small errors build up and the final digit is wrong. | Keep full accuracy, round only on the last line. |
| Guessing over or under | Half the class gets it right by luck and loses the explanation mark anyway. | Check which way the curve bends and say so. |
| Calling the answer exact | It is an estimate unless the curve is a straight line. | Write ≈, not =, and use the word “estimate”. |
| You may see | Where | Is it different? |
|---|---|---|
| ½h[…] | IB formula booklet, many A-level texts | No. |
| (h/2)[…] | Other A-level texts | No — identical, written differently. |
| “trapezoidal rule” | IB, and American books | No. “Trapezium” and “trapezoid” are the same shape here. |
| f(x0), f(x1)… | Instead of y0, y1 | No. Two names for the same heights. |
Work each one on paper, then tap to check. Every answer below was verified.
h = (2 − 0) / 4 = 0.5
x: 0, 0.5, 1, 1.5, 2
y: 0, 0.125, 1, 3.375, 8
Ends 0 + 8 = 8, middles 2(0.125 + 1 + 3.375) = 9
½ × 0.5 × (8 + 9)
estimate = 4.25
Exact value is 4, so this is an overestimate — as expected, since f″ = 6x is positive here.
h = (5 − 1) / 4 = 1
x: 1, 2, 3, 4, 5
y: 1, 0.5, 0.3333, 0.25, 0.2
Ends 1.2, middles 2(0.5 + 0.3333 + 0.25) = 2.1667
½ × 1 × 3.3667
estimate = 1.6833
True value is ln 5 = 1.6094, so again an overestimate.
Nine ordinates means eight strips, because ordinates are always one more than strips. So h = (5 − 1) ÷ 8 = 0.5. Dividing by 9 would give 0.444… and wreck the whole question.
h = π/4, and the calculator must be in radians
y: 0, 0.7071, 1, 0.7071, 0
Ends 0, middles 2(0.7071 + 1 + 0.7071) = 4.8284
½ × (π/4) × 4.8284
estimate = 1.8961
Exact value is 2. An UNDERestimate, because f″ = −sin x is negative on this interval, so the curve bends downwards.
Overestimate. Differentiate twice: f″(x) = ex, which is positive everywhere. The curve bends upwards, so every chord lies above it.
Underestimate. f″(x) = −1/x2, which is negative for every x. The curve bends downwards, so the chords sit below it.
Doubling the strips divides the error by four. So eight strips gives about 0.2, and sixteen strips about 0.05. Each doubling buys you another factor of four.
Seven readings means six strips, and h = 5 seconds
Ends 0 + 10 = 10
Middles 2(4 + 9 + 13 + 15 + 14) = 2 × 55 = 110
½ × 5 × (10 + 110)
distance ≈ 300 m
Area under a speed-time graph is distance, so the units are metres, not m/s.
It is a way of estimating the area under a curve. You cut the region into vertical strips of equal width, treat the top of each strip as a straight line so that it becomes a trapezium, work out those areas and add them up. It is used when a function is awkward or impossible to integrate, or when you only have measured data.
∫ab y dx ≈ ½h[(y0 + yn) + 2(y1 + y2 + … + yn−1)], where h = (b − a) ÷ n and n is the number of strips. The first and last heights count once, everything in between counts twice.
It depends on which way the curve bends. If it bends upwards (convex), the chords lie above it and you get an overestimate. If it bends downwards (concave), the chords lie below and you get an underestimate. The quick test is the second derivative: f″(x) > 0 means over, f″(x) < 0 means under.
n + 1. Always one more ordinate than strips, because neighbouring strips share an edge. Five strips need six y-values. Think of a fence: five panels, six posts. Getting this backwards is the most common mistake on the topic.
h = (b − a) ÷ n, where n is the number of strips. If a question hands you a table rather than a value of n, count the entries and subtract one to get the strips first.
Because each inside ordinate is a side of two trapezia — the one to its left and the one to its right — so it gets used twice when you add all the areas up. The two outside ordinates belong to one trapezium each, so they are counted once.
No. It is on Maths AI, syllabus point SL 5.8, which means AI SL and AI HL. It does not appear in the AA formula booklet at either level and AA papers do not ask for it. If you are on AA you do not need this topic.
Yes, on AQA, Edexcel and OCR, in the numerical methods part of Pure. Questions usually ask for an estimate with a stated number of strips and then whether it is an over- or underestimate.
It depends on how much the curve bends and how thin the strips are. The error is proportional to h2, which means doubling the number of strips divides the error by four. In our example, going from 4 strips to 8 cut the error from 0.667 to 0.167.
When the “curve” is a straight line. Then the chord lies exactly on it and there is no gap at all, so the estimate equals the true area however few strips you use.
Nothing. “Trapezium” is the British name and “trapezoid” the American one, and the IB uses “trapezoidal rule”. Same shape, same formula, same method.
Then part of the estimate is too big and part too small, and you cannot say which way the total goes without splitting the interval at the point of inflection. Exam questions are set to avoid this, so if your interval looks like it contains one, check you have read the limits correctly.
Yes, and this is the most common version on IB Maths AI. With measured data there is no function to integrate, so the trapezium rule is not an approximation to a better method — it is the only method. Count the readings, subtract one for the number of strips, and read h off the spacing.
You have almost certainly multiplied by h instead of ½h. It is the single most common arithmetic slip on this topic, and the giveaway is that your answer is exactly twice the right one.
Use ≈. The trapezium rule gives an estimate, not an exact value, and some mark schemes expect to see that you know the difference. Use the word “estimate” in any sentence you write too.
Most students can get the number out. Far fewer pick up the marks for saying whether it is too big, why, and what would make it better — and those are the marks that separate a six from a seven. Our tutors teach IB and A‑level maths and work through your own past papers, so we fix the habits actually costing you marks rather than reteaching what you can already do.
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