Paper 2 only (not examined in Paper 3). Some definite integrals cannot be found exactly, and some functions are only known from a table of data. The trapezium rule estimates the area by joining points on the curve with straight lines, and a sketch tells you whether the estimate is too big or too small.
With \(n\) intervals of width \(h = \dfrac{b-a}{n}\) and ordinates \(y_0, y_1, \ldots, y_n\): \(\displaystyle\int_a^b y\,dx \approx \dfrac{h}{2}\left[y_0 + 2(y_1 + \cdots + y_{n-1}) + y_n\right]\). If the curve bends upwards (lies below its chords) the rule over-estimates; if it bends downwards it under-estimates.
Fig. 1: three trapezia of width h = 1 under y = x² + 1 from x = 0 to x = 3; the curve bends upwards, so every chord lies above it and the estimate 12.5 is an over-estimate of the exact area 12
Section 01
The rule: widths, ordinates, weights.
Split \([a, b]\) into \(n\) equal intervals (strips). Each strip is approximately a trapezium with parallel sides \(y_{r}\) and \(y_{r+1}\), area \(\dfrac{h}{2}(y_{r} + y_{r+1})\). Adding them counts each inside ordinate twice and each end ordinate once.
Width: \(h = \dfrac{b-a}{n}\), where \(n\) is the number of intervals, NOT the number of ordinates.
Ordinates: \(n\) intervals need \(n + 1\) values \(y_0, \ldots, y_n\). Lay them out in a table, keeping 4 or 5 significant figures.
Weights: ends \(\times 1\), every middle ordinate \(\times 2\), then multiply the total by \(\dfrac{h}{2}\).
Give the final answer to the accuracy asked, usually 3 significant figures or 2 decimal places.
Worked example A table of values
Use the trapezium rule with 4 intervals to estimate \(\displaystyle\int_0^2 \sqrt{1+x^3}\,dx\), giving your answer to 3 significant figures.
A car accelerates from rest. Its speed \(v\) m s-1 is recorded every 10 seconds: \(t = 0, 10, 20, 30, 40, 50\) gives \(v = 0, 6, 10, 12, 13, 13.5\). Estimate the distance travelled in the 50 seconds.
Distance \(= \displaystyle\int_0^{50} v\,dt\), with 5 intervals and \(h = 10\).
Use the trapezium rule with 3 intervals to estimate \(\displaystyle\int_0^3 \sqrt{x+1}\,dx\), to 3 significant figures.
\(h = 1\); \(y = 1, 1.41421, 1.73205, 2\). \(\dfrac{1}{2}\left[1 + 2 + 2(1.41421 + 1.73205)\right] = \dfrac{1}{2}(9.29253) = 4.65\) (3 s.f.). The exact value is \(\dfrac{14}{3} = 4.67\).
Examiner trap
\(n\) intervals have \(n + 1\) ordinates. With 4 intervals on \([0, 2]\), \(h = 0.5\), not \(\dfrac{2}{5}\). Equally, do not forget the \(\dfrac{h}{2}\) at the end, or the answer comes out twice as big.
Section 02
Over-estimate or under-estimate?
Sketch the curve over the interval and look at the chords. The syllabus expects this from a sketch graph in simple cases.
Shape over \([a, b]\)
Chords lie
Trapezium rule gives
Examples
bends upwards (convex)
above the curve
an over-estimate
\(e^{x}\), \(x^2\), \(\dfrac{1}{x}\) for \(x > 0\)
bends downwards (concave)
below the curve
an under-estimate
\(\ln x\), \(\sqrt{x}\), \(\sin x\) on \([0, \pi]\)
Worked example Compare with the exact value
(a) Use the trapezium rule with 4 intervals to estimate \(\displaystyle\int_1^3 \ln x\,dx\). (b) State, with a reason, whether this is an over-estimate or an under-estimate. (c) Given that the exact value is \(3\ln 3 - 2\), find the percentage error.
Answer: 1.28, an under-estimate, about 1.06% too small
Examiner trap
Do not judge by whether the curve is going up or down: \(y = \ln x\) is increasing, but it bends downwards, so the rule under-estimates. It is the bending (concavity) that matters, not the slope.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
The trapezium rule is used with 4 intervals to estimate \(\displaystyle\int_1^3 f(x)\,dx\). What is \(h\)?
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How many ordinates (\(y\)-values) are needed for the trapezium rule with 6 intervals?
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The trapezium rule is used to estimate \(\displaystyle\int_0^1 e^{x}\,dx\). The estimate is
Quick check+10 XP
Which is the trapezium rule with ordinates \(y_0, y_1, y_2, y_3\) and width \(h\)?
Section 03
Radians and more strips.
For trigonometric functions every \(x\)-value is in radians, so set the calculator to radian mode first. Doubling the number of intervals makes the chords hug the curve more closely and the error shrinks.
Worked example Halving the width
Estimate \(\displaystyle\int_0^{\pi} \sin x\,dx\) using the trapezium rule with (a) 2 intervals and (b) 4 intervals. Compare with the exact value 2.
Both are below 2 (the sine curve bends downwards on \([0, \pi]\)); the error falls from 0.429 to 0.104.
Answer: (a) 1.57 (b) 1.90; both under-estimates, and 4 intervals is far closer
Examiner trap
In degree mode \(\sin(0.785\ldots)\) is about 0.0137, not 0.707, and the estimate collapses. If a trigonometric estimate looks tiny, check the calculator mode.
Stage 2 · Exam
Exam-style questions
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Finish the four checks above to unlock the exam questions
Exam style+20 XP
Use the trapezium rule with 2 intervals to estimate \(\displaystyle\int_0^2 2^{x}\,dx\).
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For which integral does the trapezium rule give an over-estimate?
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Use the trapezium rule with 4 intervals to estimate \(\displaystyle\int_0^1 \sqrt{1+x^2}\,dx\), giving your answer to 3 significant figures.
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Use the trapezium rule with 2 intervals to estimate \(\displaystyle\int_0^{\pi/2} x\cos x\,dx\), giving your answer to 3 significant figures.
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