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A Level · Cambridge (CIE) · 9709

Use curve shape to judge trapezium overestimates and underestimates

Compare straight chords with a curve, distinguish curvature from whether it rises or falls, and preserve the sign of the estimated integral.

Pure Mathematics 2 · Integration · 2.5
Exam scope and assessment notes

Pure Mathematics 2 (Paper 2, AS only) · Mathematics 9709, 2026–2027. Paper 1 integration and Paper 2 derivatives and identities are assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Section 2.5 covers reverse differentiation of linear-composite exponentials, reciprocals and trigonometric functions, integration using trigonometric relationships, and trapezium estimates with sketch-based over/under-estimation. The general method of integration by substitution is not required here. Use radians for trigonometric calculus. Written practice is self-checked.

Before you begin

  • Calculate a trapezium estimate and an elementary definite integral.
  • Read a graph's bending direction and distinguish integral from geometric area.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

Does an increasing graph alone tell you the sign of its trapezium error?

What you will learn

  • Use a sketch to decide whether trapezium chords lie above or below a curve.
  • Explain signed over/under-estimation, exact straight-line cases and the limitation when curvature changes.
Compare chords with the curveUpward bendy = x²024012T = 3 > 8/3Downward bendy = 4 − x²024012T = 5 < 16/3Blue: curve · Red dashed: chords
Both panels use two width-1 intervals on [0, 2], with identical drawing scales. The left blue curve y = x² bends upward, so the red chords lie above it: estimate 3 is greater than exact integral 8/3. The right blue curve y = 4 − x² bends downward, so the red chords lie below it: estimate 5 is less than exact integral 16/3. The pale regions show the gaps between chords and curves. It is their bending direction, rather than increasing or decreasing behaviour alone, that justifies these comparisons. Straight lines would have no gap.

The chord's position decides the estimate direction

When a curve bends upward throughout an interval, like y = x², its chords lie on or above the curve, so the trapezium estimate is at least the signed integral. When it bends downward, like y = 4 − x² on [0, 2], the chords lie on or below it, giving an underestimate. For smooth examples, the signs f″ ≥ 0 and f″ ≤ 0 respectively confirm those shapes, but the simple sketch comparison itself explains the area relationship.

With two equal intervals on [0, 2], x² gives T = 3 while its exact integral is 8/3, an overestimate by 1/3. For 4 − x², the corresponding T is 5 while the exact integral is 16/3, an underestimate by 1/3. A straight line agrees exactly with its chord, so the trapezium rule is exact on each such strip. Rising or falling alone does not determine the direction: an increasing curve can bend either upward or downward.

A signed overestimate need not mean a larger geometric area

For x² − 4 on [0, 2], the curve is below the axis and bends upward. Its two-strip estimate is −5, greater than exact signed integral −16/3, so it is a signed overestimate. The corresponding positive geometric areas are 5 and 16/3: the straight-sided area magnitude is smaller. State which quantity is being compared rather than replace every negative ordinate by its magnitude without explaining the changed question.

If the curve changes bending direction, some chords may lie above and others below. A single whole-interval sketch label need not establish the total error's sign; inspect the separate strips or compare with an independently known integral. For sin x on [0, 2π] with four equal intervals, symmetry gives signed trapezium estimate 0 and exact integral 0 even though the curve bends differently in its two halves. This cancellation does not imply that its total geometric area is zero: that exact area is 4. A smaller mesh often improves these smooth examples, but an observed change between estimates is not a guaranteed error bound.

Pause and explain

For a below-axis upward-bending curve, what does a chord above it imply?

Put the idea to work

Worked example

Compare two-strip trapezium estimates with exact integrals for x² and 4 − x² on [0, 2], explaining the direction from their sketches.

Show the worked solution
  1. For x², ordinates 0, 1, 4 give T = 3, above exact integral 8/3; the chords lie above the upward-bending curve.
  2. For 4 − x², ordinates 4, 3, 0 give T = 5, below exact integral 16/3; the chords lie below the downward-bending curve.
  3. Both comparisons use the signed integral. Their exact differences are +1/3 and −1/3 respectively.

Answer x²: overestimate 3 versus 8/3. 4 − x²: underestimate 5 versus 16/3.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

For f(x) = x² − 4 on [0, 2], find its two-strip trapezium estimate and exact integral. Explain both the signed and geometric comparisons.

Give me a hint

Use ordinates −4, −3, 0, retaining their signs, then integrate the polynomial.

Compare my reasoning
  1. T = (1/2)[−4 + 0 + 2(−3)] = −5.
  2. The exact integral is [x³/3 − 4x] from 0 to 2 = −16/3, so −5 is a signed overestimate.
  3. The positive straight-sided area magnitude 5 is below exact geometric area 16/3; the curve is non-positive throughout.

Signed estimate −5 versus exact −16/3: overestimate. Positive area magnitude 5 versus 16/3: underestimate.

Look for these in your work

  • Keep the ordinates signed when estimating the integral.
  • Explain why the geometric magnitude comparison differs.
2 · Independent

Use two equal intervals to estimate ∫ from 0 to 4 of √x dx and compare with its exact value. Explain why an increasing graph gives an underestimate here.

Give me a hint

The width is 2 and the graph bends downward; use ordinates 0, √2, 2.

Compare my reasoning
  1. The estimate is (2/2)[0 + 2 + 2√2] = 2 + 2√2.
  2. The exact integral is [(2/3)x^(3/2)] from 0 to 4 = 16/3.
  3. Since 2 + 2√2 < 16/3, the estimate is low: the increasing but downward-bending curve lies above its chords.

Estimate 2 + 2√2 < exact 16/3; an underestimate despite the curve increasing.

Look for these in your work

  • Use the correct width and exact ordinates.
  • Justify the estimate direction by chord position rather than increasing behaviour alone.
3 · Transfer

For sin x on [0, 2π], use four equal trapezium intervals. Compare the signed estimate with the exact integral and explain why it does not describe total area.

Give me a hint

Use the symmetric ordinates 0, 1, 0, −1, 0 and inspect both lobes.

Compare my reasoning
  1. With h = π/2, opposite interior ordinates cancel, giving T = 0.
  2. The exact signed integral is [−cos x] from 0 to 2π = 0; the agreement comes from cancellation in this case, not one fixed bending direction.
  3. The exact total area is 2 + 2 = 4, found by splitting at π; a zero signed integral does not mean no area.

Signed estimate and exact integral are both 0; exact total geometric area is 4.

Look for these in your work

  • Use signed ordinates and recognise symmetry-based cancellation.
  • Separate a signed estimate from geometric area and avoid a whole-interval curvature claim.

Common mistakes

  • Increasing does not automatically mean overestimate, and decreasing does not automatically mean underestimate.
  • For a below-axis curve, compare signed integrals separately from positive area magnitudes.
  • Changed curvature or changed mesh alone does not establish a universal total-error bound.
Recall without your notes

What can you explain now?

Is the trapezium rule exact for a straight-line function on each strip?

Compare with the explanation

Yes.

The straight interpolation coincides with the original function, so there is no gap between chord and curve.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Sketch one upward-bending and one downward-bending example and compare their chords with the curves. Explain a below-axis case and why mixed curvature needs more care.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Mathematics

Course scope and assessment

312 sequenced lessons provide 936 guided, independent and application problems covering the 160 objectives in sections 1.1–1.8, 2.1–2.6, 3.1–3.9, 4.1–4.5, 5.1–5.5 and 6.1–6.5. The map lists all 38 sections across six alternative components; all eight Pure Mathematics 1 sections and all six Pure Mathematics 2 sections, from Paper 2 Algebra through Numerical solution of equations, currently have full teaching sequences. Paper 3 Algebra, section 3.1, links the six shared foundation lessons plus six lessons on partial fractions and rational binomial expansions; Paper 3 logarithms, trigonometry and numerical equations use the shared lessons, with the Paper 1 graph foundation linked into trigonometry. Paper 3 differentiation combines the shared methods with its inverse-tangent derivative lesson. Paper 3 Integration links four shared standard and identity lessons plus seven advanced lessons, covering all six section 3.5 requirements. Eight Vectors lessons cover all six section 3.7 requirements, with 2D and 3D line and solid applications. Nine Differential equations lessons cover all four section 3.8 requirements, including Paper 3 integration techniques, initial conditions, equilibrium solutions and conditional model interpretations. Eleven Complex numbers lessons cover all eight section 3.9 requirements, with seven Argand diagrams and explicit argument and boundary checks. All nine Pure Mathematics 3 sections now have teaching linked to every mapped requirement. Nine Forces and equilibrium lessons cover all seven section 4.1 requirements, with six force diagrams, signed incline-friction intervals, reaction-dependent pull and push limits and separate-body third-law checks. Nine Kinematics lessons cover the four official section 4.2 requirements, with the graph requirement mapped to three relationships. 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Nine Energy, work and power lessons cover all five section 4.5 requirements, with angle-dependent work, kinetic and reference potential energy, signed net-work balances, smooth curved slides, rough and driven slopes, ideal connected-system energy and instantaneous power on hills. Five diagrams distinguish the energy inputs, losses and valid positive-speed force model. All five Mechanics sections now have teaching and staged practice linked to their mapped requirements. Thirteen Representation of data lessons cover all five section 5.1 requirements, including choice and limitations of displays, ordered and back-to-back stems, raw summaries and box plots, unequal-width density histograms, cumulative-frequency interpolation, descriptive moment calculations, grouped estimates, coding, combining two datasets and correcting totals. Six accessible diagrams retain repeated observations and distinguish numerical spread, frequency area and estimated readings. Ten Permutations and combinations lessons cover both official section 5.2 requirements, with selection and ordered roles, repeated objects, adjacency blocks, non-adjacency gaps, leading-zero cases, labelled multi-row seating and mixed restrictions. Six accessible diagrams show outcome trees, ordered versus unordered choices, repeated copies, blocks, gaps and distinct row seats. Eleven Probability lessons cover all four section 5.3 requirements, with equiprobable elementary outcomes, selection-based probabilities, complements and disjoint regions, independence tests, restricted sample spaces, two-way tables, replacement and removal trees, reversed conditions and unknown parameters. Seven accessible diagrams show dice outcomes, random subsets, event regions, a conditioned grid, a count table and two probability trees. Twelve Discrete random variables lessons cover all three official section 5.4 requirements, with complete probability tables, moment and parameter checks, binomial assumptions, exact probabilities and moments, geometric first-success probabilities, tails, expectation and integer reliability thresholds. Six accessible diagrams distinguish outcome grouping, weighted moments, fixed trials, discrete probability bars, waiting conventions and integer tail deadlines. Twelve Normal distribution lessons cover all three official section 5.5 requirements, with continuous-area interpretation, MF19 cumulative tables, full standardisation, tail and interval regions, conditional probabilities, percentiles, one- and two-parameter inference, binomial suitability, continuity corrections and approximate integer thresholds. Seven accessible diagrams show model areas, table reading, aligned scales, intervals, quantiles, matched binomial and normal shapes and whole integer cells. 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Fourteen Linear combinations lessons cover the six listed results in official section 6.2, with affine moments, signed scales and constant cases, expectation without independence, independent squared variance contributions, fixed and random fees, independent copies versus reused values, affine normal events, independent normal sums and differences, weighted and conditional areas, coefficient inference, exact independent Poisson sums and overlapping exposure limits. Nine accessible diagrams distinguish scale changes, variance contributions, component costs, copy mechanisms, aligned normal scales, summed areas, Poisson count pairs and unsupported distribution claims. 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Written practice is self-checked, not officially marked. Check the 2028–2030 syllabus separately if sitting an examination from 2028; this teaching map uses 2026–2027 requirements.

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