Choose exact integration or data estimates and explain their limits
Connect methods to the evidence available, keep signed and accumulated quantities distinct and avoid treating agreement or refinement as proof of accuracy.
Section 2.5 · Integration
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
- Integrate the section's linear natural and trigonometric forms and use identities.
- Calculate trapezium estimates and judge chord positions using curve sketches.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What information permits an exact actual integral to be inferred from a finite data table alone?
What you will learn
- Choose a suitable exact primitive or trapezium estimate for a supplied function or measurement table.
- Interpret units, initial values, sign changes and what the calculation does or does not establish.
Use an exact primitive when the function and method permit it
A supplied expression may give a directly checkable exact integral, such as ∫ from 0 to ln 2 of (2eᵗ + 1) dt = [2eᵗ + t] from 0 to ln 2 = 2 + ln 2. If a question requests a trapezium approximation, use its stated partition even when an exact primitive also exists; an exact value can then check the approximation. If only measurements are supplied, the unknown between-point curve generally prevents an exact actual integral from being inferred from the table.
A velocity integral gives signed displacement, while distance requires splitting at velocity sign changes and adding the magnitudes of the separate displacements. For v(t) = cos t cm/s on [0, π] seconds, the displacement is 0 but distance is 2 cm, split at t = π/2. A two-strip signed trapezium estimate also gives 0 from ordinates 1, 0, −1. Its agreement with the displacement does not prove the particle stayed still or that an area-magnitude estimate is exact.
Refinement is useful evidence, not an automatic guarantee
Suppose measured inflow rates at t = 0, 1, 2, 3, 4 minutes are 2, 5, 6, 4, 1 litres per minute. The four-strip estimate is (1/2)[2 + 1 + 2(5 + 6 + 4)] = 16.5 litres. Using only times 0, 2, 4 gives a two-strip estimate 15 litres. The difference 1.5 litres shows sensitivity to the sampled shape; it is not a proven maximum error relative to the actual unknown integral. An unmeasured bump can keep every recorded ordinate the same while changing the integral.
If the initial stock is 20 litres and there is no outflow, the finer estimate gives final total 36.5 litres. Report the interpolation-based estimate, its time interval and units. State any model or sampling limitation relevant to the requested conclusion. Do not label a data estimate as an exact measured total, or claim a target accuracy solely because two successive estimates are close. For power data in kW sampled in minutes, convert widths to hours before reporting kWh.
Pause and explain
For v(t) = cos t on [0, π], does displacement 0 imply distance 0?
Worked example
Measured inflow rates at 0, 1, 2, 3, 4 minutes are 2, 5, 6, 4, 1 litres per minute. A vessel starts with 20 litres and has no outflow. Estimate its final volume and compare with the two-strip estimate from times 0, 2, 4.
Show the worked solution
- Four width-1 strips give estimated increase (1/2)[2 + 1 + 2(5 + 6 + 4)] = 16.5 litres.
- The final total estimate is 20 + 16.5 = 36.5 litres. Two width-2 strips from ordinates 2, 6, 1 give increase 15 litres.
- The difference is 1.5 litres; it indicates partition sensitivity but supplies no guaranteed error bound on the unknown actual rate integral.
Answer Estimated increase 16.5 litres; final total 36.5 litres. Coarser increase 15 litres; difference 1.5 litres is not a guaranteed error bound.
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Test this topic →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.
A supplied flow is q(t) = 2eᵗ + 1 litres per minute for 0 ≤ t ≤ ln 2. Find the added volume exactly and explain why an exact method is available.
Give me a hint
The given function has a direct exponential-plus-constant primitive.
Compare my reasoning
- A primitive of q is 2eᵗ + t.
- Upper-minus-lower evaluation gives (4 + ln 2) − 2 = 2 + ln 2.
- The increase is exactly 2 + ln 2 litres within the supplied model; a measured table alone would not provide this same between-point information.
Exact modelled increase 2 + ln 2 litres.
Look for these in your work
- Use a verified primitive and both endpoints.
- Explain which supplied function information supports an exact modelled integral.
A particle has supplied velocity cos t cm/s on 0 ≤ t ≤ π seconds. Find its exact displacement and distance, and explain a two-strip trapezium estimate of zero.
Give me a hint
Split the motion at the velocity zero t = π/2 before finding distance.
Compare my reasoning
- The signed displacement is [sin t] from 0 to π = 0.
- The positive half contributes +1 cm and the negative half −1 cm, giving total distance 2 cm.
- Two trapezium strips with ordinates 1, 0, −1 also cancel to zero; that estimates signed displacement and does not mean no motion.
Displacement 0 cm; distance 2 cm; the zero signed trapezium estimate hides cancelling motion.
Look for these in your work
- Separate displacement from distance using the velocity sign change.
- Interpret the numerical rule as signed accumulation rather than proof of no movement.
Measured power at 0, 2, 4, 6 minutes is 1, 3, 2, 4 kW. Estimate energy use in kWh with three trapezia and state whether the measurements prove an exact energy total.
Give me a hint
Convert the common two-minute width to 1/30 hour before weighting the power ordinates.
Compare my reasoning
- Three strips have width h = 1/30 hour, with endpoints 1 and 4 and interior ordinates 3 and 2 kW.
- The estimate is (1/60)[1 + 4 + 2(3 + 2)] = 15/60 = 1/4 kWh.
- Report approximately 0.25 kWh from straight interpolation; the measurements do not establish an exact actual total or a guaranteed accuracy.
Estimated energy 1/4 = 0.25 kWh; the measured table alone does not prove an exact total.
Look for these in your work
- Match kW to hour-based widths before accumulating energy.
- State the interpolation estimate and its evidence limitation.
Common mistakes
- A data table does not determine the true between-point function unless an interpolation model is explicitly given.
- Equal signed integrals can hide non-zero distance or area.
- Convert time units, add initial stocks separately and avoid unsupported accuracy claims.
What can you explain now?
Does a small change between successive trapezium estimates prove a specified accuracy for an unknown measured-rate integral?
Compare with the explanation
No.
The difference only compares those approximations; unobserved between-point behaviour can change the actual integral.
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Make it stick
Choose an exact or numerical method for a function and for a table. Explain signed cancellation, unit conversion, initial stock and why estimate refinement alone does not certify an accuracy.
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.