Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 2: Pure Mathematics 2 · official syllabus
INTEGRATION LAB
Learn it by playing it
Four short missions on reversing the new derivatives, the 1/a shrink from the chain rule, double-angle identities for squares, and the trapezium rule. When a mission glows green, read why it worked, then take the next one.
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Mission 1 of 4
Area under the reciprocal
The dashed bar is the exact area under y = 1/x between x = 1 and x = 3. Your solid bar is your estimate. Reverse the derivative of ln x to find the area, then slide a until your bar closely matches the target.
Before you slide: what is the antiderivative of 1/x?
Your bar reads 1.00; the dashed target sits at the exact area ln 3 = 1.0986.... Work out ln 3 - ln 1, then slide to its nearest hundredth.
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
Reverse differentiation with the 1/a factor 2.5
Integration undoes differentiation, and the chain rule's multiplier comes back out as a division. The core table: integral of e^(ax + b) is (1/a) e^(ax + b) + c, of sin(ax + b) is -(1/a) cos(ax + b) + c, of cos(ax + b) is (1/a) sin(ax + b) + c, and of sec^2(ax + b) is (1/a) tan(ax + b) + c.
The reciprocal joins the table too: the integral of 1/(ax + b) is (1/a) ln|ax + b| + c, because differentiating ln(ax + b) gives a/(ax + b).
Always check by differentiating back: d/dx [(1/3) sin(3x)] = cos(3x), so that integral is right; forgetting the 1/a (writing 3 sin(3x)) is the slip this topic is built to catch.
Only a linear inner function works this way. For anything like sin(x^2) or e^(x^2) the 1/a trick does not apply - Paper 2 does not require the general substitution method.
Definite integrals as area 2.5
A definite integral is the antiderivative evaluated top limit minus bottom limit: write it in square brackets, substitute, subtract. The constant c cancels, so leave it out of definite work.
Worked example: the area under y = 1/x from 1 to 3 is [ln x] from 1 to 3 = ln 3 - ln 1 = ln 3 = 1.099 to three decimal places.
Worked example: the area under y = cos(3x) from 0 to pi/6 is [(1/3) sin(3x)] between the limits = (1/3)(sin(pi/2) - sin 0) = 1/3 exactly.
More from the table: the integral of e^(2x) from 0 to 1 is (e^2 - 1)/2 = 3.195 to three decimal places, and the integral of sec^2(2x) from 0 to pi/8 is (1/2) tan(pi/4) - 0 = 1/2 exactly.
Keep exact forms (ln 3, 1/3, e^2 - 1 over 2) until the final line; round only when the question asks for a decimal.
Using identities to integrate sin^2 and cos^2 2.5
There is no direct rule for sin^2 x or cos^2 x. The double-angle formula cos 2x = 1 - 2 sin^2 x rearranges to sin^2 x = (1 - cos 2x)/2, and cos 2x = 2 cos^2 x - 1 gives cos^2 x = (1 + cos 2x)/2.
Worked example: the area under y = sin^2 x from 0 to pi is [x/2 - (sin 2x)/4] between the limits = pi/2 - 0 = pi/2 = 1.571 to three decimal places. The sine term vanishes because sin 0 = sin 2pi = 0.
A scaled version: cos^2(2x) = (1 + cos 4x)/2, so its integral is x/2 + (sin 4x)/8. From 0 to pi/8 that gives pi/16 + 1/8 = 0.321 to three decimal places.
Spot the pattern: over a whole number of half-periods the sine term contributes nothing and the answer is just x/2 across the interval.
The trapezium rule 2.5
When an integral has no easy antiderivative, slice the region into n strips of width h and replace each curve top with a straight chord: T = (h/2)[y0 + yn + 2(y1 + y2 + ... + y(n-1))].
The ordinates come from the table of values: for y = 1/(1 + x) on [0, 2] with n = 4 and h = 0.5, they are 1, 2/3, 1/2, 0.4 and 1/3.
Worked example: T = 0.25 x (1 + 1/3 + 2(2/3 + 1/2 + 0.4)) = 67/60 = 1.117 to three decimal places. First plus last once each, everything in between doubled, all multiplied by h/2.
More strips shrink the gap between chord and curve, so the estimate improves - but every strip count gives a value you should sanity-check against a sketch.
Over-estimate or under-estimate? 2.5
Sketch the curve on the strips. If it is convex (curving upwards, like y = 1/(1 + x) on [0, 2]), each chord lies above the curve and the trapezium rule over-estimates the area.
If the curve is concave (curving downwards), each chord lies below the curve and the rule under-estimates.
The worked example above proves it: the estimate 67/60 = 1.1166... sits above the exact area ln 3 = 1.0986..., exactly as the convex sketch predicts.
Exam answers should state the direction and the reason in one line: 'over-estimate, because the curve is convex so each trapezium lies above it' earns the mark.
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