Pure Mathematics 2

Deep Dive: Definite Integrals and Areas

Provenance (2026-09-29): written after reading research/veridian/WMA12-verified-facts.md in full (§1–§9, including the two independent mark-scheme-bullet audit passes on this exact lesson at §9), content/veridian/curriculum/wma12/definite-integrals-and-areas.ts in full (979 lines — every block, not just the header), and the three existing technique-hub documents for WMA12 (codex-formula-and-method-bank.md, forge-exam-technique-guide.md, signal-frequency-and-boundaries.md), plus a keyword sweep of content/veridian/question-bank/wma12.ts (no integration/area items currently present, confirmed by grep — nothing to cross-check there yet). All arithmetic in Part 2 below is original to this document and was worked by hand twice, independently, before being written in; none of it reproduces a real exam question's numbers.

#CODEX Deep-Dive: Definite Integrals and Areas (WMA12, spec 8.1–8.2)

#Scope, held deliberately

This document covers exactly what definite-integrals-and-areas.ts covers and no more: spec items 8.1 (evaluation of definite integrals) and 8.2 (the definite integral as an area — under a curve, between a curve and a line, between two curves). It does not cover the trapezium rule (spec 8.3) — that is the-trapezium-rule.ts's own topic, with its own verified error taxonomy in the facts bank's §6.9, and folding it in here would blur a spec line the lesson file itself deliberately keeps separate (see that file's own header note).

What already exists and is not repeated here. The CODEX formula bank (§8 of codex-formula-and-method-bank.md) already states the three governing formulas (power-rule antiderivative; area under a curve; area between curve/line or two curves) and their memorised/given status. The FORGE guide already names this lesson's six trap-taxonomy items and one A1ft example. SIGNAL already gives the topic's confirmed frequency floor (3 series) and mark range (5–6). The lesson file itself already derives, from a Riemann-sum limit, exactly why "integral = area" only when the integrand keeps one sign, and already carries a fully worked axis-boundary example (curve + y-axis + line, one limit solved, one given), a three-method comparison for curve-and-line area, and a marked solution splitting a region at a sign change. None of that is reproduced below. What this document adds: a compact statement of the whole model in one place; short original micro-examples for the two sub-cases the lesson gives the thinnest treatment (area between two curves, and multi-region decomposition strategy); and a synthesis of the three real exam anchors' own mark-scheme skeletons side by side, which exists piecemeal across the facts bank's audit log but not as a single comparison table anywhere else in this course's materials.

#Part 1 — The real model, compactly

#1.1 What a definite integral is, mechanically

$\int_a^b f(x)\,dx = F(b) - F(a)$ , where $F'(x) = f(x)$ . Find an antiderivative term by term (power up by one, divide by the new power — spec 8.1 has no formula printed for this anywhere in the Mathematical Formulae and Statistical Tables booklet; it is 100% memorise-and-apply), substitute the upper bound, substitute the lower bound, subtract. There is no +c to carry, because whatever constant an antiderivative has cancels identically in the subtraction. This is definitional bookkeeping — it produces a number regardless of what that number means geometrically, which is exactly the gap spec 8.2 exists to close.

#1.2 When that number is an area

The definite integral is the limit of a signed Riemann sum: thin vertical strips of width $\Delta x$ , height $f(x_i)$ , summed, as $\Delta x \to 0$ . A strip's height is a genuine non-negative geometric length only when $f(x_i) \geq 0$ . So:

If $f(x) \geq 0$ throughout $[a,b]$ : the integral is the area, exactly.
If $f(x) \leq 0$ throughout $[a,b]$ : every strip's signed contribution is negative, but its size still equals the true geometric area of that strip (only the sign flipped, not the magnitude) — so the area is $\left|\int_a^b f(x)\,dx\right|$ .
If $f(x)$ changes sign inside $(a,b)$ : a single continuous sum mixes positive contributions from one side of the crossing with negative contributions from the other before either can be isolated — the result is neither the true area (which needs both pieces counted positive) nor an honest signed total (already corrupted by the partial cancellation). The only fix that matches the geometry: split at every sign change, take $|{\cdot}|$ of the integral over each single-sign piece, add.

This is the entire content of spec 8.2's "interpretation" clause. Everything below is this one condition applied to three progressively more complex boundary shapes.

#1.3 The three named sub-cases, and why they're really one calculation

Spec 8.2 names three cases explicitly: area under a curve; area of a region bounded by a curve and given straight lines; area of a region bounded by two curves. Each is $\int_a^b (\text{upper} - \text{lower})\,dx$ between the correct limits — the first is the special case where the lower boundary is $y=0$ (the x-axis, itself a straight line); the second generalises the lower (or upper) boundary to any line; the third drops the requirement that either boundary be straight at all. The limits $a$ and $b$ come from wherever the two boundaries actually meet — solved by setting the two expressions equal — except where a boundary is an axis, which supplies its own limit directly with no equation to solve at all (the lesson's own worked-chain on this is the fullest treatment; see Part 3 below for how that shows up in a real mark scheme).

#1.4 The one explicit restriction

The spec states, verbatim: "∫x dy will not be required." Every region in this topic is sliced into vertical strips and integrated with respect to $x$ — re-expressing a boundary as $x$ in terms of $y$ and integrating with respect to $y$ is a real technique that belongs to a later unit, not this one. A WMA12 area question never requires it.

#Part 2 — Sub-cases, with short original micro-examples

Two sub-cases below are genuinely new content relative to the lesson file; the rest are stated briefly, with a pointer, specifically so this section reads as a complete map of the topic rather than a partial one.

#2.1 Area under one curve, single sign — pointer only

Fully covered by the lesson's own worked-chain ( $\int_1^4(3x^2-2x+5)\,dx$ , discriminant check confirms no real roots) and chain-drill ( $y=x^2+1$ on $[0,3]$ ). Nothing to add: check the sign first (discriminant, or inspection of a term that can't be negative), integrate, evaluate, state the area because the check makes it safe to.

#2.2 Area under one curve requiring a split — a short original example

$f(x) = 2x - 4$ on $[0,4]$ . Root at $x=2$ ; negative on $(0,2)$ , positive on $(2,4)$ .

$\int_0^2 (2x-4)\,dx = \left[x^2-4x\right]_0^2 = (4-8)-0 = -4 \Rightarrow$ area $=4$ . $\int_2^4 (2x-4)\,dx = \left[x^2-4x\right]_2^4 = (16-16)-(4-8) = 0-(-4) = 4 \Rightarrow$ area $=4$ . Total area $=4+4=8$ .

(Check by geometry, since this one is linear: two right triangles, each base 2, height 4, area $\tfrac12(2)(4)=4$ — confirms both integrals independently.) The single-integral trap the lesson's marked solution already covers in depth applies identically here: $\int_0^4(2x-4)\,dx = \left[x^2-4x\right]_0^4 = 0$ , which is not the area (8) and not even an honest signed total of anything geometric — the positive and negative strips cancelled to zero before either could be isolated.

#2.3 Area between a curve and a line — a short original example

Curve $C$ : $y=x^2+2$ ; line $l$ : $y=x+4$ . Solve $x^2+2=x+4 \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0 \Rightarrow x=-1,\,2$ . Check which is on top: at $x=0$ , line $=4>$ curve $=2$ , so the line is above throughout.

\int_{-1}^{2}\left[(x+4)-(x^2+2)\right]dx = \int_{-1}^{2}(-x^2+x+2)\,dx = \left[-\frac{x^3}{3}+\frac{x^2}{2}+2x\right]_{-1}^{2} = \frac{10}{3}-\left(-\frac{7}{6}\right) = \frac{9}{2}.

Same method as the lesson's own $y=8x-x^2$ / $y=2x$ example (three creditable routes exist: combine first; integrate separately and subtract; use a named elementary shape where the line permits one) — not re-derived here since that comparison is already built out in full.

#2.4 Area with one boundary supplied directly by an axis — pointer only, with one generalisation

Fully covered by the lesson's worked-chain (curve $y=2x^2+5$ , y-axis, line $y=25$ — one limit solved at $x=\sqrt{10}$ , the other given directly as $x=0$ , each its own standalone mark, never folded into one "both limits" method mark). The one thing worth stating explicitly, because the lesson's own example only shows the y-axis case: the principle generalises unchanged to an x-axis boundary, and to a boundary pair of two curves rather than curve-and-line — an axis boundary is never solved for, regardless of which axis it is or what the other boundary happens to be, and confusing a function's value at that axis with the x-limit itself is the one documented real error this shape produces (§2.6's trap-taxonomy cross-reference below).

#2.5 Area between two curves (no line at all) — a short original example, and an honest gap

Curve $C_1$ : $y=x^2$ ; curve $C_2$ : $y=4x-x^2$ . Solve $x^2 = 4x-x^2 \Rightarrow 2x^2-4x=0 \Rightarrow 2x(x-2)=0 \Rightarrow x=0,\,2$ . At $x=1$ : $C_1=1$ , $C_2=3$ , so $C_2$ is on top.

\int_0^2\left[(4x-x^2)-x^2\right]dx = \int_0^2(4x-2x^2)\,dx = \left[2x^2-\frac{2x^3}{3}\right]_0^2 = \left(8-\frac{16}{3}\right)-0 = \frac{8}{3}.

Method identical to curve-and-line: solve for the true intersections, check which graph is on top, integrate (upper − lower). This is the lesson's own di-mcq-3 treatment (a different curve pair, $y=x^2$ and $y=8-x^2$ ) extended into a fuller worked step — genuinely thin coverage in the lesson (one MCQ, no worked chain), which this fills.

Flagged honestly: all three real exam anchors this course has confirmed for this topic (Part 3 below) are curve-and-line questions — a grep of the facts bank for "two curves" or "curve and a curve" surfaces only the general spec-list mention (§7, §2), never a named, dated WMA12 series testing a pure two-curve region with no line involved. Spec 8.2 names the case explicitly, so it is real, examinable content and is taught here on that basis — but no confirmed real WMA12 anchor for it exists in this course's research to date. This is a genuine gap in the archive, not a claim that the case is untested; a future pass re-reading the six series flagged as "grepped but not read in full depth" in the facts bank's own §8 (October 2019, October 2021, October 2024, June 2023, June 2024, January 2025) might close it.

#2.6 Multi-region decomposition strategy — a short original example

The real January 2023 Q9 anchor (Part 3 below) defines two related regions, $R_1$ and $R_2$ , from the same curve and line, and its examiner report confirms multiple independently creditable strategies for the second region: subtracting $R_1$ from a triangle, or finding one larger bounded region and subtracting a smaller one. That strategy — reconstruct a second region from a named elementary shape rather than writing a fresh integral for it — is real, spec-relevant technique that the lesson's own method-comparison block doesn't model (it compares three routes to one region, not the relationship between two).

Curve $C$ : $y=6x-x^2$ ( $x\geq0$ ); line $l$ : $y=5$ . Solve $6x-x^2=5 \Rightarrow x^2-6x+5=0 \Rightarrow (x-1)(x-5)=0 \Rightarrow x=1,\,5$ .

$R_1$ (between curve and line, $1\leq x\leq5$ , curve on top — check $x=3$ : curve $=9>5$ ):

\int_1^5\left[(6x-x^2)-5\right]dx = \left[3x^2-\frac{x^3}{3}-5x\right]_1^5 = \frac{25}{3}-\left(-\frac{7}{3}\right) = \frac{32}{3}.

$R_2$ (bounded by curve, line, and the y-axis, $0\leq x\leq1$ , line on top — curve $(0)=0<5=$ line, and the two meet at $x=1$ ): instead of a fresh integral, use the rectangle formed by the line, the y-axis and the x-axis over $[0,1]$ (area $5\times1=5$ ) minus the area under the curve over the same interval ( $\int_0^1(6x-x^2)\,dx=\left[3x^2-\frac{x^3}{3}\right]_0^1=\frac{8}{3}$ ):

R_2 = 5-\frac{8}{3} = \frac{7}{3}.

(Check by direct integration: $\int_0^1\left[5-(6x-x^2)\right]dx=\left[5x-3x^2+\frac{x^3}{3}\right]_0^1=\frac{7}{3}$ — matches.) The decomposition route and the direct route agree, as they must; the decomposition route is the one worth reaching for when the "large shape minus already-found piece" is a rectangle or triangle, exactly as the real anchor's examiner report confirms is independently creditable.

#Part 3 — Synthesis: every real exam angle this topic has actually been tested from

Three confirmed real WMA12 anchors exist for spec 8.2 in this course's archive (20 series with question paper + mark scheme read; this topic verified present in 3 of 20 — a floor on individually-cited series, not a rate, since full per-question topic-tagging of all 20 has not been completed, per SIGNAL's own stated caveat):

Series & Q	Setup	Marks	Real mark-scheme skeleton (verified)
Jan 2020, Q4	Curve $y=2x^2+7$ ( $x\geq0$ ); region bounded by curve, y-axis, and line $y=17$	6	"Line and curve separate" route: M1 A1 B1 M1 M1 A1 — the B1 is the axis-given limit ( $x=0$ ), credited standalone and separately from the solved intersection ( $x=\sqrt5$ , from $2x^2+7=17$ )
Jan 2023, Q9	Curve $C$ : $y=x^2-4x+5$ ; line $l$ : $y=2$ ; two regions $R_1$ , $R_2$	Part (b) 2, part (c) 5	Part (b): M1 (sets $C=l$ , solves) A1 ( $x=1,3$ ). Part (c): multiple independently creditable strategies for $R_2$ (subtract $R_1$ from a triangle; large region minus small one)
June 2025, Q3	Curve $C$ : $y=3x+\frac{16}{x^2}-8$ ( $x>0$ ); region bounded by $C$ and a line segment	5	M1 A1 A1 (antiderivative of a 3-term non-polynomial expression, split into two separate accuracy marks) dM1 A1 (final value 13.5, cao)

(The Jan 2020 Q4 anchor's own correct final total area was not transcribed beyond this M/A/B skeleton and the confirmed solved limit $x=\sqrt5$ in the facts bank behind this document — only a wrong-path value, $\frac{31\sqrt5}{3}$ , the area under the curve alone with the line ignored, is on record, explicitly labelled as incorrect. It is not restated here as if it were the answer, and the real total is not derived in this document either, to avoid introducing a number this course has not itself verified against the primary mark scheme.)

What this table shows that no single existing document states in one place: all three real anchors split "find the antiderivative" from "apply the limits and evaluate" into separate marks, never one combined mark — confirmed independently across three different series and two different curve types (polynomial and non-polynomial). And exactly one of the three anchors (Jan 2020) tests the axis-boundary mixed-limit case Part 2.4 describes; the other two are both curve-and-line questions with two solved intersections, following the "both limits, M1 A1" convention the lesson's method-comparison block already models. Two of three anchors are on record as producing a specific, named error when the line boundary is dropped and only the curve is integrated — a genuinely recurring cross-series trap (already in the lesson's trap-taxonomy as line-boundary-ignored-entirely, sourced to Jan 2020, Jan 2023, and June 2025 across three separate examiner reports).

The efficiency angle, confirmed rather than assumed. June 2025 Q3's examiner report explicitly ranks the three creditable strategies for area-between-curve-and-line by relative speed, not just correctness — combining the two expressions before integrating once is faster than integrating each separately, because the latter carries more terms through the working and creates more chances for a sign slip when subtracting at the end. This is a real, quotable strategy point, not a general study-skills platitude: it is Pearson's own examiners comparing mathematically equivalent methods and finding them unequally fast under time pressure.

The discriminator pattern. Jan 2023 Q9's two-region structure is the one confirmed real instance in this archive where spec 8.2 rewards strategic decomposition — recognising that a second region relates to the first by subtraction from a named shape — rather than mechanical integration alone. This is a genuinely different skill from "integrate the right thing between the right limits," and it is where Part 2.6 above adds a worked illustration the lesson itself does not carry.

#Part 4 — What remains genuinely unverified

No confirmed real WMA12 anchor for a pure two-curve (no line) area question — see 2.5. Spec-real, exam-status unconfirmed for this specific unit.
The sign-change-inside-the-interval split (Part 1.2, third case) has no attached real examiner-report citation anywhere in this course's archive — the lesson's own trap-taxonomy already flags this honestly (sign-change-inside-the-interval-not-split), and this document does not manufacture one. It remains taught because it is a direct, checkable consequence of the Riemann-sum mechanism (Part 1.2), not because a graded question testing it has been found.
October 2025 and January 2026 examiner reports were not found in any of the three local archives this course has searched (per the facts bank's §8) — commentary from those two series, if it exists, has not yet been checked for this topic.
Six series (Oct 2019, Oct 2021, Oct 2024, Jun 2023, Jun 2024 + regional, Jan 2025) were confirmed present and keyword-grepped but not read start-to-finish for this topic specifically — the "3 of 20" frequency floor above could rise on a fuller re-read of those six, per the facts bank's own §8 note.