Definite Integration and Areas Bounded by Curves and Lines

~55 min · WMA12 · 8.2

WMA12 · 8.2 · 55 min

A is a number you can compute mechanically in three lines — the entire topic is really asking whether you know when that number is allowed to be called an area. Evaluating $\int_a^b f(x)\,dx$ is bookkeeping: integrate, substitute, subtract. Whether the result you get is the enclosed area, the negative of it, or a number that isn't the area of anything at all depends on one thing the bookkeeping never checks for you — whether the curve stays on one side of the x-axis, or the correct boundary, for the whole interval. The paper builds almost every mark in this topic around exactly that gap.

Key terms in this lesson

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

Evaluating a definite integral — the mechanics, before the meaning

A definite integral $\int_a^b f(x)\,dx$ is evaluated in three steps, always in this order: find an $F(x)$ with $F'(x) = f(x)$ ; substitute the upper bound to get $F(b)$ ; substitute the lower bound to get $F(a)$ ; subtract, $F(b) - F(a)$ . The notation $\left[F(x)\right]_a^b$ is shorthand for exactly that difference. There is no $+c$ to write, and this is not a rule to remember separately — whatever constant an antiderivative carries appears in both $F(b)$ and $F(a)$ identically, so it cancels the moment the subtraction happens: $\left(F(b)+c\right) - \left(F(a)+c\right) = F(b) - F(a)$ .

Finding $F(x)$ from $f(x)$ term by term uses the reverse of the power rule: the power of $x$ in each term goes up by one, and the whole term is divided by that new power. $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}$ , applied to each term of a polynomial separately and the results added. This is genuinely memorise-and-apply skill at P2 — spec item 8.1 has no formula printed for it anywhere in the Mathematical Formulae and Statistical Tables booklet, unlike the arithmetic and geometric series formulae or the trapezium rule, which are printed in full.

Two of the general mark-scheme principles that apply to every integration question on this paper are worth stating before any worked example, because they explain what the marks are actually for. The method mark for an attempt at integration is defined as increasing the power of at least one term by 1 — the mark is for doing that step, independent of whether the rest of the working is later correct. And where an exact answer is required, working with a rounded decimal instead of the exact fraction or surd normally loses marks — permitted calculators on this paper can perform numerical integration, which gives a decimal approximation, but cannot perform the symbolic integration a written method has to show.

The definite integral as area — and exactly what the spec restricts it to

Spec item 8.2 asks for the interpretation of the definite integral as the area under a curve, extended to two further named cases: the area of a region bounded by a curve and given straight lines, and the area of a region bounded by two curves. All three are the same underlying calculation — integrate the right expression between the right limits — but the first is a special case of the second (the x-axis is itself a straight line, $y = 0$ ), and both are special cases of the third once you allow one of the two boundaries to be a horizontal line.

One restriction is stated explicitly in the spec's own guidance and is worth carrying forward deliberately: '∫x dy will not be required.' Every region in this topic — under a curve, between a curve and a line, between two curves — is found by slicing the region into thin vertical strips and integrating with respect to $x$ . Re-expressing a boundary as $x$ in terms of $y$ and integrating with respect to $y$ instead is a real technique, but it belongs to a later unit; nothing on this paper needs it, and reaching for it here is solving a harder problem than the one being asked.

The mechanism block below derives, from first principles, exactly which condition has to hold for 'the integral' and 'the area' to be the same number — it is not always true, and knowing precisely when it fails is most of what separates a fluent answer from a lucky one on this topic.

Mechanism

Where 'area equals the integral' actually comes from — thin strips, not a definition to accept on faith

Cut the interval $[a,b]$ into $n$ equal strips of width $\Delta x = \frac{b-a}{n}$ . Approximate the area of the $i$ -th strip by a rectangle of width $\Delta x$ and height $f(x_i)$ , where $x_i$ is some point inside that strip. The total area under the curve is approximately the sum of these rectangles, $\sum_{i=1}^{n} f(x_i)\,\Delta x$ . As $n \to \infty$ the strips narrow, the rectangles hug the curve more and more closely, and the error in the approximation shrinks to zero — the limit of that sum is, by definition, the definite integral $\int_a^b f(x)\,dx$ . So far this only says the integral is the LIMIT of a sum of signed rectangle areas — it does not yet say that limit equals the geometric area, because a rectangle's height $f(x_i)$ is only a genuine, non-negative geometric length when $f(x_i) \geq 0$ . When $f(x) \geq 0$ across the whole of $[a,b]$ , every rectangle in the sum has a non-negative height, the sum really is approximating physical area, and the limit — the integral — equals that area exactly. The two other pieces of the topic follow directly from the same picture. First, why the mechanical evaluation method (find $F$ , substitute, subtract) gives the same number as this geometric limit: define $A(x) = \int_a^x f(t)\,dt$ , the accumulated signed area up to $x$ . Widening the interval by a sliver $\delta x$ adds a sliver of area approximately $f(x)\,\delta x$ , so $A'(x) = f(x)$ — $A$ is itself an antiderivative of $f$ . Any other antiderivative $F$ differs from $A$ by a constant, and since $A(a) = 0$ , that constant is fixed by $F(a)$ : $A(x) = F(x) - F(a)$ , so $A(b) = F(b) - F(a)$ — exactly the evaluation rule from the teach block above, now derived rather than asserted. Second, why the whole thing breaks the moment $f$ goes negative somewhere in $(a,b)$ — covered as its own mechanism next, because it is where almost every mark in this topic is actually lost.

Mechanism

Why a region below the axis needs its own integral — not a minus sign bolted onto the end

Run the same rectangle argument where $f(x) < 0$ . Each rectangle still has width $\Delta x > 0$ , but its height $f(x_i)$ is now negative, so its signed contribution $f(x_i)\,\Delta x$ is negative too. Summing negative contributions over an interval where $f$ stays negative throughout gives a negative total whose SIZE — its modulus — equals the true geometric area of that region, because the geometric shape being measured (a strip between the curve and the axis) is exactly as large as it would be if the curve were reflected above the axis; only the sign of the height flipped, not its magnitude. So on a single-sign interval, area = $\left|\int_a^b f(x)\,dx\right|$ , and that modulus is safe to take. The reasoning breaks down the moment the interval contains a point where $f$ changes sign. A single continuous sum $\sum f(x_i)\,\Delta x$ run across that point adds positive-height rectangles from one side and negative-height rectangles from the other INTO THE SAME RUNNING TOTAL — the positive and negative contributions partially cancel each other before the sum is ever finished, let alone before a modulus could be taken. Whatever number comes out at the end is neither the true combined area (which needs both pieces counted as positive) nor an honest signed total (which the cancellation has already corrupted) — it is a third quantity with no direct geometric meaning. The fix follows directly from the mechanism itself, not from a rule to memorise: split the interval at every point where $f$ changes sign, so that each piece is genuinely single-sign, take the modulus of the integral over each piece separately — each one is now safe, by the argument two sentences ago — and add the results. That is the one procedure that actually matches the geometry.

Diagram — The region enclosed between the curve y = 8x − x² and the line y = 2x

x-axis: x · y-axis: y

Curve C: y = 8x − x²: A downward-opening parabola through the origin and through (8, 0), rising to a maximum in between. Strictly above the line for every x strictly between the two points where the two graphs meet.
Line l: y = 2x: A straight line through the origin with gradient 2. Meets the curve at exactly the two points where 8x − x² = 2x.

Intersection (0, 0): Both curve and line pass through the origin. Worth checking deliberately rather than assumed: x = 0 is a genuine root of curve − line here, not a coincidence of the sketch.
Intersection (6, 12): The second solution of 8x − x² = 2x, i.e. of 6x − x² = 0. Found by solving the equation, not read off a rough sketch — the sketch's job is only to confirm which graph sits on top between the two solutions, not to supply the coordinates.
Shaded region R: Bounded above by the curve and below by the line for 0 < x < 6. Its area is the integral of (curve − line), upper minus lower — reversing that order does not just flip the sign of an otherwise-correct number, it changes what is actually being calculated.

Common error: Integrating (line − curve) instead of (curve − line), because the subtraction order was written down without first checking which graph is actually on top over the interval.

Correct: Check which function is greater at one interior point before subtracting — here x = 3 gives curve = 15, line = 6, so the curve is on top — and subtract upper minus lower. Get the order wrong here and the integral evaluates to −36 instead of +36; a real examiner report on this exact question type records that candidates who made this error would 'generally realise the final answer must be positive at the end and recover the marks' — i.e. the sign is self-correcting for anyone who checks it, but only if they do.

examiner-report · Jan 2020 · Q4

Worked, in full

Evaluating $\int_1^4 (3x^2 - 2x + 5)\,dx$ , and reading the result as an area because the curve never dips below the axis on this interval

01
Check the sign before integrating anything. $3x^2 - 2x + 5$ is a quadratic with $a = 3$ , $b = -2$ , $c = 5$ ; its is $(-2)^2 - 4(3)(5) = 4 - 60 = -56 < 0$ , and $a > 0$ , so this quadratic has no real roots and its curve never crosses the x-axis at all — it is strictly positive for every real $x$ , in particular throughout $[1, 4]$ .
Earns: Nothing — not itself a scored step on a real mark scheme, but skipping it is exactly how a student ends up trying to explain away a negative number at the very end instead of never introducing the possibility in the first place.
02
Integrate term by term: $\int (3x^2 - 2x + 5)\,dx = x^3 - x^2 + 5x$ (the $+c$ is dropped, since this is a definite integral and it would cancel in the subtraction regardless).
Earns: M1 — attempts to integrate, increasing the power of at least one term by 1. This is the general marking principle's own definition of the method mark for an integration attempt, so it is earned by the structure of the working, independently of whether the arithmetic that follows is correct.
03
Substitute both bounds into that antiderivative. At $x=4$ : $64 - 16 + 20 = 68$ . At $x=1$ : $1 - 1 + 5 = 5$ .
Earns: A1 — correct antiderivative, substituted correctly at both bounds. Dependent on the method mark above.
04
Subtract, upper minus lower: $68 - 5 = 63$ .
Earns: A1 — correct final value, cao. The most common way to lose exactly this mark is subtracting the bounds the wrong way round — $F(1) - F(4)$ instead of $F(4) - F(1)$ — which would give $-63$ here. On a question that only asks for the value of the integral, with no area language attached, a negative result is a prompt to check the subtraction order immediately, not a sign the earlier working must be wrong.
05
Interpret the result. Because stage 1 confirmed $f(x) > 0$ across the whole of $[1,4]$ , the number $63$ is not only the value of the integral — it is exactly the area enclosed between the curve, the x-axis, and the lines $x=1$ and $x=4$ , with no modulus and no splitting required.
Earns: B1 — states the area interpretation, which is only safe to state because the sign check in stage 1 was actually done. The same number 63, produced without that check, could only honestly be called 'the value of the integral' — not yet 'the area'.

Source — Examiner report, Jan 2020

"it is noteworthy that there were a number of students who relied on a calculator to evaluate the definite integral for them, which is not advisable when an exact answer is required as permitted calculators are supposed to only be able to carry out numerical integration, which will commonly give a decimal answer only... full method must be shown for full credit"

Complete it yourself

Complete the chain — the area enclosed between y = x² + 1 and the x-axis, between x = 0 and x = 3

01
Check the sign first. $x^2 + 1 \geq 1 > 0$ for every real $x$ — the $+1$ alone guarantees the curve never touches, let alone crosses, the x-axis, so on $[0,3]$ the integral will equal the area directly, with no splitting or modulus required.
02
Integrate term by term: $\int (x^2+1)\,dx = \dfrac{x^3}{3} + x$ (the $+c$ dropped, since this is a definite integral).

Same question, every valid method

Find the area of the finite region R enclosed by the curve C with equation y = 8x − x² and the line l with equation y = 2x. (VERIDIAN-original question. It is not a reproduction of any real past-paper question — the curve and line are original to this lesson — but the three methods below genuinely mirror the pattern the June 2025 Q3 examiner report confirms are all independently creditable for an area-between-curve-and-line question of this type.)

3 valid methods · every one reaches 36 · 6 marks available

01
$8x - x^2 = 2x \Rightarrow 6x - x^2 = 0 \Rightarrow x(6-x) = 0$
Method mark for attempting to solve curve = line to find the limits of integration.
M1
02
$x = 0 \text{ or } x = 6$
Accuracy mark for both correct limit values, dependent on the method mark above — matching the real convention for finding the x-coordinates of an intersection (a genuine solve, credited M1 A1, not folded into a single mark). Both values are needed — a single solution gives no interval to integrate over at all.
A1
03
Subtract upper minus lower: $(8x - x^2) - 2x = 6x - x^2$ . Checking $x=3$ confirms curve (15) is above line (6) on this interval, so this is the right order.
Method mark for forming the single combined integrand curve − line, with the subtraction correctly ordered and an attempt at integration under way. Reversing the order produces −36 rather than +36 — a sign fault, not a typo, under the general marking principles' exact-answer discipline.
M1
04
$\displaystyle\int (6x - x^2)\,dx = 3x^2 - \dfrac{x^3}{3}$
Accuracy mark for the correct antiderivative (need not be simplified further), dependent on the method mark immediately above — the antiderivative itself earns its own mark, separate from substituting any limits into it.
A1
05
$\left[3x^2 - \dfrac{x^3}{3}\right]_0^6 = (108 - 72) - 0$
Method mark for substituting both limits found above into the antiderivative — applying the limits is its own step, separate from finding the antiderivative itself.
M1
06
$= 36$
Accuracy mark, correct exact value, cao, dependent on every mark above.
A1

Fastest to write down — one antiderivative, one pair of bounds, one subtraction. The entire risk is concentrated in the single line where curve − line is written: get that one sign wrong and every later number is wrong too, with nothing else in the working to flag it until the final answer comes out negative.

Worked, in full

Area bounded by a curve, the y-axis, and a line — where the y-axis supplies one limit directly and only the OTHER one needs solving

01
Identify what each named boundary actually gives you, before writing down any integral. Curve $C$ : $y = 2x^2+5$ , for $x \geq 0$ . The finite region $R$ is bounded by $C$ , the y-axis, and the line $y=25$ . Two of these three boundaries — the curve and the line — meet somewhere that has to be found by solving an equation. The third, the y-axis, does not: the y-axis IS the vertical line $x=0$ , by definition, not a value to derive from anything else.
Earns: Nothing — an orientation step, not itself scored on a real mark scheme, but skipping it is exactly how a student ends up hunting for an equation to solve for a limit the question already gave away.
02
Solve for the ONE limit that genuinely needs finding — where curve meets line. $2x^2+5 = 25 \Rightarrow 2x^2=20 \Rightarrow x^2=10 \Rightarrow x=\sqrt{10}$ (rejecting $x=-\sqrt{10}$ : $C$ is only defined for $x \geq 0$ ).
Earns: B1 — achieves the value of the limit found by solving curve = line, as its own standalone fact-finding mark. This is deliberately NOT folded into a combined 'both limits' method mark the way this lesson's method-comparison block models it for a region bounded by two solved curve/line intersections ('Method mark for finding both limits of integration by solving curve = line. Both values are needed — a single solution gives no interval to integrate over at all.'). That criterion describes the case where BOTH boundaries come from solving an intersection — it does not generalise to every area question on this spec. Here one boundary is an axis, given directly by the question's own wording, so finding this single value is already the whole job.
03
State the other limit directly. No equation to form, nothing to solve: the y-axis boundary gives $x=0$ . This is NOT the curve's value at $x=0$ (which is $y=5$ ), and it is NOT the line's own value ( $y=25$ ) either — those are y-coordinates, heights read off the diagram, and an integral taken with respect to $x$ only ever takes x-values as its limits. Writing $\int_5^{25}$ or $\int_5^{\sqrt{10}}$ in place of $\int_0^{\sqrt{10}}$ mixes a y-value into an x-limit slot — a real, examiner-documented error on exactly this shape of question (see this lesson's trap-taxonomy).
Earns: B1 — states the second limit correctly as x = 0. Independent of the marks around it: getting this boundary right is a matter of correctly reading what 'bounded by the y-axis' means on the question, not a step inside the integration method itself.
04
Check which graph is on top before combining. At $x=0$ : line $=25$ , curve $=5$ , so the line is above. At $x=\sqrt{10}$ the two meet. Between the two limits the curve is increasing from 5 up to 25 while the line stays constant at 25, so the line stays on top throughout — the integrand is (line − curve), never the reverse.
Earns: Nothing scored on its own, but this is where a reversed subtraction gets caught before it corrupts every later line, exactly as the diagram's own commonError models for the curve-and-line case.
05
Form the combined integrand and integrate: $\displaystyle\int_0^{\sqrt{10}} \left[25-(2x^2+5)\right]dx = \int_0^{\sqrt{10}} (20-2x^2)\,dx = \left[20x-\dfrac{2x^3}{3}\right]_0^{\sqrt{10}}$ .
Earns: M1 — forms the correct (upper − lower) integrand and attempts to integrate it, increasing the power of at least one term by 1.
06
Substitute both limits — using $(\sqrt{10})^3 = 10\sqrt{10}$ — and subtract: $\left(20\sqrt{10}-\dfrac{2(10\sqrt{10})}{3}\right) - 0 = 20\sqrt{10}-\dfrac{20\sqrt{10}}{3} = \dfrac{60\sqrt{10}-20\sqrt{10}}{3} = \dfrac{40\sqrt{10}}{3}$ .
Earns: A1 — correct exact area, cao, dependent on both limits (stages 2 and 3) and the integration (stage 5) all being right. Left as an exact surd, never rounded to a decimal — the same exact-answer discipline the general marking principles apply everywhere else on this paper.

Source — Mark scheme, Jan 2020

"Achieves a limit of √5 either attached to an integral or used to find the area of the rectangle"

In your own words

In one sentence: why does the number 19/3 never appear if you integrate x² − 5x + 4 from x = 0 to x = 4 in a single calculation, rather than splitting at the root?

Marked, line by line

f(x) = x² − 5x + 4. (a) Show that the curve with equation y = f(x) crosses the x-axis at x = 1 and at x = 4. (2) (b) Find the exact value of $\int_1^4 f(x)\,dx$ . (3) (c) Given also that $\int_0^1 f(x)\,dx = \frac{11}{6}$ , find the total area of the finite region R bounded by the curve, the x-axis, and the lines x = 0 and x = 4. (3) — VERIDIAN-original question, not a reproduction of any past-paper question; the per-line mark allocations are modelled on verified mark-scheme conventions rather than transcribed from a real scheme, which for an original question does not exist.

8 marks available

(a) — 2 marks

01
$x^2 - 5x + 4 = (x-1)(x-4)$
Method mark for attempting to factorise the quadratic. The general principles credit factorisation, the formula or completing the square equally for a three-term quadratic; factorisation is fastest here since $1 \times 4 = 4$ and $1+4=5$ are visible by inspection.
M1
02
$(x-1)(x-4) = 0 \Rightarrow x = 1 \text{ or } x = 4$ , both roots of $f$
Accuracy mark for stating both values. A 'show that' part needs both roots identified explicitly, not just the factorised form left unevaluated.
A1

(b) — 3 marks

101
$\int (x^2 - 5x + 4)\,dx = \dfrac{x^3}{3} - \dfrac{5x^2}{2} + 4x$ (+c, dropped for a definite integral)
Method mark for integrating: the power of every term is increased by 1, which is the general marking principle's own definition of an attempt at integration.
M1
102
At $x=4$ : $\dfrac{64}{3} - 40 + 16 = \dfrac{64}{3} - 24 = -\dfrac{8}{3}$ . At $x=1$ : $\dfrac{1}{3} - \dfrac{5}{2} + 4 = \dfrac{11}{6}$ .
Accuracy mark for correct substitution at both limits, kept as exact fractions rather than rounded decimals — matching the general principle that marks are normally lost for resorting to decimals when an exact value is asked for.
A1
103
$\int_1^4 f(x)\,dx = -\dfrac{8}{3} - \dfrac{11}{6} = -\dfrac{16}{6} - \dfrac{11}{6} = -\dfrac{27}{6} = -\dfrac{9}{2}$
Accuracy mark for the correct exact value, cao, dependent on both marks above. A negative value here is not a mistake to fix — it is the correct answer to the question exactly as asked, because f(x) is below the axis for every x strictly between 1 and 4.
A1

201
On $[0,1]$ , $f(x) \geq 0$ (confirmed by the sign of the given value $\frac{11}{6}$ ), so that part of R has area exactly $\frac{11}{6}$ . On $[1,4]$ , $f(x) \leq 0$ , so the integral found in (b), $-\frac{9}{2}$ , is the negative of that part's area: its area is $\left|-\frac{9}{2}\right| = \frac{9}{2}$ .
Method mark for recognising that the region must be split at the root x = 1 into two single-sign pieces before either can be read off as an area — the same principle this lesson's own mechanism derives: a signed integral spanning a sign change is not a geometric area until it is split.
M1
202
Total area $= \dfrac{11}{6} + \dfrac{9}{2}$
Method mark, dependent on the split above, for adding the two pieces as positive areas — not adding the raw signed values, which would silently let part (b)'s $-\frac{9}{2}$ cancel against part of the $\frac{11}{6}$ instead of combining with it.
M1
203
$\dfrac{11}{6} + \dfrac{27}{6} = \dfrac{38}{6} = \dfrac{19}{3}$
Accuracy mark for the correct exact total, cao, dependent on both method marks above.
A1

Named traps

answer-produced-with-no-method-shown: Confirmed on a real area-between-a-line-and-a-curve question: "it is noteworthy that there were a number of students who relied on a calculator to evaluate the definite integral for them, which is not advisable when an exact answer is required as permitted calculators are supposed to only be able to carry out numerical integration, which will commonly give a decimal answer only... full method must be shown for full credit" (Jan 2020, Q4). Permitted calculators on this paper can do numerical integration — approximate, decimal — but not the symbolic integration a written method has to show; the two are not interchangeable when the question asks for an exact value.
curve-and-line-subtracted-in-the-wrong-order: Confirmed on the same question: "it was not uncommon to see the difference the wrong way round, though students would generally realise the final answer must be positive at the end and recover the marks" (Jan 2020, Q4). The mechanism is exactly this lesson's diagram: subtracting (line − curve) instead of (upper − lower) flips the sign of the whole answer. It is self-correcting only for a student who actually checks the sign of the final number against "area must be positive" — silently accepting a negative result loses the mark this trap describes.
line-boundary-ignored-entirely: Confirmed independently on two different real series, both testing this exact spec point — a genuinely recurring trap, not a one-off. On a two-region area question with a curve and a line: "Many attempted solutions simply integrated the equation of the curve and evaluated between the limits 0 and 1, thus completely ignoring the requirement to subtract the area under the line l. This gained no credit as it was not a correct strategy for R1" (Jan 2023, Q9(c)). And on a separate curve-and-line area question: "One common error, aside from bracketing mistakes, made in this question was that only the area under the curve being found, 31√5⁄3 being given as the area of the region R" (Jan 2020, Q4) — echoed again on a third series, "Some candidates, however, did not know how to deal with the additional line and just found the area under the curve" (June 2025, Q3). Reading the question fully before starting — is the region bounded by the curve ALONE, or by the curve and something else — is the entire fix, and it has to happen before any integral is written down, not after.
sign-change-inside-the-interval-not-split: Not attached to a real examiner-report quote in the facts bank behind this course — the archived area questions reviewed for WMA12 all use curve-and-line or curve-and-curve regions that stay one sign throughout, not a region where the curve itself crosses the axis. It is included anyway because it is genuine, checkable spec-8.2 content, and it follows directly from the mechanism derived earlier in this lesson: integrating straight across a root mixes a positive and a negative contribution into one running total before either can be made positive, producing a number that is neither the honest signed integral nor the true area. See the marked-solution's common wrong path for exactly what number that mistake produces.
slowest-valid-method-chosen-under-time-pressure: Confirmed as a real, quoted efficiency comparison between mathematically equivalent methods on an area-between-curve-and-line question: one valid strategy "was a more time-consuming approach requiring either five terms to be integrated or terms to be collected resulting in occasional sign errors" (June 2025, Q3) than the other creditable strategies for the same question. All three methods in this lesson's method-comparison earn full marks — but under a 90-minute paper with 75 marks across the whole thing, picking the slower of two equally valid routes is a real cost even when it produces the right answer.
y-values-used-as-x-limits: Confirmed on the same real anchor question as this lesson's y-axis-boundary worked example above: "Another mistake seen on numerous occasions was to use the y values of 7 and 17 as their limits to an integration with respect to x, which was incorrect and usually meant only the first two marks were scored" (Jan 2020, Q4). The two numbers named there are the curve's value at x = 0 and the height of the bounding line — real numbers on the diagram, but y-values, never x-values, and ∫(…)dx only ever takes x-values as its limits. It is exactly the shape the worked-chain above is built to guard against: one limit found by solving curve = line, the other supplied directly by the y-axis as x = 0 — never by reading off the curve's or the line's own y-coordinate at that point.

Retrieval — with feedback on every choice

Question 1

3 marks

What is the exact value of $\displaystyle\int_{2}^{5} (4x-3)\,dx$ ?

Question 2

2 marks

A curve lies entirely below the x-axis for every x in $[0,3]$ , and $\int_0^3 g(x)\,dx = -18$ . What is the area of the region enclosed between the curve and the x-axis on this interval?

Question 3

4 marks

Find the exact area of the finite region enclosed between the curves y = x² and y = 8 − x².

Question 4

2 marks

Before evaluating $\int_a^b f(x)\,dx$ and calling the result 'the area of the region bounded by the curve, the x-axis, and the lines x = a and x = b', what must be checked first?

Question 5

1 mark

The instructions on a WMA12 paper state that permitted calculators must not have 'the facility for symbolic algebra manipulation, differentiation and integration.' What does this mean in practice for a definite-integral question that asks for an exact answer?

Reference — not a study method, a lookup

FTC: ∫ₐᵇf(x)dx = F(b) − F(a), F′(x)=f(x). Integrate termwise: power up by 1, divide by the new power.
Area = the integral ONLY where the curve doesn't cross the axis in (a,b). If it does, split at each root, take |value| per piece, add.
Curve/line or curve/curve: integrate (upper − lower) between the true intersection points — solve for them, don't guess.
Exact answer required: quote the method. A calculator's numerical integration checks a value; it doesn't earn the marks.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation and figure attributed to a mark scheme or examiner report in this lesson was independently verified against the primary Pearson document, not carried over from prior course material.

Question 13 marks

What is the exact value of $\displaystyle\int_{2}^{5} (4x-3)\,dx$ ?

$33$
Correct. Antiderivative $2x^2 - 3x$ . At $x=5$ : $50-15=35$ . At $x=2$ : $8-6=2$ . $35-2=33$ .
B $35$
This is $F(5)$ alone, with nothing subtracted at the lower bound. Every definite integral is a difference of two values of the antiderivative, never a single evaluated value on its own.
C $-33$
The bounds have been subtracted the wrong way round: $F(2) - F(5) = 2 - 35 = -33$ . It is always upper minus lower for $\int_a^b$ , i.e. $F(b) - F(a)$ .
D $12$
This evaluates the original function $4x-3$ at the two bounds and subtracts — $(4(5)-3) - (4(2)-3) = 17 - 5 = 12$ — with no integration step at all. The definite integral is built from the antiderivative, not from the original function evaluated at the endpoints.

Traps tested: Lower bound not subtracted · Bounds subtracted in wrong order · Original function evaluated not antiderivative

Question 22 marks

A curve lies entirely below the x-axis for every x in $[0,3]$ , and $\int_0^3 g(x)\,dx = -18$ . What is the area of the region enclosed between the curve and the x-axis on this interval?

$18$
Correct — every rectangle used to build this integral has negative height, since g(x) < 0 throughout, so the sum comes out negative, and the geometric area is its modulus: |−18| = 18.
B $-18$
This reports the signed integral as though it were the area itself. A geometric area can never be negative; the integral, by contrast, can.
C $0$
A curve below the axis still encloses a genuine region between itself and the axis — being below the axis does not make the enclosed area vanish.
DCannot be found without the equation of g
Knowing the value of the integral together with the fact that g keeps one sign throughout the interval is already sufficient: the area is the modulus of that one number. The explicit formula for g contributes nothing further once both of those are known.

Traps tested: Negative integral reported as area · Below axis region treated as zero area · Overclaims uncertainty

Question 34 marks

Find the exact area of the finite region enclosed between the curves y = x² and y = 8 − x².

$\frac{64}{3}$
Correct. Solving $x^2 = 8-x^2$ gives $2x^2=8$ , so $x=\pm 2$ . On $(-2,2)$ , $8-x^2$ is the upper curve (check $x=0$ : $8-0=8 > 0$ ). $\int_{-2}^{2}\left[(8-x^2)-x^2\right]dx = \int_{-2}^{2}(8-2x^2)\,dx = \left[8x - \frac{2x^3}{3}\right]_{-2}^{2} = \left(16-\frac{16}{3}\right)-\left(-16+\frac{16}{3}\right) = \frac{64}{3}$ .
B $-\frac{64}{3}$
This subtracts the two curves in the wrong order — lower minus upper instead of upper minus lower. Between two curves, exactly as between a curve and a line, the boundary that is actually on top has to be identified first, by checking a value inside the interval.
C $\frac{32}{3}$
This is exactly half the correct area — the value you get by integrating only over $[0,2]$ instead of the full interval $[-2,2]$ found from solving $x^2=8-x^2$ . Using half the true interval, whether from an incomplete solve or a misapplied symmetry argument, halves the answer.
D $32$
This comes from integrating $8 - 2x^2$ as though it were $8x - 2x^2$ — leaving the power on the $x^2$ term unchanged instead of increasing it by 1. The method mark for an integration attempt is specifically defined as increasing the power of at least one term by 1; a term left at its original power hasn't been integrated at all.

Traps tested: Curves subtracted in wrong order · Only half the interval integrated · Power not increased when integrating

Question 42 marks

Before evaluating $\int_a^b f(x)\,dx$ and calling the result 'the area of the region bounded by the curve, the x-axis, and the lines x = a and x = b', what must be checked first?

Whether f(x) changes sign anywhere in the open interval (a, b)
Correct. This is the one condition the whole topic turns on: the integral equals the area exactly when the curve stays on one side of the x-axis throughout the interval. If it crosses, the region has to be split at the crossing point before either piece can be called an area.
BWhether f(x) is a polynomial
The form of the function is irrelevant to whether the integral equals the area — the condition is purely about sign, and applies identically to polynomial and non-polynomial functions alike.
CWhether a and b are both positive
The signs of the bounds themselves have no bearing on whether the integral equals the area — what matters is the sign of f(x) across the interval between them, not the sign of the interval's endpoints.
DWhether the curve is symmetric about the y-axis
Symmetry has nothing to do with whether the integral equals the area — a symmetric curve can still cross the x-axis inside the interval, and an asymmetric one can stay on one side throughout.

Traps tested: Polynomial form treated as required · Bounds sign confused with function sign · Symmetry assumed necessary

Question 51 mark

A calculator can check a numeric approximation, but cannot itself produce the exact fraction your written method has to derive
Correct. The permitted calculators can carry out numerical integration — an approximate, decimal evaluation — which is genuinely useful for checking a final answer, but they cannot perform the symbolic integration a written method has to show to earn the method marks.
BCalculators cannot be used at all on any WMA12 question
WMA12 is a calculator-permitted paper throughout — there is no non-calculator section. The restriction quoted is specifically about symbolic manipulation, not about calculator use in general.
CThe restriction only applies to trigonometry questions
The rule is printed once, on the cover of the whole paper, and applies to every question — it is not scoped to any particular spec topic.
DSince a calculator can approximate the integral numerically, a decimal answer is always acceptable
The general marking principles state the opposite: where an exact answer is asked for, marks are normally lost for resorting to a rounded decimal, even if that decimal is numerically close to the exact value.

Traps tested: Calculators banned entirely · Rule scoped too narrowly · Decimal accepted for exact request

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Jan 2020 · Q4 — cited directly in this lesson
Mark scheme: Jan 2020 · Q4 — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA12.

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Up next

The Trapezium Rule and Justifying an Over- or Underestimate

Two-thirds of this topic is arithmetic you cannot get wrong if you build it correctly, and the exam's own examiners record the last third — the one-mark reasoning question — as the single hardest mark to earn on the whole paper. The trapezium rule replaces a curve you may not be able to integrate exactly with a chain of straight lines you can measure exactly, and the formula for doing that is printed in front of you in the exam. What isn't printed is the sentence that explains WHY your answer came out too big or too small — and "it's less than the true value" is not that sentence, however true it is, because it restates the conclusion instead of giving the reason for it.

50 min