Why is "Integrating across a root, so a below-axis piece cancels an above-axis piece" a common mistake in Area under a curve?

Why it happens: Treating \int_a^b y\,dx as 'the area' without checking the sign of y; not sketching the curve. How to avoid it: Find the roots in [a,b], integrate each sub-interval separately, take the modulus of any negative piece, then add. Source: 9231 Paper 2 Examiner Reports — recurring comment

Why is "Writing $\int\frac{dx}{a^2+x^2} = \tan^{-1}\frac{x}{a}$, dropping the $\frac1a$" a common mistake in Area under a curve?

Why it happens: Memorising the a=1 case /1+x^2=^-1x and not generalising the \frac1a factor. How to avoid it: Quote the full form \frac1a^-1x/a; check it by differentiating mentally, or read it directly from MF19. Source: 9231 Paper 2 mark schemes

Why is "Using $a^2$ as $a$ (e.g. taking $a=9$ instead of $a=3$ from $x^2+9$)" a common mistake in Area under a curve?

Why it happens: Confusing the constant in the denominator (a^2) with a itself. How to avoid it: Always write 'so a^2 = 9, hence a = 3' as a separate line before substituting into the standard form. Source: 9231 Paper 2 Examiner Reports

Why is "Giving a decimal when an exact answer ($\pi$, $\ln$, surd) was required" a common mistake in Area under a curve?

Why it happens: Reaching for the calculator out of habit; not reading the word 'exact' in the question. How to avoid it: Leave \pi4, 3, (2+√3) in exact form. Use the log conversions for inverse-hyperbolic values. Source: 9231 Paper 2 Examiner Reports

Why is "Not factoring out the leading coefficient before matching a standard form" a common mistake in Area under a curve?

Why it happens: An integrand like 1/9+4x^2 does not literally read a^2+x^2 until the 4 is taken out. How to avoid it: Factor the coefficient of x^2 out of the denominator: 9+4x^2 = 4(x^2+\frac94), then a=\frac32. Source: 9231 Paper 2 — integration questions

Why is "Integrating $y\,dx$ when the area is bounded by the $y$-axis" a common mistake in Area under a curve?

Why it happens: Applying y\,dx automatically without checking which axis bounds the region. How to avoid it: For area to the y-axis, rearrange to x=g(y) and integrate \int_c^d x\,dy — the variables swap roles. Source: 9231 Paper 2 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Area under a curve

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Detailed Study Notes

Detailed notes on Integration for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Area under a Curve — Cambridge International AS & A Level Further Mathematics 9231 Paper 2 (Further Pure Mathematics 2, 2026-2027 syllabus)

The definite integral $\int_a^b y\,dx$ measures the signed area between a curve and the $x$ -axis. See why a sum of thin rectangles becomes an integral, then evaluate real FPM2 areas using the hyperbolic and inverse-trig integration techniques.

At a glance

Area $= \int_a^b y\,dx$ — the definite integral is the limit of a sum of thin rectangle areas.
Riemann idea: split $[a,b]$ into strips of width $\delta x$ ; area $\approx \sum y\,\delta x$ ; let $\delta x \to 0$ .
Signed area: where the curve dips below the $x$ -axis the integral is negative. Split the range at the roots and add the magnitudes for true geometric area.
FPM2 toolkit: many areas need $\int \frac{dx}{\sqrt{x^2+a^2}}=\sinh^{-1}\frac{x}{a}$ , $\int\frac{dx}{a^2+x^2}=\frac1a\tan^{-1}\frac{x}{a}$ , etc. (these standard forms ARE in MF19).
Quote exact answers ( $\ln$ , $\pi$ , $\sinh^{-1}$ ) where possible; otherwise round to 3 significant figures.

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

2.4a — Understand the definite integral $\int_a^b y\,dx$ as the limit of a sum and as the (signed) area between the curve $y=f(x)$ and the $x$ -axis.
2.4b — Find the area of a region bounded by a curve and the $x$ -axis, splitting the range where the curve crosses the axis so that parts below the axis are not cancelled.
2.4c — Evaluate such areas using the FPM2 integration techniques (hyperbolic, inverse-trigonometric and inverse-hyperbolic standard forms).

Area as the limit of a sum of rectangles

Thin rectangles approximate the area; their total tends to the integral as the width shrinks to zero.

The starting picture. To find the area between a curve $y=f(x)$ and the $x$ -axis from $x=a$ to $x=b$ , we approximate it with $n$ thin vertical rectangles (strips).

Divide $[a,b]$ into $n$ strips each of width $\delta x = \dfrac{b-a}{n}$ . The $i$ -th strip sits at $x=x_i$ and has height $y_i = f(x_i)$ , so its area is $y_i\,\delta x$ . Adding them all:

$\text{Area} \approx \sum_{i=1}^{n} y_i\,\delta x$

This is a Riemann sum. Each rectangle slightly over- or under-shoots the true region, but as we use more, thinner strips ( $\delta x \to 0$ , equivalently $n \to \infty$ ) the staircase of rectangles squeezes onto the curve and the error vanishes:

$\text{Area} = \lim_{\delta x \to 0} \sum y\,\delta x = \int_a^b y\,dx$

This limit is the definition of the definite integral. The elongated "S" symbol $\int$ is literally a stretched "sum".

Each strip has area (height × width) ≈ y·δx. Summing them gives the Riemann sum; the limit as δx→0 is the integral ∫y dx.

Cambridge tip. A short "explain what $\int_a^b y\,dx$ represents" part wants the words limit of a sum of rectangle areas and area between the curve and the $x$ -axis — not just "the area". State the limit explicitly.

Area $\approx \sum y\,\delta x$ (rectangles of width $\delta x$ ).
Let $\delta x \to 0$ : the sum becomes $\int_a^b y\,dx$ .
$\int$ is a stretched S — it literally means 'sum'.

See the full worked example for area under a curve →

Evaluating a definite integral as an area

Antidifferentiate, then apply the limits with the Fundamental Theorem of Calculus.

Once we accept area $= \int_a^b y\,dx$ , we evaluate it with the Fundamental Theorem of Calculus: find any antiderivative $F$ of $f$ , then

$\int_a^b f(x)\,dx = \Big[\,F(x)\,\Big]_a^b = F(b) - F(a).$

The routine, every time:

Integrate $y$ to get $F(x)$ (ignore the $+c$ — it cancels).
Substitute the upper limit, then subtract the value at the lower limit.
Simplify to an exact form ( $\ln$ , $\pi$ , surds) where possible, else 3 s.f.

Worked sketch. The area under $y = \dfrac{1}{x}$ from $x=1$ to $x=e$ is $\int_1^{e} \frac{1}{x}\,dx = \Big[\ln|x|\Big]_1^{e} = \ln e - \ln 1 = 1 - 0 = 1.$

Why the height must be $y$ . The strip height is the function value $y=f(x)$ , measured from the $x$ -axis up to the curve. So you always integrate $y$ with respect to $x$ . If a question gives the curve as $x = g(y)$ and asks for area to the $y$ -axis, you instead integrate $x$ with respect to $y$ : $\int_c^d x\,dy$ — the roles of the variables swap.

Cambridge tip. Write the square-bracket line $\big[F(x)\big]_a^b$ explicitly. It earns the method mark even if you later mis-evaluate, and it is the line examiners look for first.

$\int_a^b f\,dx = F(b)-F(a)$ where $F'=f$ (drop the $+c$ ).
Strip height is $y=f(x)$ , so integrate $y$ with respect to $x$ .
Area to the $y$ -axis instead uses $\int_c^d x\,dy$ .

See the full worked example for area under a curve →

Signed area — handling regions below the axis

Below the x-axis, y is negative, so the integral is negative. Split at the roots and add magnitudes.

The integral measures signed area. Where the curve lies above the $x$ -axis, $y>0$ and the contribution is positive; where it lies below, $y<0$ and the contribution is negative.

The danger. If you integrate straight across a root, a positive piece and a negative piece can partly (or wholly) cancel, giving a number that is not the geometric area.

The fix — split at the roots:

Solve $f(x)=0$ to find where the curve crosses the $x$ -axis inside $[a,b]$ .
Integrate over each sub-interval separately.
Take the modulus of any negative piece, then add the magnitudes: $\text{Area} = \left|\int_a^c y\,dx\right| + \left|\int_c^b y\,dx\right|.$

The integral over the red region is negative. For total area, take its modulus and add it to the green region's area.

By symmetry, watch out: $\int_{-a}^{a} (\text{odd function})\,dx = 0$ , but the area enclosed is certainly not zero — it is twice the magnitude of either half.

Cambridge tip. Always sketch (or at least check the sign of $y$ ) before integrating across the interval. A negative final "area" is the tell-tale sign you forgot to split at a root.

Below the axis $\Rightarrow y<0 \Rightarrow$ integral negative.
Split at every root, take $|\,\cdot\,|$ of each piece, then add.
An odd function integrates to $0$ but still encloses area.

See the full worked example for area under a curve →

Using the FPM2 techniques to evaluate areas

Inverse-trig and inverse-hyperbolic standard forms turn awkward areas into clean exact answers.

What makes an FPM2 area question is the integrand. Many regions are bounded by curves such as $y=\dfrac{1}{\sqrt{x^2+a^2}}$ or $y=\dfrac{1}{a^2+x^2}$ , whose integrals are the inverse-hyperbolic and inverse-trigonometric standard forms.

The forms you reach for (all in MF19):

$\displaystyle\int \frac{1}{\sqrt{x^2+a^2}}\,dx = \sinh^{-1}\frac{x}{a}\;(+c)$
$\displaystyle\int \frac{1}{\sqrt{x^2-a^2}}\,dx = \cosh^{-1}\frac{x}{a}\;(+c)$
$\displaystyle\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a}\;(+c)$
$\displaystyle\int \frac{1}{\sqrt{a^2-x^2}}\,dx = \sin^{-1}\frac{x}{a}\;(+c)$

The recognise-and-match routine:

Identify $a^2$ from the constant term (e.g. $x^2+9 \Rightarrow a=3$ ).
Match the integrand to one of the four forms above.
Antidifferentiate, then apply the limits.
Convert any inverse-hyperbolic value to a logarithm if an exact decimal-free answer is wanted, using $\sinh^{-1}t = \ln\!\big(t+\sqrt{t^2+1}\big)$ (also in MF19).

Worked sketch. Area under $y=\dfrac{1}{4+x^2}$ from $x=0$ to $x=2$ : $\int_0^{2}\frac{dx}{4+x^2} = \Big[\tfrac12\tan^{-1}\tfrac{x}{2}\Big]_0^{2} = \tfrac12\tan^{-1}1 - 0 = \tfrac12\cdot\tfrac{\pi}{4} = \frac{\pi}{8}.$

Cambridge tip. Spot the $a$ before you write anything: the difference between $\tan^{-1}x$ and $\frac1a\tan^{-1}\frac{x}{a}$ is a whole accuracy mark. And keep answers exact — leave the $\pi$ or $\ln$ in.

Identify $a^2$ from the constant, then match a standard form.
$\int\frac{dx}{a^2+x^2}=\frac1a\tan^{-1}\frac{x}{a}$ — don't drop the $\frac1a$ .
Convert $\sinh^{-1}t = \ln(t+\sqrt{t^2+1})$ for an exact log answer.

See the full worked example for area under a curve →

Quick recap

Area $= \int_a^b y\,dx$ is the limit of a sum of rectangle areas $\sum y\,\delta x$ .
Evaluate via the Fundamental Theorem: $\big[F(x)\big]_a^b = F(b)-F(a)$ .
Below the axis $y<0$ , so split at the roots and add the magnitudes for true area.
FPM2 areas use the inverse-trig / inverse-hyperbolic standard forms (in MF19).
Identify $a^2$ first; keep the $\frac1a$ in $\frac1a\tan^{-1}\frac{x}{a}$ .
Give exact answers ( $\pi$ , $\ln$ , $\sinh^{-1}$ ) where possible, else 3 s.f.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Area $= \int_a^b y\,dx$ = $\lim_{\delta x\to0}\sum y\,\delta x$ (limit of a Riemann sum).
Split at roots and add magnitudes for a region partly below the axis.
$\int\frac{dx}{a^2+x^2}=\frac1a\tan^{-1}\frac{x}{a}$ and $\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a}$ .
$\int\frac{dx}{\sqrt{x^2+a^2}}=\sinh^{-1}\frac{x}{a}$ and $\int\frac{dx}{\sqrt{x^2-a^2}}=\cosh^{-1}\frac{x}{a}$ .
$\sinh^{-1}t=\ln\!\big(t+\sqrt{t^2+1}\big)$ (to turn an inverse-hyperbolic value into a log).

How it’s examined

Area appears in Paper 2 (FPM2) most often as the application part of an integration question (4-7 marks). A typical part (a) sets up or evaluates $\int_a^b y\,dx$ where the integrand is a standard inverse-trig/inverse-hyperbolic form; a follow-on part may ask for the exact value as a logarithm. Marks are lost through (i) integrating across a root without splitting, so a below-axis piece cancels part of the area, (ii) dropping the $\frac1a$ factor in $\frac1a\tan^{-1}\frac{x}{a}$ , and (iii) giving a decimal where an exact $\pi$ or $\ln$ answer was expected. The standard forms are in MF19, but recognising which one applies is the examined skill.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 2, Further Pure Mathematics 2); List of formulae (MF19) — standard integrals; 9231/21, 9231/23 past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Area under a curve

Step-by-step solutions to past-paper-style questions on area under a curve, written exactly the way a tutor would explain them at the board.

Question type:

1Estimating area with rectangles (Riemann sum)
Getting startedDirect calculation• riemann, limit-of-sum, easy
Question
Estimate the area under $y = x^2$ from $x=0$ to $x=2$ using 4 rectangles of equal width, taking the height of each rectangle at its right-hand edge.
Step-by-step solution
1. Step 1
  Strip width. Four strips over $[0,2]$ gives $\delta x = \dfrac{2-0}{4} = 0.5$ . Right edges are at $x = 0.5,\,1,\,1.5,\,2$ .
2. Step 2
  Heights $y=x^2$ at those edges:
  $0.5^2=0.25,\;\; 1^2=1,\;\; 1.5^2=2.25,\;\; 2^2=4$
3. Step 3
  Sum the rectangle areas $\sum y\,\delta x = \delta x \sum y$ .
  $\text{Area} \approx 0.5(0.25 + 1 + 2.25 + 4) = 0.5 \times 7.5 = 3.75$
Answer
Estimate $\approx 3.75$ (the exact value is $\int_0^2 x^2\,dx = \frac{8}{3} \approx 2.67$ ; right-edge rectangles overestimate here).
Examiner tip
The estimate overshoots because $y=x^2$ is increasing, so the right-hand height is the tallest part of each strip. As $\delta x \to 0$ the estimate converges to the integral.
2Area under y = 1/x as a logarithm
Getting startedDirect calculation• definite-integral, logarithm, easy
Question
Find the exact area of the region bounded by $y = \dfrac{1}{x}$ , the $x$ -axis, and the lines $x=2$ and $x=6$ .
Step-by-step solution
1. Step 1
  Set up the definite integral. The curve lies above the axis for $x>0$ , so the area is simply $\int_2^6 y\,dx$ .
  $\text{Area} = \int_2^{6} \frac{1}{x}\,dx$
2. Step 2
  Integrate ( $\int \frac1x\,dx = \ln|x|$ ) and write the bracket line.
  $= \Big[\ln|x|\Big]_2^{6}$
3. Step 3
  Apply the limits and use $\ln 6 - \ln 2 = \ln\frac{6}{2}$ .
  $= \ln 6 - \ln 2 = \ln\frac{6}{2} = \ln 3$
Answer
Area $= \ln 3 \approx 1.10$ .
Examiner tip
Leave the answer as the exact $\ln 3$ . Combining $\ln 6 - \ln 2 = \ln 3$ shows control of log laws and is the cleanest form.
3Region partly below the x-axis
Building confidenceMulti-step problem• signed-area, split-at-roots, medium
Question
Find the total area enclosed between the curve $y = x^3 - 4x$ and the $x$ -axis.
Step-by-step solution
1. Step 1
  Find the roots. $x^3 - 4x = x(x^2-4) = x(x-2)(x+2) = 0$ , so the curve cuts the axis at $x=-2,\,0,\,2$ . The enclosed region runs from $-2$ to $2$ .
2. Step 2
  Check the sign. On $(-2,0)$ the curve is above the axis ( $y>0$ ); on $(0,2)$ it is below ( $y<0$ ). Integrate each piece separately. Antiderivative: $\frac{x^4}{4} - 2x^2$ .
  $\int_{-2}^{0}(x^3-4x)\,dx = \Big[\tfrac{x^4}{4}-2x^2\Big]_{-2}^{0} = 0 - (4-8) = 4$
3. Step 3
  Second piece over $(0,2)$ — this comes out negative.
  $\int_{0}^{2}(x^3-4x)\,dx = \Big[\tfrac{x^4}{4}-2x^2\Big]_{0}^{2} = (4-8) - 0 = -4$
4. Step 4
  Add the magnitudes. Take $|{-4}|$ for the below-axis piece.
  $\text{Area} = |4| + |-4| = 8$
Answer
Total area $= 8$ .
Examiner tip
Integrating straight across from $-2$ to $2$ gives $0$ (the function is odd) — a classic trap. Splitting at $x=0$ and adding magnitudes gives the true area of $8$ .
4Area using an inverse-hyperbolic standard form
Building confidenceMulti-step problem• inverse-hyperbolic, standard-form, medium
Question
Find the exact area under $y = \dfrac{1}{\sqrt{x^2+9}}$ from $x=0$ to $x=4$ , giving your answer as a logarithm.
Step-by-step solution
1. Step 1
  Identify $a$ . The integrand matches $\frac{1}{\sqrt{x^2+a^2}}$ with $a^2=9$ , so $a=3$ .
  $\int_0^{4}\frac{dx}{\sqrt{x^2+9}} = \Big[\sinh^{-1}\tfrac{x}{3}\Big]_0^{4}$
2. Step 2
  Apply the limits ( $\sinh^{-1}0 = 0$ ).
  $= \sinh^{-1}\tfrac{4}{3} - 0 = \sinh^{-1}\tfrac{4}{3}$
3. Step 3
  Convert to a logarithm with $\sinh^{-1}t = \ln\!\big(t+\sqrt{t^2+1}\big)$ , $t=\frac43$ : $\sqrt{\frac{16}{9}+1} = \sqrt{\frac{25}{9}} = \frac53$ .
  $\sinh^{-1}\tfrac{4}{3} = \ln\!\Big(\tfrac{4}{3}+\tfrac{5}{3}\Big) = \ln 3$
Answer
Area $= \sinh^{-1}\frac{4}{3} = \ln 3 \approx 1.10$ .
Examiner tip
Recognising $a=3$ (not $a=9$ ) is the key step. The neat $\sqrt{16/9+1}=5/3$ is engineered, so the logarithm collapses to $\ln 3$ .
5Exact area giving a multiple of π
StretchMulti-step problem• inverse-trig, exact, hard
Question
Find the exact area of the region bounded by $y = \dfrac{6}{9+4x^2}$ , the $x$ -axis, and the lines $x=0$ and $x=\dfrac{3}{2}$ .
Step-by-step solution
1. Step 1
  Force the $a^2+u^2$ shape. Factor the $4$ : $9+4x^2 = 4\!\left(\frac94 + x^2\right)$ , so $a^2=\frac94$ , $a=\frac32$ .
  $\int_0^{3/2}\frac{6\,dx}{9+4x^2} = \int_0^{3/2}\frac{6\,dx}{4\!\left(x^2+\frac94\right)} = \frac{3}{2}\int_0^{3/2}\frac{dx}{x^2+\frac94}$
2. Step 2
  Apply $\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\frac{x}{a}$ with $a=\frac32$ (so $\frac1a=\frac23$ ).
  $= \frac{3}{2}\cdot\frac{2}{3}\Big[\tan^{-1}\tfrac{x}{3/2}\Big]_0^{3/2} = \Big[\tan^{-1}\tfrac{2x}{3}\Big]_0^{3/2}$
3. Step 3
  Evaluate. At $x=\frac32$ : $\frac{2x}{3}=1$ , and $\tan^{-1}1 = \frac{\pi}{4}$ .
  $= \tan^{-1}1 - \tan^{-1}0 = \frac{\pi}{4} - 0 = \frac{\pi}{4}$
Answer
Area $= \dfrac{\pi}{4} \approx 0.785$ .
Examiner tip
Factoring the leading coefficient out of the denominator before matching the standard form is the make-or-break move. The constants are chosen so the $\frac32$ and $\frac23$ cancel cleanly.
6Area measured to the y-axis
StretchMulti-step problem• x-dy, inverse-trig, hard
Question
The curve $x = \dfrac{4}{\sqrt{4-y^2}}$ is defined for $-1 \le y \le 1$ ... actually consider $x = \sqrt{4 - y^2}$ . Find the exact area of the region between this curve and the $y$ -axis from $y=0$ to $y=2$ .
Step-by-step solution
1. Step 1
  Area to the $y$ -axis uses $\int x\,dy$ . Here $x = \sqrt{4-y^2}$ (the right half of the circle $x^2+y^2=4$ ), so the area is $\int_0^2 \sqrt{4-y^2}\,dy$ .
  $\text{Area} = \int_0^{2}\sqrt{4-y^2}\,dy$
2. Step 2
  Recognise the quarter-circle. $x=\sqrt{4-y^2}$ with radius $2$ , $0\le y\le 2$ , is exactly one quarter of a circle of radius $2$ . (Equivalently use the standard reduction $\int\sqrt{a^2-y^2}\,dy = \frac{y}{2}\sqrt{a^2-y^2}+\frac{a^2}{2}\sin^{-1}\frac{y}{a}$ .)
  $\int_0^{2}\sqrt{4-y^2}\,dy = \Big[\tfrac{y}{2}\sqrt{4-y^2} + 2\sin^{-1}\tfrac{y}{2}\Big]_0^{2}$
3. Step 3
  Evaluate. At $y=2$ : $\sqrt{4-4}=0$ and $\sin^{-1}1 = \frac{\pi}{2}$ .
  $= \big(0 + 2\cdot\tfrac{\pi}{2}\big) - 0 = \pi$
Answer
Area $= \pi$ (a quarter of the area $\pi r^2 = 4\pi$ of the radius-2 circle).
Examiner tip
Two valid routes: the antiderivative with $\sin^{-1}$ , or recognising the quarter-circle so Area $=\frac14\pi(2)^2=\pi$ . Switching to $\int x\,dy$ for area against the $y$ -axis is the examined idea.

Model Answers — Area under a curve

High-scoring sample answers for area under a curve on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
Explain what is meant by the definite integral $\int_a^b y\,dx$ as an area, and state how it is obtained as a limit.
Model answer
Divide the interval $a \le x \le b$ into thin vertical strips, each of width $\delta x$ . A strip at $x$ has height $y = f(x)$ , so its area is approximately $y\,\delta x$ (the area of a rectangle).

The total area between the curve $y=f(x)$ and the $x$ -axis is approximated by the sum of all these rectangle areas: $\text{Area} \approx \sum y\,\delta x.$

As the strips are made thinner, $\delta x \to 0$ , this sum tends to a limit which is defined to be the definite integral: $\int_a^b y\,dx = \lim_{\delta x \to 0}\sum y\,\delta x.$

So $\int_a^b y\,dx$ is the (signed) area between the curve and the $x$ -axis from $x=a$ to $x=b$ .
Why this scores
B1 for 'sum of rectangle areas $\sum y\,\delta x$ ', B1 for the limit as $\delta x \to 0$ , B1 for identifying it as the area between curve and $x$ -axis. A bare 'it's the area' scores only one mark.
Question 2
4 marks
Find the exact area of the region bounded by the curve $y = \dfrac{1}{1+x^2}$ , the $x$ -axis, and the lines $x = 0$ and $x = 1$ .
Model answer
The curve is positive throughout $0 \le x \le 1$ , so the area is $\text{Area} = \int_0^{1}\frac{1}{1+x^2}\,dx.$

This is the standard form $\int \frac{dx}{a^2+x^2} = \frac1a\tan^{-1}\frac{x}{a}$ with $a=1$ : $= \Big[\tan^{-1} x\Big]_0^{1} = \tan^{-1} 1 - \tan^{-1} 0.$

Therefore $\text{Area} = \frac{\pi}{4} - 0 = \frac{\pi}{4}.$
Why this scores
M1 for the correct standard form $\tan^{-1}x$ , M1 for inserting the limits, A1 for $\tan^{-1}1 = \frac\pi4$ , A1 for the exact $\frac\pi4$ . A decimal answer loses the final A1 when 'exact' is asked.
Question 3
6 marks
The curve $y = x^2 - 3x$ meets the $x$ -axis at $x=0$ and $x=3$ . The line $x = 4$ is also drawn. Find the total area enclosed between the curve and the $x$ -axis from $x=0$ to $x=4$ .
Model answer
Sign of the curve. $y = x(x-3)$ , so $y \le 0$ on $0 \le x \le 3$ (below the axis) and $y \ge 0$ on $3 \le x \le 4$ (above). Split at $x=3$ . Antiderivative: $\frac{x^3}{3} - \frac{3x^2}{2}$ .

Piece 1 ( $0$ to $3$ , below the axis): $\int_0^{3}(x^2-3x)\,dx = \Big[\tfrac{x^3}{3}-\tfrac{3x^2}{2}\Big]_0^{3} = \big(9 - \tfrac{27}{2}\big) - 0 = -\tfrac{9}{2}.$

Piece 2 ( $3$ to $4$ , above the axis): $\int_3^{4}(x^2-3x)\,dx = \Big[\tfrac{x^3}{3}-\tfrac{3x^2}{2}\Big]_3^{4} = \big(\tfrac{64}{3}-24\big) - \big(9-\tfrac{27}{2}\big) = -\tfrac{8}{3} + \tfrac{9}{2} = \tfrac{11}{6}.$

Total area (take the modulus of the first piece): $\text{Area} = \left|-\tfrac{9}{2}\right| + \tfrac{11}{6} = \tfrac{27}{6} + \tfrac{11}{6} = \frac{38}{6} = \frac{19}{3}.$
Why this scores
M1 for splitting at $x=3$ , A1 A1 for the two integrals $-\frac92$ and $\frac{11}{6}$ , M1 for taking the modulus of the negative piece, A1 for $\frac{19}{3}$ . Adding the signed values ( $-\frac92+\frac{11}{6}$ ) is the standard error and scores no accuracy marks for area.
Question 4
5 marks
Find the exact area of the region bounded by the curve $y = \dfrac{1}{\sqrt{x^2-1}}$ , the $x$ -axis, and the lines $x = 1$ and $x = 2$ . Give your answer as a logarithm.
Model answer
The integrand matches the standard form $\int \frac{dx}{\sqrt{x^2-a^2}} = \cosh^{-1}\frac{x}{a}$ with $a=1$ : $\text{Area} = \int_1^{2}\frac{dx}{\sqrt{x^2-1}} = \Big[\cosh^{-1}x\Big]_1^{2}.$

Applying the limits ( $\cosh^{-1}1 = 0$ ): $= \cosh^{-1} 2 - 0 = \cosh^{-1} 2.$

Convert using $\cosh^{-1}t = \ln\!\big(t + \sqrt{t^2-1}\big)$ with $t=2$ : $\cosh^{-1} 2 = \ln\!\big(2 + \sqrt{4-1}\big) = \ln\!\big(2 + \sqrt{3}\big).$

Therefore $\text{Area} = \ln\!\big(2+\sqrt{3}\big) \approx 1.32.$
Why this scores
M1 for the $\cosh^{-1}$ standard form, A1 for $\cosh^{-1}2$ , M1 for the log conversion formula, A1 for $\ln(2+\sqrt3)$ , B1 for noting $\cosh^{-1}1=0$ . The integral is improper at the lower limit but converges; a clean evaluation earns full marks.
Question 5
7 marks
Find the exact area of the region bounded by the curve $y = \dfrac{x}{\sqrt{x^2+4}}$ , the $x$ -axis, and the lines $x=0$ and $x=2\sqrt{3}$ .
Model answer
The curve is positive on $0 \le x \le 2\sqrt3$ , so $\text{Area} = \int_0^{2\sqrt3}\frac{x}{\sqrt{x^2+4}}\,dx.$

Recognise the form. The numerator $x$ is (half) the derivative of $x^2+4$ , so this is a $\frac{f'}{2\sqrt{f}}$ pattern: $\frac{d}{dx}\sqrt{x^2+4} = \frac{x}{\sqrt{x^2+4}}$ . Hence the antiderivative is $\sqrt{x^2+4}$ : $\int \frac{x}{\sqrt{x^2+4}}\,dx = \sqrt{x^2+4} \;(+c).$

Apply the limits: $\text{Area} = \Big[\sqrt{x^2+4}\Big]_0^{2\sqrt3} = \sqrt{(2\sqrt3)^2 + 4} - \sqrt{0+4}.$

Now $(2\sqrt3)^2 = 12$ , so: $= \sqrt{12+4} - \sqrt{4} = \sqrt{16} - 2 = 4 - 2 = 2.$
Why this scores
M1 for spotting that the integrand is $\frac{d}{dx}\sqrt{x^2+4}$ (or for the substitution $u=x^2+4$ ), A1 for the antiderivative $\sqrt{x^2+4}$ , M1 A1 for the limits, A1 for $(2\sqrt3)^2=12$ , A1 for the final $2$ . Students who reach for $\sinh^{-1}$ here have misread the numerator — the extra $x$ changes everything.
Question 6
6 marks
The region $R$ is bounded by the curve $y = \dfrac{4}{4+x^2}$ and the $x$ -axis, between $x=-2$ and $x=2$ . Find the exact area of $R$ .
Model answer
The curve is positive everywhere and is symmetric about the $y$ -axis ( $y$ depends only on $x^2$ ). So $\text{Area} = \int_{-2}^{2}\frac{4}{4+x^2}\,dx = 2\int_0^{2}\frac{4}{4+x^2}\,dx.$

Standard form $\int\frac{dx}{a^2+x^2} = \frac1a\tan^{-1}\frac{x}{a}$ with $a^2=4$ , $a=2$ : $\int_0^{2}\frac{4\,dx}{4+x^2} = 4\cdot\frac{1}{2}\Big[\tan^{-1}\tfrac{x}{2}\Big]_0^{2} = 2\big(\tan^{-1}1 - \tan^{-1}0\big).$

Now $\tan^{-1}1 = \frac\pi4$ , so $\int_0^2 = 2\cdot\frac\pi4 = \frac\pi2$ . Doubling for symmetry: $\text{Area} = 2 \times \frac{\pi}{2} = \pi.$
Why this scores
B1 for using symmetry (or integrating $-2$ to $2$ directly), M1 for the $\frac1a\tan^{-1}\frac{x}{a}$ form with $a=2$ , A1 for keeping the $\frac12$ factor, M1 for the limits, A1 A1 for $\frac\pi2$ per half and $\pi$ total. Dropping the $\frac1a$ gives $2\pi$ — a common slip.

Key Formulae — Area under a curve

The formulae you need to memorise for area under a curve on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Area as a definite integral
$\text{Area} = \int_a^b y\,dx = \lim_{\delta x \to 0}\sum y\,\delta x$
$y=f(x)$
height of the strip above the $x$ -axis
$\delta x$
width of each thin strip (rectangle)
$a, b$
lower and upper limits on $x$
When to use
Area between a curve and the $x$ -axis. For area against the $y$ -axis use $\int_c^d x\,dy$ .
Fundamental Theorem of Calculus
$\int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a),\quad F'=f$
$F$
any antiderivative of $f$ (the $+c$ cancels)
When to use
Evaluating any definite integral once you have an antiderivative.
Inverse-tangent standard form
$\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a} + c$
$a$
$a^2$ is the constant term in the denominator
When to use
Integrand $\frac{1}{a^2+x^2}$ — keep the $\frac1a$ factor. In MF19.
Inverse-sine standard form
$\int \frac{1}{\sqrt{a^2-x^2}}\,dx = \sin^{-1}\frac{x}{a} + c$
$a$
$a^2$ is the constant under the root, with $|x|<a$
When to use
Integrand $\frac{1}{\sqrt{a^2-x^2}}$ (a finite-domain root). In MF19.
Inverse-hyperbolic-sine standard form
$\int \frac{1}{\sqrt{x^2+a^2}}\,dx = \sinh^{-1}\frac{x}{a} + c$
$a$
$a^2$ is the constant added under the root
When to use
Integrand $\frac{1}{\sqrt{x^2+a^2}}$ . Convert with $\sinh^{-1}t=\ln(t+\sqrt{t^2+1})$ . In MF19.
Inverse-hyperbolic-cosine standard form
$\int \frac{1}{\sqrt{x^2-a^2}}\,dx = \cosh^{-1}\frac{x}{a} + c$
$a$
$a^2$ subtracted under the root, with $x>a$
When to use
Integrand $\frac{1}{\sqrt{x^2-a^2}}$ . Convert with $\cosh^{-1}t=\ln(t+\sqrt{t^2-1})$ . In MF19.

Key Definitions and Keywords — Area under a curve

Definitions to memorise and the exact keywords mark schemes credit for area under a curve answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Definite integral
Examiner keyword
$\int_a^b f(x)\,dx$ , the limit of the Riemann sum $\sum y\,\delta x$ as $\delta x \to 0$ ; numerically equal to the signed area between $y=f(x)$ and the $x$ -axis from $x=a$ to $x=b$ .
Riemann sum
Examiner keyword
An approximation $\sum y_i\,\delta x$ to an area, formed by summing the areas of thin rectangles (strips) under the curve. Its limit as $\delta x\to0$ is the definite integral.
Signed area
Examiner keyword
The value of $\int_a^b y\,dx$ counts area above the $x$ -axis as positive and area below as negative. The two can cancel, so it is not always the geometric area.
Fundamental Theorem of Calculus
Examiner keyword
The result that $\int_a^b f\,dx = F(b) - F(a)$ for any antiderivative $F$ of $f$ ; it links the area (integration) to antidifferentiation.
Standard integral form
One of the recognised integrals in MF19 (e.g. $\int\frac{dx}{a^2+x^2}=\frac1a\tan^{-1}\frac{x}{a}$ ) that you match an integrand to rather than deriving from scratch.
Improper integral
A definite integral whose integrand is unbounded at a limit (e.g. $\frac{1}{\sqrt{x^2-1}}$ at $x=1$ ) or whose range is infinite; here treated as the limiting value, which may still be finite.

Common Mistakes and Misconceptions — Area under a curve

The traps other students keep falling into on area under a curve questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Integrating across a root, so a below-axis piece cancels an above-axis piece
9231 Paper 2 Examiner Reports — recurring comment
Why it happens
Treating $\int_a^b y\,dx$ as 'the area' without checking the sign of $y$ ; not sketching the curve.
How to avoid it
Find the roots in $[a,b]$ , integrate each sub-interval separately, take the modulus of any negative piece, then add.
✕Writing $\int\frac{dx}{a^2+x^2} = \tan^{-1}\frac{x}{a}$ , dropping the $\frac1a$
9231 Paper 2 mark schemes
Why it happens
Memorising the $a=1$ case $\int\frac{dx}{1+x^2}=\tan^{-1}x$ and not generalising the $\frac1a$ factor.
How to avoid it
Quote the full form $\frac1a\tan^{-1}\frac{x}{a}$ ; check it by differentiating mentally, or read it directly from MF19.
✕Using $a^2$ as $a$ (e.g. taking $a=9$ instead of $a=3$ from $x^2+9$ )
9231 Paper 2 Examiner Reports
Why it happens
Confusing the constant in the denominator ( $a^2$ ) with $a$ itself.
How to avoid it
Always write 'so $a^2 = 9$ , hence $a = 3$ ' as a separate line before substituting into the standard form.
✕Giving a decimal when an exact answer ( $\pi$ , $\ln$ , surd) was required
9231 Paper 2 Examiner Reports
Why it happens
Reaching for the calculator out of habit; not reading the word 'exact' in the question.
How to avoid it
Leave $\frac\pi4$ , $\ln 3$ , $\ln(2+\sqrt3)$ in exact form. Use the log conversions for inverse-hyperbolic values.
✕Not factoring out the leading coefficient before matching a standard form
9231 Paper 2 — integration questions
Why it happens
An integrand like $\frac{1}{9+4x^2}$ does not literally read $a^2+x^2$ until the $4$ is taken out.
How to avoid it
Factor the coefficient of $x^2$ out of the denominator: $9+4x^2 = 4(x^2+\frac94)$ , then $a=\frac32$ .
✕Integrating $y\,dx$ when the area is bounded by the $y$ -axis
9231 Paper 2 Examiner Reports
Why it happens
Applying $\int y\,dx$ automatically without checking which axis bounds the region.
How to avoid it
For area to the $y$ -axis, rearrange to $x=g(y)$ and integrate $\int_c^d x\,dy$ — the variables swap roles.

Area under a curve

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Estimating area with rectangles (Riemann sum)

2Area under y = 1/x as a logarithm

3Region partly below the x-axis

4Area using an inverse-hyperbolic standard form

5Exact area giving a multiple of π

6Area measured to the y-axis

Area as a definite integral

Fundamental Theorem of Calculus

Inverse-tangent standard form

Inverse-sine standard form

Inverse-hyperbolic-sine standard form

Inverse-hyperbolic-cosine standard form

Definite integral

Riemann sum

Signed area

Fundamental Theorem of Calculus

Standard integral form

Improper integral

✕Integrating across a root, so a below-axis piece cancels an above-axis piece

✕Writing ∫dxa2+x2=tan⁡−1xa\int\frac{dx}{a^2+x^2} = \tan^{-1}\frac{x}{a}∫a2+x2dx​=tan−1ax​, dropping the 1a\frac1aa1​

✕Using a2a^2a2 as aaa (e.g. taking a=9a=9a=9 instead of a=3a=3a=3 from x2+9x^2+9x2+9)

✕Giving a decimal when an exact answer (π\piπ, ln⁡\lnln, surd) was required

✕Not factoring out the leading coefficient before matching a standard form

✕Integrating y dxy\,dxydx when the area is bounded by the yyy-axis

✕Writing $\int\frac{dx}{a^2+x^2} = \tan^{-1}\frac{x}{a}$ , dropping the $\frac1a$

✕Using $a^2$ as $a$ (e.g. taking $a=9$ instead of $a=3$ from $x^2+9$ )

✕Giving a decimal when an exact answer ( $\pi$ , $\ln$ , surd) was required

✕Integrating $y\,dx$ when the area is bounded by the $y$ -axis