MathsISC Class 12 · Application of Integrals

Area Under a Curve by Integration: Parabolas, Circles, Ellipses and More

Q: Why can the integral be negative when area is positive?

The integral counts parts below the x-axis as negative. Area is a positive quantity, so for regions below the axis you take the absolute value, and for curves that cross the axis you split at the crossing points.

Q: When should I integrate with respect to y?

When the region is bounded by the y-axis and horizontal lines, or when the curve is easier to write as x = g(y), such as a parabola opening sideways between two horizontal lines.

Q: Do I have to sketch the curve?

It is strongly recommended. Examiners reported that missing or wrong sketches were the main reason candidates could not find the limits or the region, and a correct sketch often earns a mark itself.

Q: Which curves are in the ISC 2027 syllabus for area?

Simple curves only: lines, circles, parabolas, ellipses, polynomial functions, the modulus function, and exponential and logarithmic functions, together with the area between two curves.

Q: How many marks does area by integration carry in ISC?

In 2024 and 2025 it appeared as 4-mark questions with internal choice, including a case-based question on the parabola y = 4 − x² in 2025.

Updated 15 October 2026 · Fact-checked

The area between a curve y = f(x), the x-axis and the lines x = a and x = b is ∫ₐᵇ |f(x)| dx. Where the curve dips below the axis, split at the x-intercepts and add absolute values. Use symmetry for parabolas, circles and ellipses, and integrate in y for regions against the y-axis.

Above the axis the integral is positive, below it is negative, so split at the x-intercept c and add the absolute values to get the area.

Illustration drawn for QuestionBuddy.

Understand Area Under a Curve

The basic idea. If $f (x) \geq 0$ on $[a, b]$ , the area between the curve $y = f (x)$ , the x-axis and the vertical lines $x = a$ and $x = b$ is
$A = \int_{a}^{b} f (x) d x = \int_{a}^{b} y d x .$
Think of the region as made of thin vertical strips of height $y$ and width $d x$ ; the integral adds them up.

Always sketch first. The ISC 2024 and 2025 examiners both reported candidates who did not sketch, drew wrong sketches, and so could not find the limits or the bounded region. A rough sketch shows where the curve crosses the axes, which part is shaded, and whether symmetry can halve the work.

Regions below the axis. If $f (x) < 0$ on part of the interval, the integral over that part is negative. Area is never negative, so split at the points where $f (x) = 0$ and add the absolute values:
$A = \int_{a}^{b} ∣ f (x) ∣ d x .$
For $y = x^{2} - 4 x$ between its roots 0 and 4, the integral is $- \frac{32}{3}$ , so the area is $\frac{32}{3}$ .

Area against the y-axis. If the region is bounded by $x = g (y)$ , the y-axis and the horizontal lines $y = c$ and $y = d$ , use horizontal strips:
$A = \int_{c}^{d} x d y = \int_{c}^{d} g (y) d y .$
This is often easier for curves like $x^{2} = 4 y$ between two horizontal lines.

Symmetric curves. Use symmetry to integrate over a convenient part and multiply.

Parabola $y^{2} = 4 a x$ cut off by $x = k$ : the region is symmetric about the x-axis, so $A = 2 \int_{0}^{k} 2 a x d x$ . (ISC 2024 asked this with $k = a$ , giving $\frac{8 a ^{2}}{3}$ .)
Circle $x^{2} + y^{2} = r^{2}$ : $A = 4 \int_{0}^{r} r^{2} - x^{2} d x = π r^{2}$ , using the standard integral $\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a}$ .
Ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ : $A = 4 \cdot \frac{b}{a} \int_{0}^{a} a^{2} - x^{2} d x = π ab$ .

Other simple curves in the syllabus. Polynomials, the modulus function (split where the inside is zero), $e^{x}$ (area under $e^{x}$ from 0 to $k$ is $e^{k} - 1$ ) and $lo g x$ (area under $lo g x$ from 1 to $e$ is 1, by parts).

Regions cut by a line. When a straight line and a curve together bound part of the region (for example the circle and the line $y = x$ in the first quadrant), split at the point where the upper boundary changes from one to the other, and integrate each piece separately. The topic page on the area between two curves develops this further.

Area Under a Curve: formulas to remember

Area with the x-axis: $A = \int_{a}^{b} ∣ y ∣ d x$; Split at points where y = 0.
Area with the y-axis: $A = \int_{c}^{d} ∣ x ∣ d y$; Use horizontal strips.
Standard integral: $\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a} + C$; Needed for circles and ellipses.
Circle: $x^{2} + y^{2} = r^{2} : A = π r^{2}$; By integration: 4∫₀ʳ √(r² − x²) dx.
Ellipse: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 : A = π ab$; 4(b/a)∫₀ᵃ √(a² − x²) dx.
Parabola segment: $y^{2} = 4 a x, x = k : A = \frac{8}{3} a k^{3/2}$; Twice ∫₀ᵏ 2√(ax) dx.

How to solve Area Under a Curve questions

Use this routine for every area question; each step can earn a mark.

1Sketch the curve and the boundary lines. Mark the intercepts and shade the required region.
2Decide whether vertical strips ( $d x$ ) or horizontal strips ( $d y$ ) give a simpler integral.
3Find the limits from the sketch: intersections with the axes or with the given lines.
4Use symmetry if the region is symmetric about an axis, and multiply.
5If the curve crosses the axis inside the interval, split there and take absolute values.
6Integrate, using the standard integral for $a^{2} - x^{2}$ where needed.
7Write the answer in square units.

Area Under a Curve in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Area of an ellipse by integration

Suppose a garden pond is in the shape of an ellipse 20 m long and 12 m wide. With its centre at the origin, its edge is $\frac{x ^{2}}{100} + \frac{y ^{2}}{36} = 1$ .

Find the surface area of the pond by integration.

Try it first, then show the solutionSolution

In the first quadrant, $y = \frac{6}{10} 100 - x^{2}$ . By symmetry, $A = 4 \int_{0}^{10} \frac{3}{5} 100 - x^{2} d x$ .
$\int_{0}^{10} 100 - x^{2} d x = [\frac{x}{2} 100 - x^{2} + 50 sin^{- 1} \frac{x}{10}]_{0}^{10} = 50 \cdot \frac{π}{2} = 25 π$ .
$A = \frac{12}{5} \times 25 π = 60 π \approx 188.5$ m².

Answer: $60 π \approx 188.5$ m², which matches $π ab = π \times 10 \times 6$ .

Memory hook: Ellipse area = π × semi-axis × semi-axis, and integration proves it.

Area under a square-root curve

Imagine a plot of land between a straight road along the x-axis and a stream whose bank follows $y = 2 x$ on a site plan (in metres), from $x = 0$ to a fence at $x = 9$ .

Find the area of the plot.

Try it first, then show the solutionSolution

$A = \int_{0}^{9} 2 x d x = 2 \cdot \frac{2}{3} x^{3/2}_{0}^{9}$ .
$= \frac{4}{3} \times 27 = 36$ m².

Answer: 36 m².

Memory hook: Area under a curve: add thin strips of height y from one boundary to the other.

Tricks to remember Area Under a Curve

Short memory aids. Say each one aloud once, then use it on a question the same day.

Sketch, shade, limits
Never integrate before you have sketched the curve and shaded the region.
Use it when: Avoiding the commonest error reported by examiners.
Area is never negative
If part of the curve is below the axis, split there and take the modulus.
Use it when: Curves that cross the x-axis.
Four quarters
Circles and ellipses: integrate the first-quadrant quarter and multiply by 4.
Use it when: Saving time and avoiding sign errors.
Strip direction
Region against the y-axis or curve given as x = g(y)? Use dy strips.
Use it when: Choosing the easier integral.

Area Under a Curve: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A basic area · 1 mark · MCQBoard level

The area bounded by $y = x^{2}$ , the x-axis and the lines $x = 0$ , $x = 3$ is:
(a) 3 sq units
(b) 9 sq units
(c) 27 sq units
(d) 6 sq units

Try it first, then show the solutionSolution

$\int_{0}^{3} x^{2} d x = \frac{27}{3} = 9$ .

Answer: (b) 9 sq units.

Example 2: A parabola cut by a line · 3 marks · Short answerBoard level

Using integration, find the area bounded by the parabola $y^{2} = 9 x$ and the line $x = 4$ .

Try it first, then show the solutionSolution

Sketch: the parabola opens to the right; the region is symmetric about the x-axis. Upper half: $y = 3 x$ . (1 mark)
$A = 2 \int_{0}^{4} 3 x d x = 6 \cdot \frac{2}{3} x^{3/2}_{0}^{4}$ . (1 mark)
$= 4 \times 8 = 32$ sq units. (1 mark)

Answer: 32 sq units.

Example 3: Area of an ellipse · 3 marks · Short answerHarder board level

Find, by integration, the area enclosed by the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{4} = 1$ .

Try it first, then show the solutionSolution

First quadrant: $y = \frac{2}{5} 25 - x^{2}$ . By symmetry $A = 4 \cdot \frac{2}{5} \int_{0}^{5} 25 - x^{2} d x$ . (1 mark)
$\int_{0}^{5} 25 - x^{2} d x = [\frac{x}{2} 25 - x^{2} + \frac{25}{2} sin^{- 1} \frac{x}{5}]_{0}^{5} = \frac{25}{2} \cdot \frac{π}{2} = \frac{25 π}{4}$ . (1 mark)
$A = \frac{8}{5} \times \frac{25 π}{4} = 10 π$ sq units, which agrees with $π ab = π \times 5 \times 2$ . (1 mark)

Answer: $10 π$ sq units.

Example 4: A region below the x-axis · 3 marks · Short answerHarder board level

Find the area enclosed between the curve $y = x^{2} - 4 x$ and the x-axis.

Try it first, then show the solutionSolution

The curve meets the x-axis at $x = 0$ and $x = 4$ , and lies below it between them. (1 mark)
$\int_{0}^{4} (x^{2} - 4 x) d x = \frac{64}{3} - 32 = - \frac{32}{3}$ . (1 mark)
Area $= - \frac{32}{3} = \frac{32}{3}$ sq units. (1 mark)

Answer: $\frac{32}{3}$ sq units.

Example 5: A modulus curve · 2 marks · Short answerHarder board level

Find the area bounded by $y = ∣ x - 1∣$ , the x-axis and the lines $x = - 1$ and $x = 3$ .

Try it first, then show the solutionSolution

Split at $x = 1$ : $\int_{- 1}^{1} (1 - x) d x + \int_{1}^{3} (x - 1) d x$ . (1 mark)
$= [x - \frac{x ^{2}}{2}]_{- 1}^{1} + [\frac{x ^{2}}{2} - x]_{1}^{3} = 2 + 2 = 4$ sq units. (1 mark)

Answer: 4 sq units.

Example 6: A circle sector cut by a line · 5 marks · Long answerHOTS / competency

Find, by integration, the area of the region in the first quadrant enclosed by the x-axis, the line $y = x$ and the circle $x^{2} + y^{2} = 16$ .

Try it first, then show the solutionSolution

Sketch: the line meets the circle where $2 x^{2} = 16$ , so at $(22, 22)$ . The upper boundary is the line from $x = 0$ to $22$ , then the circle from $22$ to 4. (1 mark)
$A = \int_{0}^{22} x d x + \int_{22}^{4} 16 - x^{2} d x$ . (1 mark)
First part: $\frac{x ^{2}}{2}_{0}^{22} = 4$ . (1 mark)
Second part: $[\frac{x}{2} 16 - x^{2} + 8 sin^{- 1} \frac{x}{4}]_{22}^{4} = 4 π - (\frac{2 2}{2} \cdot 22 + 8 \cdot \frac{π}{4}) = 4 π - 4 - 2 π = 2 π - 4$ . (1 mark)
$A = 4 + 2 π - 4 = 2 π$ sq units, which is one-eighth of the circle, as expected. (1 mark)

Answer: $2 π$ sq units.

Where students lose marks in Area Under a Curve

The slips examiners see most often, and what to write instead.

Mistake: No sketch, wrong limits
Both the 2024 and 2025 examiners reported candidates who did not sketch or drew wrong sketches, and so could not find the limits or the bounded region.
Do this: Draw a quick labelled sketch, mark intercepts and shade the region before writing any integral.
Usually costs: Most of a 4-mark question.
Mistake: Negative area
Integrating across a part where the curve is below the axis gives a negative or cancelled value.
Wrong working
Area under $cos x$ from 0 to $2 π$ $= \int_{0}^{2 π} cos x d x = 0$ .
Right working
Split at the zeros: $1 + 2 + 1 = 4$ .
Do this: Find where the curve crosses the axis and take absolute values of each part.
Usually costs: 1 to 2 marks.
Mistake: Forgetting to double for symmetry
For y² = 4ax, y = 2√(ax) gives only the upper half of the region.
Do this: Check the sketch: if the region extends below the axis too, multiply by 2.
Usually costs: 1 mark.
Mistake: Not using the given information
In the 2025 case question, candidates did not use the given equation to find the intercepts and limits.
Do this: Set y = 0 (or x = 0) in the given equation to find where the region begins and ends.
Usually costs: 1 to 2 marks.

Exam tips

Include the sketch in your answer; examiners often give a mark for a correct, shaded diagram.
Write the area integral before evaluating it, with the limits clearly shown.
Give answers in "sq units" (or m² in context).
For circles and ellipses, quote the standard integral of $a^{2} - x^{2}$ ; you do not need to derive it.
Check your answer against a known formula ( $π r^{2}$ , $π ab$ ) or a rough estimate from the sketch.

Practice questions on Area Under a Curve

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The area bounded by $y = sin 2 x$ and the x-axis from $x = 0$ to $x = \frac{π}{2}$ is:
(a) $\frac{1}{2}$
(b) 1
(c) 2
(d) $\frac{π}{2}$
Hint
$[- cos x]_{0}^{π}$ .
Answer
(b) 1: $\int_{0}^{π /2} sin 2 x d x = [- \frac{c o s 2 x}{2}]_{0}^{π /2} = \frac{1}{2} + \frac{1}{2} = 1$ .
Q2 · 2 marks · Short answerBoard level
Find the area under the line $y = 2 x + 1$ between $x = 1$ and $x = 3$ .
Hint
The line stays above the axis.
Answer
$[x^{2} + x]_{1}^{3} = 12 - 2 = 10$ sq units.
Q3 · 3 marks · Short answerHarder board level
Find the area in the first quadrant bounded by $x^{2} = 4 y$ , the y-axis and the lines $y = 1$ and $y = 4$ .
Hint
Use horizontal strips: $x = 2 y$ .
Answer
$\int_{1}^{4} 2 y d y = \frac{4}{3} [y^{3/2}]_{1}^{4} = \frac{4}{3} (8 - 1) = \frac{28}{3}$ sq units.
Q4 · 3 marks · Short answerHarder board level
Find, by integration, the area of the circle $x^{2} + y^{2} = 25$ .
Hint
Four times the first-quadrant area.
Answer
$4 \int_{0}^{5} 25 - x^{2} d x = 4 [\frac{x}{2} 25 - x^{2} + \frac{25}{2} sin^{- 1} \frac{x}{5}]_{0}^{5} = 4 \cdot \frac{25 π}{4} = 25 π$ sq units.
Q5 · 2 marks · Short answerHarder board level
Find the area under $y = lo g x$ from $x = 1$ to $x = e$ .
Hint
$\int lo g x d x = x lo g x - x$ .
Answer
$[x lo g x - x]_{1}^{e} = (e - e) - (0 - 1) = 1$ sq unit.
Q6 · 3 marks · Short answerHOTS / competency
Find the area bounded by $y = cos x$ and the x-axis between $x = 0$ and $x = 2 π$ .
Hint
Split at $\frac{π}{2}$ and $\frac{3 π}{2}$ .
Answer
$\int_{0}^{π /2} cos x d x + \int_{π /2}^{3 π /2} cos x d x + \int_{3 π /2}^{2 π} cos x d x = 1 + 2 + 1 = 4$ sq units. (The plain integral from 0 to $2 π$ is 0.)
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose the front of a greenhouse is a parabolic arch $y = 4 - \frac{x ^{2}}{4}$ (in metres) standing on the ground $y = 0$ .
(i) Find the width of the arch at ground level. (1 mark)
(ii) Find the area of the front by integration. (1 mark)
(iii) Glass costs Rs 500 per m². Find the cost of glazing the front. (1 mark)
(iv) Find the area of the part of the front above the height 3 m. (1 mark)
Hint
Use symmetry about the y-axis.
Answer
(i) $y = 0$ at $x = \pm 4$ : width 8 m. (ii) $2 \int_{0}^{4} (4 - \frac{x ^{2}}{4}) d x = 2 (16 - \frac{16}{3}) = \frac{64}{3} \approx 21.33$ m². (iii) $\frac{64}{3} \times 500 \approx$ Rs 10 666.67. (iv) $4 - \frac{x ^{2}}{4} \geq 3$ for $∣ x ∣ \leq 2$ : $2 \int_{0}^{2} (1 - \frac{x ^{2}}{4}) d x = \frac{8}{3}$ m².

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write the area between $y = f (x)$ , the x-axis and $x = a$ , $x = b$ when $f \geq 0$ .Show answer
$\int_{a}^{b} f (x) d x$ .
2.What do you do if the curve goes below the x-axis?Show answer
Split at the x-intercepts and add the absolute values of the integrals.
3.Area of the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ ?Show answer
$6 π$ sq units.
4.Area under $y = e^{x}$ from 0 to $lo g 3$ ?Show answer
2 sq units.
5.Which strips do you use for a region bounded by $x = g (y)$ and the y-axis?Show answer
Horizontal strips: $\int x d y$ .
6.Area bounded by $y^{2} = 4 x$ and $x = 1$ ?Show answer
$\frac{8}{3}$ sq units.

Area Under a Curve: frequently asked questions

Why can the integral be negative when area is positive?

The integral counts parts below the x-axis as negative. Area is a positive quantity, so for regions below the axis you take the absolute value, and for curves that cross the axis you split at the crossing points.

When should I integrate with respect to y?

When the region is bounded by the y-axis and horizontal lines, or when the curve is easier to write as x = g(y), such as a parabola opening sideways between two horizontal lines.

Do I have to sketch the curve?

It is strongly recommended. Examiners reported that missing or wrong sketches were the main reason candidates could not find the limits or the region, and a correct sketch often earns a mark itself.

Which curves are in the ISC 2027 syllabus for area?

Simple curves only: lines, circles, parabolas, ellipses, polynomial functions, the modulus function, and exponential and logarithmic functions, together with the area between two curves.

How many marks does area by integration carry in ISC?

In 2024 and 2025 it appeared as 4-mark questions with internal choice, including a case-based question on the parabola y = 4 − x² in 2025.