MathsISC Class 12 · Application of Integrals

Area Between Two Curves: Intersections, Upper Minus Lower and Strips

Updated 15 October 2026 · Fact-checked

To find the area between two curves, solve their equations together to get the points of intersection, which give the limits. Between those limits, integrate (upper curve − lower curve) dx, or (right curve − left curve) dy for horizontal strips. If the curves swap position, split the interval at each crossing.

Understand Area Between Two Curves

Upper minus lower. If $f (x) \geq g (x)$ on $[a, b]$ , the area between $y = f (x)$ and $y = g (x)$ from $x = a$ to $x = b$ is
$A = \int_{a}^{b} [f (x) - g (x)] d x .$
Each vertical strip has height $f (x) - g (x)$ and width $d x$ . This works whether the curves are above or below the x-axis, because only the difference matters. The ISC 2024 examiners noted candidates who did not express the region as a difference of two functions.

Finding the limits. The curves usually enclose a region between their points of intersection. Solve the two equations simultaneously; the x-coordinates (or y-coordinates) are the limits. For $y^{2} = 4 x$ and $y = 2 x$ : $4 x^{2} = 4 x$ , so $x = 0$ or $1$ .

Which curve is on top? Test a point between the limits. At $x = \frac{1}{2}$ : $2 x \approx 1.41 > 1 = 2 x$ , so the parabola is above the line.

Horizontal strips. If the curves are naturally of the form $x = g (y)$ , such as parabolas opening sideways, or if vertical strips would need splitting, use
$A = \int_{c}^{d} [x_{right} - x_{left}] d y .$
For $y^{2} = 4 x$ and the line $y = 2 x - 4$ , horizontal strips give one integral from $y = - 2$ to $y = 4$ , while vertical strips would need two.

When the boundary changes. If the upper (or lower) boundary is made of different curves in different parts, split the interval where it changes. The ISC 2025 field bounded by three lines needed exactly this: one integral from the left vertex to the middle vertex, another from the middle vertex to the right vertex, each as upper line minus lower line. The examiners noted candidates who did not express the region as a sum of two regions.

Circles and lines. For a segment of a circle cut off by a chord, integrate $r^{2} - x^{2}$ (using the standard integral) minus the line, or use symmetry about an axis. The answer can always be checked geometrically: sector area minus triangle area.

Sanity checks. The area must be positive. If two curves cross inside the interval, $\int (f - g)$ over the whole interval can cancel; split at the crossing and take absolute values. Compare with a rough estimate from the sketch (count squares) to catch large errors.

Area Between Two Curves: formulas to remember

Vertical strips: $A = \int_{a}^{b} [y_{upper} - y_{lower}] d x$; Limits: x-coordinates of intersections.
Horizontal strips: $A = \int_{c}^{d} [x_{right} - x_{left}] d y$; Limits: y-coordinates of intersections.
Crossing curves: $A = \int_{a}^{b} ∣ f (x) - g (x) ∣ d x$; Split at every crossing inside [a, b].
Parabola and line: $y^{2} = 4 a x, y = m x : A = \frac{8 a ^{2}}{3 m ^{3}}$; A handy check for the standard case.
Two parabolas: $y^{2} = 4 a x, x^{2} = 4 a y : A = \frac{16 a ^{2}}{3}$; Meet at (0, 0) and (4a, 4a).
Circular segment: $A = \frac{r ^{2}}{2} (θ - sin θ)$; θ = angle at the centre in radians; useful to check an integral.

How to solve Area Between Two Curves questions

Follow these steps for every area-between-curves question.

1Sketch both curves on the same axes and shade the region asked for.
2Solve the equations together to find the points of intersection; label them on the sketch.
3Decide on vertical strips ( $d x$ ) or horizontal strips ( $d y$ ): choose the one that needs fewer integrals.
4Identify the upper and lower (or right and left) boundary on each part; test a point if unsure.
5If the boundary changes, split the interval at that point and write one integral per part.
6Integrate (upper − lower) on each part and add.
7State the answer in square units, and check it is positive and roughly matches the sketch.

Area Between Two Curves 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 between a parabola and a line

Suppose a park has a pond whose curved bank follows $y = x^{2}$ and whose straight boardwalk follows $y = x + 2$ on a plan (units of 10 m).

Find the area of the pond enclosed between the bank and the boardwalk.

Try it first, then show the solutionSolution

Intersections: $x^{2} = x + 2 \Rightarrow x^{2} - x - 2 = 0 \Rightarrow x = - 1$ or $x = 2$ .
At $x = 0$ : line $2 >$ parabola $0$ , so the line is on top.
$A = \int_{- 1}^{2} (x + 2 - x^{2}) d x = [\frac{x ^{2}}{2} + 2 x - \frac{x ^{3}}{3}]_{- 1}^{2} = \frac{10}{3} - (- \frac{7}{6}) = \frac{9}{2}$ .
Each square unit is $10 \times 10 = 100$ m², so the area is $450$ m².

Answer: $\frac{9}{2}$ square units, that is 450 m².

Memory hook: Intersections give the limits; the line on top minus the parabola below gives the height.

Area between two trigonometric curves

Imagine a decorative metal strip whose edges follow $y = sin x$ and $y = cos x$ between two consecutive crossing points, $x = \frac{π}{4}$ and $x = \frac{5 π}{4}$ (units in decimetres).

Find the area of the strip.

Try it first, then show the solutionSolution

Between $\frac{π}{4}$ and $\frac{5 π}{4}$ , $sin x \geq cos x$ (check at $x = \frac{π}{2}$ : $1 > 0$ ).
$A = \int_{π /4}^{5 π /4} (sin x - cos x) d x = [- cos x - sin x]_{π /4}^{5 π /4}$ .
$= (\frac{2}{2} + \frac{2}{2}) - (- \frac{2}{2} - \frac{2}{2}) = 22$ .

Answer: $22 \approx 2.83$ dm².

Memory hook: Between two crossings the order of the curves stays the same: integrate top minus bottom.

Tricks to remember Area Between Two Curves

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

Top minus bottom
Vertical strips: (upper curve) − (lower curve). Horizontal strips: (right) − (left).
Use it when: Setting up every integral.
Meet to find the limits
Where the curves meet, the region begins and ends: solve them together.
Use it when: Finding limits, the step examiners say is most often missed.
Sideways parabola, sideways strips
If a parabola opens left or right, try dy strips first.
Use it when: Avoiding unnecessary splitting.
Swap means split
If the top curve changes, split the interval there.
Use it when: Regions bounded by three lines or by a curve and two lines.

Area Between Two Curves: solved examples

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

Example 1: Line and parabola through the origin · 1 mark · MCQBoard level

The area between $y = x$ and $y = x^{2}$ is:
(a) $\frac{1}{2}$
(b) $\frac{1}{3}$
(c) $\frac{1}{6}$
(d) $\frac{1}{12}$

Try it first, then show the solutionSolution

They meet at $x = 0, 1$ ; $\int_{0}^{1} (x - x^{2}) d x = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ .

Answer: (c) $\frac{1}{6}$ .

Example 2: Parabola and line · 3 marks · Short answerBoard level

Find the area of the region bounded by $y^{2} = 4 x$ and the line $y = 2 x$ .

Try it first, then show the solutionSolution

Intersections: $(2 x)^{2} = 4 x \Rightarrow x = 0, 1$ ; points $(0, 0)$ and $(1, 2)$ . (1 mark)
Between them the parabola $y = 2 x$ is above the line. $A = \int_{0}^{1} (2 x - 2 x) d x$ . (1 mark)
$= \frac{4}{3} - 1 = \frac{1}{3}$ sq unit. (1 mark)

Answer: $\frac{1}{3}$ sq unit.

Example 3: Two parabolas · 3 marks · Short answerHarder board level

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

Try it first, then show the solutionSolution

$x^{2} = 4 - x^{2} \Rightarrow x = \pm 2$ . The upper curve is $y = 4 - x^{2}$ . (1 mark)
By symmetry, $A = 2 \int_{0}^{2} (4 - 2 x^{2}) d x = 2 [4 x - \frac{2 x ^{3}}{3}]_{0}^{2}$ . (1 mark)
$= 2 (42 - \frac{4 2}{3}) = \frac{16 2}{3}$ sq units. (1 mark)

Answer: $\frac{16 2}{3}$ sq units.

Example 4: A segment of a circle · 3 marks · Short answerHarder board level

Find the area of the smaller part of the circle $x^{2} + y^{2} = 4$ cut off by the line $x = 1$ .

Try it first, then show the solutionSolution

The smaller part lies between $x = 1$ and $x = 2$ , symmetric about the x-axis: $A = 2 \int_{1}^{2} 4 - x^{2} d x$ . (1 mark)
$= 2 [\frac{x}{2} 4 - x^{2} + 2 sin^{- 1} \frac{x}{2}]_{1}^{2} = 2 [π - (\frac{3}{2} + \frac{π}{3})]$ . (1 mark)
$= \frac{4 π}{3} - 3$ sq units. (1 mark)

Answer: $\frac{4 π}{3} - 3 \approx 2.46$ sq units.

Example 5: Horizontal strips · 3 marks · Short answerHarder board level

Find the area bounded by the curves $x = y^{2}$ and $x = 3 - 2 y^{2}$ .

Try it first, then show the solutionSolution

Intersections: $y^{2} = 3 - 2 y^{2} \Rightarrow y = \pm 1$ . (1 mark)
The right curve is $x = 3 - 2 y^{2}$ , the left $x = y^{2}$ . $A = \int_{- 1}^{1} (3 - 3 y^{2}) d y$ . (1 mark)
$= [3 y - y^{3}]_{- 1}^{1} = 2 - (- 2) = 4$ sq units. (1 mark)

Answer: 4 sq units.

Example 6: A circle and a chord · 5 marks · Long answerHOTS / competency

Find, by integration, the area of the region ${(x, y) : x^{2} + y^{2} \leq 4, x + y \geq 2}$ .

Try it first, then show the solutionSolution

The line $x + y = 2$ meets the circle at $(2, 0)$ and $(0, 2)$ . The region is the part of the disc above the line, in the first quadrant. (1 mark)
Upper boundary: circle $y = 4 - x^{2}$ ; lower: line $y = 2 - x$ ; $x$ from 0 to 2. (1 mark)
$A = \int_{0}^{2} 4 - x^{2} d x - \int_{0}^{2} (2 - x) d x$ . (1 mark)
$\int_{0}^{2} 4 - x^{2} d x = [\frac{x}{2} 4 - x^{2} + 2 sin^{- 1} \frac{x}{2}]_{0}^{2} = π$ ; $\int_{0}^{2} (2 - x) d x = 2$ . (1 mark)
$A = π - 2$ sq units. Check: quarter disc $π$ minus triangle of area 2. (1 mark)

Answer: $π - 2 \approx 1.14$ sq units.

Where students lose marks in Area Between Two Curves

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

Mistake: Wrong or missing limits
The ISC 2024 examiners noted candidates unfamiliar with finding the limits of a region bounded by two curves.
Do this: Always solve the two equations together and write the intersection points before integrating.
Usually costs: Most of the 4 marks.
Mistake: Not writing the area as a difference
Integrating only the upper curve gives the area down to the axis, not between the curves.
Do this: Write (upper − lower) explicitly in the integrand.
Usually costs: 2 marks.
Mistake: Not splitting a region with a changing boundary
In the 2025 three-lines question, candidates did not express the region as the sum of two parts.
Do this: Mark on the sketch where the upper or lower boundary changes and split there.
Usually costs: 2 to 3 marks.
Mistake: Wrong upper curve
Choosing the wrong curve as upper gives a negative area.
Do this: Test a point between the limits in both equations.
Usually costs: 1 mark.

Exam tips

Show the sketch with intersection points labelled; it supports your limits.
Write the area integral in full before evaluating; method marks depend on it.
For circles and ellipses, use the standard integral of $a^{2} - x^{2}$ and check with sector or triangle areas.
Choose horizontal strips when the left and right boundaries are simpler than the top and bottom.
Finish with "sq units" and, in context, the real units.

Practice questions on Area Between Two Curves

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 enclosed between $y = x^{2}$ and $y = 4$ is:
(a) $\frac{16}{3}$
(b) $\frac{32}{3}$
(c) 8
(d) 16
Hint
They meet at $x = \pm 2$ .
Answer
(b) $\frac{32}{3}$ .
Q2 · 2 marks · Short answerBoard level
Find the area enclosed by $y = x$ and $y = x^{3}$ in the first quadrant.
Hint
They meet at $x = 0$ and $x = 1$ .
Answer
$\int_{0}^{1} (x - x^{3}) d x = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$ sq unit.
Q3 · 3 marks · Short answerHarder board level
Find the area bounded by $y^{2} = 4 x$ and the line $y = 2 x - 4$ .
Hint
Use horizontal strips: $x_{right} = \frac{y + 4}{2}$ , $x_{left} = \frac{y ^{2}}{4}$ .
Answer
Intersections: $y^{2} - 2 y - 8 = 0$ , so $y = - 2, 4$ . $A = \int_{- 2}^{4} (\frac{y + 4}{2} - \frac{y ^{2}}{4}) d y = 9$ sq units.
Q4 · 3 marks · Short answerHarder board level
Find the area of the smaller region bounded by the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ and the line $\frac{x}{3} + \frac{y}{2} = 1$ .
Hint
Quarter ellipse minus a triangle.
Answer
$\int_{0}^{3} [\frac{2}{3} 9 - x^{2} - \frac{2}{3} (3 - x)] d x = \frac{2}{3} \cdot \frac{9 π}{4} - 3 = \frac{3 π}{2} - 3$ sq units.
Q5 · 3 marks · Short answerHarder board level
Find the area bounded by $y = e^{x}$ , $y = e^{- x}$ and the line $x = 1$ .
Hint
The curves meet at $x = 0$ .
Answer
$\int_{0}^{1} (e^{x} - e^{- x}) d x = (e - 1) + (e^{- 1} - 1) = e + \frac{1}{e} - 2$ sq units.
Q6 · 3 marks · Short answerHOTS / competency
Find the area of the region enclosed by $y = ∣ x ∣$ and $y = 2 - x^{2}$ .
Hint
For $x \geq 0$ , $x = 2 - x^{2}$ gives $x = 1$ ; use symmetry.
Answer
$2 \int_{0}^{1} (2 - x^{2} - x) d x = 2 (2 - \frac{1}{3} - \frac{1}{2}) = \frac{7}{3}$ sq units.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a company logo is the lens-shaped region between $y = x^{2}$ and $y = 2 x - x^{2}$ (in cm).
(i) Find the points where the curves meet. (1 mark)
(ii) Find the area of the logo. (1 mark)
(iii) The logo is enlarged by a factor of 3 in both directions for a banner. What is the new area? (1 mark)
(iv) Which curve forms the upper edge of the logo? Justify. (1 mark)
Hint
Area scales with the square of the factor.
Answer
(i) $x^{2} = 2 x - x^{2} \Rightarrow x = 0, 1$ : $(0, 0)$ and $(1, 1)$ . (ii) $\int_{0}^{1} (2 x - 2 x^{2}) d x = 1 - \frac{2}{3} = \frac{1}{3}$ cm². (iii) $9 \times \frac{1}{3} = 3$ cm². (iv) $y = 2 x - x^{2}$ : at $x = \frac{1}{2}$ it gives $\frac{3}{4} > \frac{1}{4}$ .

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 $y = f (x)$ (upper) and $y = g (x)$ (lower) from $a$ to $b$ .Show answer
$\int_{a}^{b} [f (x) - g (x)] d x$ .
2.How do you find the limits?Show answer
Solve the two equations simultaneously for their intersection points.
3.Area between $y = x^{2}$ and $y = x$ ?Show answer
$\frac{1}{6}$ sq unit.
4.Area between $y^{2} = 4 x$ and $x^{2} = 4 y$ ?Show answer
$\frac{16}{3}$ sq units.
5.When should you use horizontal strips?Show answer
When the boundaries are easier as $x = g (y)$ , or when vertical strips would need splitting.
6.What do you do if the curves cross inside the interval?Show answer
Split at the crossing and add the absolute values.

Area Between Two Curves: frequently asked questions

Does it matter if the region is below the x-axis?

No. For the area between two curves, only the difference (upper − lower) matters, and it is positive wherever the upper curve is above the lower one, whatever their position relative to the axis.

How do I know which curve is on top?

Pick any x between the limits and compute both y-values. The larger one belongs to the upper curve on that interval. A sketch makes this obvious.

When do I need to split the integral?

Whenever the upper or lower boundary changes from one curve or line to another, or the two curves cross inside the interval. Each part then has a single upper and lower curve.

Can I use horizontal strips instead of vertical ones?

Yes. Use (right − left) dy with y-limits. It is often simpler for parabolas opening sideways, and it can avoid splitting the region.

Is the area between two curves in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus includes the area enclosed between two curves, with lines, circles, parabolas, ellipses and other simple curves, and it was asked in 2024 (the region between y² = 4x and x² = 4y) and in 2025 (a region bounded by three lines).