MathsISC Class 12 · Application of Integrals

Area of a Region Bounded by Lines Using Integration

Q: Why use integration for a triangle when there is a formula?

ISC questions often require integration to test your ability to set up area integrals with correct limits and boundaries. Use the formula only to check your answer.

Q: How do I know where to split the integral?

Sketch the region. Split wherever the line forming the top edge or the bottom edge changes, which for a triangle is at the middle vertex.

Q: What if the triangle is partly below the x-axis?

It does not matter. Integrating (upper line − lower line) measures the vertical width of the region at each x, so the answer is correct wherever the triangle lies.

Q: Can I use horizontal strips instead?

Yes. Write the left and right lines as x in terms of y and integrate (right − left) dy. This is sometimes a single integral when vertical strips need two.

Q: Was this asked in recent ISC papers?

Yes. ISC 2025 asked for the area of a field bounded by three lines using integration, as a 4-mark question.

Updated 15 October 2026 · Fact-checked

To find the area of a region bounded by straight lines, such as a triangle, find the vertices by solving the lines in pairs. Then integrate (upper line − lower line) between consecutive vertex x-coordinates, splitting the interval wherever the upper or lower boundary changes, and add the parts.

Split the triangle at the middle vertex's x-value, integrate top minus bottom on each part, and add: 5/6 + 5/3 = 5/2.

Illustration drawn for QuestionBuddy.

Understand Area of a Region Bounded by Lines

A triangle's area is easy by formula, but ISC questions often insist "using integration". The purpose is to test whether you can turn a region into the right integrals, a skill that matters for curved regions too. The ISC 2025 paper asked for the area of a farmer's field bounded by three lines, and the examiners noted candidates who could not find the limits, did not split the region into two parts, and made errors writing each part as a difference of two functions.

Step 1: equations in the form y = mx + c. Rewrite each line, for example $x + 2 y = 2$ as $y = 1 - \frac{x}{2}$ .

Step 2: vertices. Solve the lines in pairs. Three lines give three vertices; sort them by x-coordinate, say $x_{1} < x_{2} < x_{3}$ .

Step 3: identify upper and lower boundaries. For vertical strips between $x_{1}$ and $x_{2}$ , one line forms the top edge and another the bottom; between $x_{2}$ and $x_{3}$ , one of these changes. Which one changes depends on whether the middle vertex is above or below the line joining the other two. A sketch settles it at once.

Step 4: integrate in parts.
$A = \int_{x_{1}}^{x_{2}} (upper - lower) d x + \int_{x_{2}}^{x_{3}} (upper - lower) d x .$
Each integrand is linear, so the integrals are quick.

Step 5: check. For a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ , the area is also $\frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$ (the determinant formula from the Determinants chapter). If your integral disagrees, look for a wrong line or a missing split.

Horizontal strips. Sometimes the region is simpler with $d y$ : if the left and right boundaries are single lines over the whole height, write each as $x = \dots$ and integrate (right − left) $d y$ in one go.

**Regions with $∣ x ∣$ .** The graph of $y = ∣ x ∣$ is a pair of lines meeting at the origin; $y = ∣ x - a ∣$ is the same shape shifted. Regions such as "bounded by $y = ∣ x ∣$ and $y = 3$ " are symmetric, so integrate over the right half and double.

Regions above or below the axis. If a line crosses the x-axis inside the region, the "upper minus lower" form still works, because it measures the vertical distance between boundaries rather than height above the axis.

Area of a Region Bounded by Lines: formulas to remember

Area by vertical strips: $A = \int_{x_{1}}^{x_{2}} (y_{u} - y_{l}) d x + \int_{x_{2}}^{x_{3}} (y_{u} - y_{l}) d x$; Split at the middle vertex.
Area by horizontal strips: $A = \int_{c}^{d} (x_{r} - x_{l}) d y$; Use when left and right boundaries do not change.
Line through two points: $y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1})$; Needed when vertices are given instead of lines.
Determinant check: $Δ = \frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$; Confirms the integral for a triangle.
Integral of a line: $\int_{a}^{b} (m x + c) d x = \frac{m}{2} (b^{2} - a^{2}) + c (b - a)$; Equals the trapezium area when the line is above the axis.

How to solve Area of a Region Bounded by Lines questions

Use this sequence for every "area bounded by lines, using integration" question.

1Rewrite every line in the form $y = m x + c$ (or $x = m y + c$ if using horizontal strips).
2Find the vertices by solving the lines in pairs, and list them in order of x-coordinate.
3Sketch the triangle (or polygon) and shade it; mark which line forms the top and which the bottom on each part.
4Write one integral of (upper − lower) for each x-interval between consecutive vertices.
5Evaluate each integral and add.
6Check with the determinant (shoelace) formula for a triangle.
7State the answer in square units (or m² in context).

Area of a Region Bounded by Lines 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 a symmetric region bounded by lines

Suppose a garden bed is shaped like an arrowhead bounded by the lines $y = ∣ x ∣$ and $y = 4$ on a plan in metres.

Find the area of the bed by integration.

Try it first, then show the solutionSolution

The lines $y = ∣ x ∣$ meet $y = 4$ at $x = \pm 4$ .
By symmetry, $A = 2 \int_{0}^{4} (4 - x) d x = 2 [4 x - \frac{x ^{2}}{2}]_{0}^{4} = 2 \times 8 = 16$ .

Answer: 16 m².

Memory hook: For |x|, work on the right half and double.

Triangle bounded by three lines

Imagine a triangular traffic island at a junction, bounded by the roads $y = 0$ , $y = x$ and $x + y = 6$ on a plan in metres.

Find the area of the island by integration.

Try it first, then show the solutionSolution

Vertices: $y = 0$ and $y = x$ meet at $(0, 0)$ ; $y = 0$ and $x + y = 6$ at $(6, 0)$ ; $y = x$ and $x + y = 6$ at $(3, 3)$ .
From $x = 0$ to 3 the top edge is $y = x$ ; from 3 to 6 it is $y = 6 - x$ ; the bottom is $y = 0$ throughout.
$A = \int_{0}^{3} x d x + \int_{3}^{6} (6 - x) d x = 4.5 + 4.5 = 9$ .
Check: $\frac{1}{2} \times 6 \times 3 = 9$ .

Answer: 9 m².

Memory hook: When the top edge changes at a vertex, the integral splits there.

Tricks to remember Area of a Region Bounded by Lines

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

Vertices first
No limits without vertices: solve the lines in pairs before anything else.
Use it when: Finding limits, the most common failure.
Split at the middle
For a triangle with vertical strips, split at the middle x-coordinate.
Use it when: Setting up two integrals.
Top minus bottom, line by line
On each part, name the top line and the bottom line before integrating.
Use it when: Avoiding sign errors.
Check with ½|det|
The determinant formula gives the triangle area in one line; use it to check.
Use it when: Catching wrong lines or splits.

Area of a Region Bounded by Lines: solved examples

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

Example 1: Area under a straight line · 1 mark · MCQBoard level

The area bounded by $y = 2 x$ , the x-axis and the line $x = 3$ is:
(a) 3
(b) 6
(c) 9
(d) 18

Try it first, then show the solutionSolution

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

Answer: (c) 9 sq units.

Example 2: A triangle on the x-axis · 3 marks · Short answerBoard level

Using integration, find the area of the triangle bounded by $y = x$ , $y = 4 - x$ and the x-axis.

Try it first, then show the solutionSolution

Vertices: $(0, 0)$ , $(4, 0)$ and $(2, 2)$ . (1 mark)
$A = \int_{0}^{2} x d x + \int_{2}^{4} (4 - x) d x$ . (1 mark)
$= 2 + 2 = 4$ sq units. (1 mark)

Answer: 4 sq units.

Example 3: A triangle from its vertices · 5 marks · Long answerHarder board level

Using integration, find the area of the triangle with vertices $A (2, 0)$ , $B (4, 5)$ and $C (6, 3)$ .

Try it first, then show the solutionSolution

Lines: AB: $y = \frac{5}{2} (x - 2)$ ; BC: $y = 9 - x$ ; AC: $y = \frac{3}{4} (x - 2)$ . (1½ marks)
From $x = 2$ to 4: top AB, bottom AC. From 4 to 6: top BC, bottom AC. (½ mark)
$\int_{2}^{4} (\frac{5}{2} (x - 2) - \frac{3}{4} (x - 2)) d x = \int_{2}^{4} \frac{7 ( x - 2 )}{4} d x = \frac{7}{2}$ . (1 mark)
$\int_{4}^{6} (9 - x - \frac{3}{4} (x - 2)) d x = \int_{4}^{6} \frac{7 ( 6 - x )}{4} d x = \frac{7}{2}$ . (1 mark)
Area $= 7$ sq units. Check: $\frac{1}{2} ∣2 (5 - 3) + 4 (3 - 0) + 6 (0 - 5) ∣ = \frac{1}{2} ∣4 + 12 - 30∣ = 7$ . (1 mark)

Answer: 7 sq units.

Example 4: Two lines and the x-axis · 3 marks · Short answerHarder board level

Find, by integration, the area of the region bounded by $y = x + 1$ , $y = 3 - x$ and the x-axis.

Try it first, then show the solutionSolution

Vertices: $(- 1, 0)$ , $(3, 0)$ , and the lines meet at $(1, 2)$ . (1 mark)
$A = \int_{- 1}^{1} (x + 1) d x + \int_{1}^{3} (3 - x) d x$ . (1 mark)
$= 2 + 2 = 4$ sq units. (1 mark)

Answer: 4 sq units.

Example 5: A region bounded by a modulus graph · 2 marks · Short answerHarder board level

Find the area of the region bounded by $y = ∣ x ∣$ and $y = 3$ .

Try it first, then show the solutionSolution

The region is symmetric about the y-axis, between $x = - 3$ and $x = 3$ . (1 mark)
$A = 2 \int_{0}^{3} (3 - x) d x = 2 \times \frac{9}{2} = 9$ sq units. (1 mark)

Answer: 9 sq units.

Example 6: A field bounded by three lines · 5 marks · Case-basedHOTS / competency

Suppose a farmer's field is bounded by the lines $y - x = 1$ , $3 x + 2 y = 12$ and $x + 4 y = 4$ .
(i) Find the vertices of the field. (2 marks)
(ii) Using integration, find its area. (2 marks)
(iii) Verify your answer using determinants. (1 mark)

Try it first, then show the solutionSolution

(i) $y = x + 1$ and $x + 4 y = 4$ : $x + 4 x + 4 = 4 \Rightarrow x = 0$ , vertex $(0, 1)$ . $y = x + 1$ and $3 x + 2 y = 12$ : $5 x + 2 = 12 \Rightarrow (2, 3)$ . $3 x + 2 y = 12$ and $x + 4 y = 4$ : $x = 4$ , $y = 0$ , vertex $(4, 0)$ .
(ii) Lower boundary throughout: $y = 1 - \frac{x}{4}$ . Upper: $y = x + 1$ for $0 \leq x \leq 2$ , and $y = 6 - \frac{3 x}{2}$ for $2 \leq x \leq 4$ .
$A = \int_{0}^{2} \frac{5 x}{4} d x + \int_{2}^{4} (5 - \frac{5 x}{4}) d x = \frac{5}{2} + [5 x - \frac{5 x ^{2}}{8}]_{2}^{4} = \frac{5}{2} + (10 - \frac{15}{2}) = 5$ .
(iii) $\frac{1}{2} ∣0 (3 - 0) + 2 (0 - 1) + 4 (1 - 3) ∣ = \frac{1}{2} ∣ - 2 - 8∣ = 5$ . Verified.

Answer: Vertices $(0, 1)$ , $(2, 3)$ , $(4, 0)$ ; area 5 sq units.

Where students lose marks in Area of a Region Bounded by Lines

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

Mistake: Not splitting the region
The ISC 2025 examiners noted candidates who did not express the triangular field as a sum of two regions.
Do this: Split at the x-coordinate of the middle vertex, where the upper or lower line changes.
Usually costs: 2 to 3 marks of a 4-mark question.
Mistake: Wrong vertices or limits
Errors in solving the lines in pairs give wrong limits for every integral.
Do this: Check each vertex in both of its lines before integrating.
Usually costs: Most of the question.
Mistake: Integrating only the upper line
This gives the area down to the x-axis, not between the lines, unless the lower boundary is the axis.
Do this: Write (upper − lower) explicitly on each part.
Usually costs: 1 to 2 marks.
Mistake: Using a formula when integration is required
The question says "using integration"; a formula-only answer earns little.
Do this: Do the integrals, then use the determinant formula only as a check.
Usually costs: Most of the marks.

Exam tips

Write each line as $y = \dots$ beside the sketch; it makes upper and lower easy to read.
List the vertices in order of x before writing the limits.
Linear integrands are quick: expand, integrate, substitute, and keep fractions exact.
End with a one-line determinant check; it reassures you and the examiner.
For symmetric regions with $∣ x ∣$ , integrate over one half and double, saying so.

Practice questions on Area of a Region Bounded by Lines

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 of the triangle bounded by $x + y = 4$ and the coordinate axes is:
(a) 4
(b) 8
(c) 16
(d) 2
Hint
$\int_{0}^{4} (4 - x) d x$ .
Answer
(b) 8 sq units.
Q2 · 2 marks · Short answerBoard level
Find, by integration, the area bounded by $y = 3 x$ , the x-axis and $x = 2$ .
Hint
One integral.
Answer
$\int_{0}^{2} 3 x d x = 6$ sq units.
Q3 · 3 marks · Short answerHarder board level
Using integration, find the area of the triangle with vertices $(- 1, 0)$ , $(1, 3)$ and $(3, 2)$ .
Hint
Lower boundary is the line through $(- 1, 0)$ and $(3, 2)$ .
Answer
Lines: $y = \frac{3}{2} (x + 1)$ , $y = \frac{7 - x}{2}$ , $y = \frac{x + 1}{2}$ . $A = \int_{- 1}^{1} (x + 1) d x + \int_{1}^{3} (3 - x) d x = 2 + 2 = 4$ sq units.
Q4 · 3 marks · Short answerHarder board level
Find the area of the region bounded by $y = ∣ x - 2∣$ and $y = 2$ .
Hint
The region lies between $x = 0$ and $x = 4$ .
Answer
$\int_{0}^{4} (2 - ∣ x - 2∣) d x = 8 - 4 = 4$ sq units.
Q5 · 2 marks · Short answerHarder board level
Find the area between the lines $y = x$ and $y = 2 x$ from $x = 0$ to $x = 3$ .
Hint
Upper minus lower is $x$ .
Answer
$\int_{0}^{3} (2 x - x) d x = \frac{9}{2}$ sq units.
Q6 · 3 marks · Short answerHOTS / competency
Find the area of the region ${(x, y) : ∣ x ∣ \leq y \leq 4 - ∣ x ∣}$ .
Hint
Sketch: it is a square with vertices $(0, 0)$ , $(2, 2)$ , $(0, 4)$ , $(- 2, 2)$ .
Answer
$2 \int_{0}^{2} [(4 - x) - x] d x = 2 [4 x - x^{2}]_{0}^{2} = 8$ sq units.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a triangular plot has corners at $O (0, 0)$ , $P (6, 0)$ and $Q (4, 4)$ on a plan where 1 unit is 10 m.
(i) Write the equations of $O Q$ and $P Q$ . (1 mark)
(ii) Find the area of the plot by integration. (1 mark)
(iii) Find its real area in m². (1 mark)
(iv) A fence along $x = 4$ splits the plot. Find the ratio of the two parts. (1 mark)
Hint
Split the integral at $x = 4$ .
Answer
(i) $O Q$ : $y = x$ ; $P Q$ : $y = 12 - 2 x$ . (ii) $\int_{0}^{4} x d x + \int_{4}^{6} (12 - 2 x) d x = 8 + 4 = 12$ sq units. (iii) $12 \times 100 = 1200$ m². (iv) $8 : 4 = 2 : 1$ .

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.How do you find the vertices of a triangle bounded by three lines?Show answer
Solve the lines in pairs.
2.Where do you split the integral for a triangle using vertical strips?Show answer
At the x-coordinate of the middle vertex.
3.Area bounded by $y = ∣ x ∣$ and $y = 2$ ?Show answer
4 sq units.
4.State the determinant formula for the area of a triangle.Show answer
$\frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$ .
5.Area under $y = x + 1$ from $x = 0$ to $x = 2$ ?Show answer
4 sq units.
6.When are horizontal strips better?Show answer
When the left and right boundaries stay the same over the whole height of the region.

Area of a Region Bounded by Lines: frequently asked questions

Why use integration for a triangle when there is a formula?

ISC questions often require integration to test your ability to set up area integrals with correct limits and boundaries. Use the formula only to check your answer.

How do I know where to split the integral?

Sketch the region. Split wherever the line forming the top edge or the bottom edge changes, which for a triangle is at the middle vertex.

What if the triangle is partly below the x-axis?

It does not matter. Integrating (upper line − lower line) measures the vertical width of the region at each x, so the answer is correct wherever the triangle lies.

Can I use horizontal strips instead?

Yes. Write the left and right lines as x in terms of y and integrate (right − left) dy. This is sometimes a single integral when vertical strips need two.

Was this asked in recent ISC papers?

Yes. ISC 2025 asked for the area of a field bounded by three lines using integration, as a 4-mark question.