MathsCBSE Class 12 · Application of Integrals

Area Under Lines and Modulus Graphs by Integration (Class 12)

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 the integral of |f(x)| from a to b. For a modulus graph such as y = |x − 2|, split where the inside changes sign. Parts below the x-axis give negative integrals; add their absolute values.

The plain integral of x − 2 on [0, 4] cancels to 0, but flipping the part below the axis (the modulus) shows the true area is 2 + 2 = 4.

Illustration drawn for QuestionBuddy.

Understand Area Under Lines and Modulus Graphs

Area as an integral. If $f (x) \geq 0$ on $[a, b]$ , the area of the region bounded by $y = f (x)$ , the x-axis and the vertical lines $x = a$ , $x = b$ is
$A = \int_{a}^{b} f (x) d x$ .
Think of the region as made of thin vertical strips of width $d x$ and height $f (x)$ ; the integral adds them up. With horizontal strips, the region bounded by $x = g (y)$ , the y-axis and the lines $y = c$ , $y = d$ has area $\int_{c}^{d} g (y) d y$ .

Below the axis. If $f (x) \leq 0$ on part of the interval, the integral there is negative, but area is never negative. So:

find where $f (x) = 0$ in $[a, b]$ ;
integrate separately on each part;
add the absolute values.

For $y = x - 2$ on $[0, 4]$ the plain integral is 0, but the area is $2 + 2 = 4$ .

Lines. For a straight line the integral can be checked with geometry: the area under $y = 2 x + 1$ from $x = 1$ to $x = 3$ is a trapezium with parallel sides 3 and 7 and width 2, so $A = 10$ , matching $\int_{1}^{3} (2 x + 1) d x$ . Board questions still require the integral, but the geometric check is a quick safety net. A triangle with given vertices is found by writing the equation of each side and integrating piece by piece over the x-ranges.

Modulus graphs. $y = ∣ x - k ∣$ is a V-shape with its vertex at $(k, 0)$ . To integrate, split at $x = k$ : $∣ x - k ∣ = k - x$ to the left, $x - k$ to the right. For regions like $∣ x ∣ + ∣ y ∣ \leq 2$ , use symmetry: find the area in the first quadrant and multiply by 4.

Always sketch. The marking scheme gives a mark for a neat sketch showing the curve, the bounding lines and the shaded region. Mark intercepts and the vertex of any modulus graph.

The 2026-27 syllabus covers the area under simple curves: lines, circles, parabolas and ellipses in standard form. This topic covers lines and modulus graphs (NCERT Section 8.2). Questions on $y = ∣ x + k ∣$ appeared in the 2025 board paper and the 2026 sample paper.

Area Under Lines and Modulus Graphs: formulas to remember

Area with vertical strips: $A = \int_{a}^{b} ∣ f (x) ∣ d x$; Equals $\int_{a}^{b} f (x) d x$ when $f \geq 0$ on $[a, b]$ .
Area with horizontal strips: $A = \int_{c}^{d} ∣ g (y) ∣ d y$; Region bounded by $x = g (y)$ , the y-axis, $y = c$ , $y = d$ .
Curve crossing the axis: $A = \int_{a}^{c} f d x + \int_{c}^{b} f d x, f (c) = 0$; Split at every zero inside $[a, b]$ .
Modulus split: $∣ x - k ∣ = {k - x, x - k, x < k x \geq k$; Split the integral at $x = k$ .
Line through two points: $y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1})$; Needed for triangles and quadrilaterals.
Symmetry: $Area of ∣ x ∣ + ∣ y ∣ \leq a = 4 \int_{0}^{a} (a - x) d x = 2 a^{2}$; Use symmetry about the axes.

How to solve Area Under Lines and Modulus Graphs questions

A sketch-first routine for every area question with lines or modulus graphs.

1Sketch the curve, the axes and the bounding lines; shade the required region and mark intercepts or vertices.
2Decide on vertical strips (integrate in $x$ ) or horizontal strips (integrate in $y$ ), whichever gives a single formula.
3Find the points where the curve meets the x-axis (or where a modulus changes form) inside the interval, and split the integral there.
4Write each piece with the correct formula (for example $2 - x$ or $x - 2$ ) and integrate.
5Add the absolute values of the pieces; give units of area (square units) and, if possible, check with a simple geometric formula.

Area Under Lines and Modulus Graphs 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 under a straight line

Suppose a builder makes a concrete wedge-shaped ramp. Its side profile is the region under $y = \frac{x}{4}$ (metres) from $x = 0$ to $x = 8$ , and the ramp is 1.5 m wide.

Find the area of the side profile using integration and the volume of concrete needed.

Try it first, then show the solutionSolution

Area $= \int_{0}^{8} \frac{x}{4} d x = [\frac{x ^{2}}{8}]_{0}^{8} = 8$ m².
Check: it is a triangle with base 8 and height 2: $\frac{1}{2} \times 8 \times 2 = 8$ .
Volume $= 8 \times 1.5 = 12$ m³.

Answer: Side area 8 m²; concrete 12 m³.

Memory hook: Under a straight line, the integral is just a triangle or a trapezium.

Area under a speed-time line is distance

Imagine a scooter whose speed increases steadily, $v (t) = 2 t + 4$ m/s, for $0 \leq t \leq 5$ seconds.

Find the distance travelled in these 5 seconds as an area under the line.

Try it first, then show the solutionSolution

Distance $= \int_{0}^{5} (2 t + 4) d t = [t^{2} + 4 t]_{0}^{5}$ .
$= 25 + 20 = 45$ m.
Check: trapezium with parallel sides $v (0) = 4$ and $v (5) = 14$ and width 5: $\frac{4 + 14}{2} \times 5 = 45$ .

Answer: 45 m

Memory hook: Area under a speed-time graph is the distance.

Tricks to remember Area Under Lines and Modulus Graphs

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

Sketch, split, sum
Sketch the region, split where the curve crosses the axis or the modulus bends, then sum absolute values.
Use it when: Every area question.
Area is never negative
A negative integral means the region is below the axis; flip its sign before adding.
Use it when: Curves crossing the x-axis.
The V of a modulus
$y = ∣ x - k ∣$ is a V with its tip at $x = k$ . Split there.
Use it when: Integrals of modulus functions.
Check with geometry
Under straight lines, the answer must match the triangle or trapezium formula.
Use it when: Catching arithmetic slips.

Area Under Lines and Modulus Graphs: solved examples

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

Example 1: Area under a line · 2 marks · Very short answerBoard level

Find the area of the region bounded by $y = 2 x + 1$ , the x-axis and the lines $x = 1$ and $x = 3$ .

Try it first, then show the solutionSolution

On $[1, 3]$ , $2 x + 1 > 0$ , so Area $= \int_{1}^{3} (2 x + 1) d x$ . (1 mark)
$= [x^{2} + x]_{1}^{3} = 12 - 2 = 10$ square units. (1 mark)

Answer: 10 square units

Example 2: A modulus graph · 3 marks · Short answerBoard level

Sketch $y = ∣ x - 2∣$ and find the area bounded by it, the x-axis and the lines $x = 0$ and $x = 5$ .

Try it first, then show the solutionSolution

Sketch: a V-shape with vertex $(2, 0)$ , meeting the y-axis at $(0, 2)$ ; at $x = 5$ , $y = 3$ . (½ mark)
$∣ x - 2∣ = 2 - x$ on $[0, 2]$ and $x - 2$ on $[2, 5]$ . (½ mark)
Area $= \int_{0}^{2} (2 - x) d x + \int_{2}^{5} (x - 2) d x = [2 x - \frac{x ^{2}}{2}]_{0}^{2} + [\frac{x ^{2}}{2} - 2 x]_{2}^{5}$ . (1 mark)
$= 2 + (\frac{5}{2} - (- 2)) = 2 + \frac{9}{2} = \frac{13}{2}$ square units. (1 mark)

Answer: $\frac{13}{2}$ square units

Example 3: A triangle by integration · 3 marks · Short answerHarder board level

Using integration, find the area of the triangle with vertices $A (1, 0)$ , $B (4, 0)$ and $C (3, 4)$ .

Try it first, then show the solutionSolution

Side AC: through $(1, 0)$ and $(3, 4)$ : $y = 2 (x - 1)$ . Side CB: through $(3, 4)$ and $(4, 0)$ : $y = - 4 (x - 4) = 4 (4 - x)$ . (1 mark)
Area $= \int_{1}^{3} 2 (x - 1) d x + \int_{3}^{4} 4 (4 - x) d x$ . (1 mark)
$= [(x - 1)^{2}]_{1}^{3} + [- 2 (4 - x)^{2}]_{3}^{4} = 4 + 2 = 6$ square units (check: $\frac{1}{2} \times 3 \times 4 = 6$ ). (1 mark)

Answer: 6 square units

Example 4: Horizontal strips · 3 marks · Short answerHarder board level

Find the area of the region bounded by the line $y = \frac{x}{2} + 1$ , the y-axis and the lines $y = 2$ and $y = 4$ .

Try it first, then show the solutionSolution

Write $x$ in terms of $y$ : $x = 2 (y - 1)$ , which is $\geq 0$ for $2 \leq y \leq 4$ . (1 mark)
Area $= \int_{2}^{4} 2 (y - 1) d y = [(y - 1)^{2}]_{2}^{4}$ . (1 mark)
$= 9 - 1 = 8$ square units. (1 mark)

Answer: 8 square units

Example 5: A region defined by moduli · 3 marks · Short answerHOTS / competency

Using integration, find the area of the region ${(x, y) : ∣ x ∣ + ∣ y ∣ \leq 2}$ .

Try it first, then show the solutionSolution

The region is symmetric about both axes; in the first quadrant it is bounded by $x + y = 2$ , i.e. $y = 2 - x$ , $0 \leq x \leq 2$ . (1 mark)
First-quadrant area $= \int_{0}^{2} (2 - x) d x = [2 x - \frac{x ^{2}}{2}]_{0}^{2} = 2$ . (1 mark)
Total area $= 4 \times 2 = 8$ square units (a square with diagonal 4). (1 mark)

Answer: 8 square units

Example 6: Case study: a triangular roof panel · 4 marks · Case-basedHOTS / competency

Suppose the front panel of a roof is the region above the x-axis and below $y = 3 - ∣ x ∣$ (in metres).
(i) Where does the roof line meet the x-axis? (1 mark)
(ii) Using integration, find the area of the panel. (1 mark)
(iii) The panel is painted on both sides at Rs 500 per m². Find the cost. (1 mark)
(iv) Find the area of the part of the panel with $x \geq 1$ . (1 mark)

Try it first, then show the solutionSolution

(i) $3 - ∣ x ∣ = 0 \Rightarrow x = \pm 3$ : at $(- 3, 0)$ and $(3, 0)$ . (1 mark)
(ii) Area $= \int_{- 3}^{3} (3 - ∣ x ∣) d x = 2 \int_{0}^{3} (3 - x) d x = 2 \times \frac{9}{2} = 9$ m² (the integrand is even). (1 mark)
(iii) Cost $= 2 \times 9 \times 500 =$ Rs 9000. (1 mark)
(iv) $\int_{1}^{3} (3 - x) d x = [3 x - \frac{x ^{2}}{2}]_{1}^{3} = \frac{9}{2} - \frac{5}{2} = 2$ m². (1 mark)

Answer: (i) $x = \pm 3$ (ii) 9 m² (iii) Rs 9000 (iv) 2 m²

Where students lose marks in Area Under Lines and Modulus Graphs

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

Mistake: Integrating straight across a sign change
Areas below and above the axis cancel in a plain integral.
Wrong working
Area $= \int_{0}^{4} (x - 2) d x = 0$ .
Right working
Area $= 2 + 2 = 4$ .
Do this: Find the zeros first and split.
Usually costs: Most of the marks.
Mistake: No sketch
The marking scheme usually gives a mark for the figure, and the sketch shows where to split.
Wrong working
Integral written without any figure.
Right working
A labelled sketch with the vertex, intercepts and shaded region.
Do this: Spend one minute on a neat sketch.
Usually costs: 1 mark.
Mistake: Wrong piece of the modulus
Mixing up $k - x$ and $x - k$ gives negative heights.
Wrong working
$\int_{0}^{2} ∣ x - 2∣ d x = \int_{0}^{2} (x - 2) d x = - 2$ .
Right working
$\int_{0}^{2} (2 - x) d x = 2$ .
Do this: Test one point in each piece: the expression must be non-negative.
Usually costs: 1 to 2 marks.
Mistake: Using x-limits with horizontal strips
When integrating in $y$ , the limits must be y-values.
Wrong working
$\int_{0}^{6} 2 (y - 1) d y$ using x-limits.
Right working
$\int_{2}^{4} 2 (y - 1) d y = 8$ .
Do this: Write 'strips parallel to the x-axis, y from c to d' before integrating.
Usually costs: The value.

Exam tips

Modulus-graph area questions appeared in the 2025 board paper and the 2026 sample paper as 3-mark questions; always sketch the V-shape.
State the type of strip you use and the limits in the variable of integration.
Write 'square units' in the final answer.
For regions symmetric about an axis, compute one part and multiply; say which symmetry you used.
Region-between-two-curves questions (NCERT 8.3) are not in the 2026-27 syllabus; standard questions use one curve with the axes and lines.

Practice questions on Area Under Lines and Modulus Graphs

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 $y = x$ , the x-axis and the lines $x = 0$ , $x = 4$ is:
(a) 4
(b) 8
(c) 16
(d) 2
Hint
$\int_{0}^{4} x d x$ .
Answer
(b) 8 square units.
Q2 · 2 marks · Very short answerBoard level
Find the area bounded by $y = x - 2$ , the x-axis and the lines $x = 0$ and $x = 4$ .
Hint
The line crosses the x-axis at $x = 2$ .
Answer
$\int_{0}^{2} (x - 2) d x + \int_{2}^{4} (x - 2) d x = ∣ - 2∣ + 2 = 4$ square units (the plain integral from 0 to 4 is 0).
Q3 · 3 marks · Short answerBoard level
Sketch $y = ∣ x + 2∣$ and find the area bounded by it, the x-axis and the lines $x = - 5$ and $x = 1$ .
Hint
Split at $x = - 2$ .
Answer
$\int_{- 5}^{- 2} - (x + 2) d x + \int_{- 2}^{1} (x + 2) d x = \frac{9}{2} + \frac{9}{2} = 9$ square units.
Q4 · 3 marks · Short answerHarder board level
Using integration, find the area of the triangle with vertices $(0, 0)$ , $(4, 0)$ and $(2, 3)$ .
Hint
The sides are $y = \frac{3 x}{2}$ and $y = \frac{3 ( 4 - x )}{2}$ .
Answer
$\int_{0}^{2} \frac{3 x}{2} d x + \int_{2}^{4} \frac{3 ( 4 - x )}{2} d x = 3 + 3 = 6$ square units.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The area bounded by $y = x - 1$ , the x-axis and the lines $x = 0$ and $x = 2$ is 0.
Reason (R): $\int_{0}^{2} (x - 1) d x = 0$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Part of the region is below the axis.
Answer
(d). The integral is 0 because the two triangles cancel, but the area is $\frac{1}{2} + \frac{1}{2} = 1$ .
Q6 · 2 marks · Very short answerHarder board level
Using horizontal strips, find the area bounded by $x = y + 2$ , the y-axis and the lines $y = 0$ and $y = 2$ .
Hint
Integrate $x = y + 2$ with respect to $y$ .
Answer
$\int_{0}^{2} (y + 2) d y = 2 + 4 = 6$ square units.
Q7 · 3 marks · Short answerHOTS / competency
Find the area of the region bounded by $y = 2 - ∣ x - 1∣$ and the x-axis.
Hint
The graph is an upside-down V with its top at $(1, 2)$ ; find where it meets the x-axis.
Answer
$2 - ∣ x - 1∣ = 0 \Rightarrow x = - 1$ or $3$ . Area $= \int_{- 1}^{1} (1 + x) d x + \int_{1}^{3} (3 - x) d x = 2 + 2 = 4$ square units.

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 under $y = f (x) \geq 0$ from $x = a$ to $x = b$ .Show answer
$\int_{a}^{b} f (x) d x$ .
2.What happens if f is negative on part of $[a, b]$ ?Show answer
Split at the zeros and add absolute values.
3.Where is the vertex of $y = ∣ x + 3∣$ ?Show answer
At $(- 3, 0)$ .
4.Area under $y = 4 - x$ from 0 to 4?Show answer
8 square units.
5.Area of $∣ x ∣ + ∣ y ∣ \leq 1$ ?Show answer
2 square units.
6.When are horizontal strips easier?Show answer
When the boundary is given as $x = g (y)$ or the region lies against the y-axis.

Area Under Lines and Modulus Graphs: frequently asked questions

Why do I need to split the integral for modulus functions?

Because |x − k| has two different formulas: k − x to the left of k and x − k to the right. An integral needs one formula on each piece, so split at x = k.

Can the area ever be negative?

No. An integral can be negative when the curve is below the x-axis, but area is the absolute value. Split at the zeros and add the absolute values.

Is the area of a triangle using integration in the syllabus?

The syllabus covers areas under lines and the standard curves. Finding a triangle's area by integrating its sides is an application of area under lines and is good practice.

Which NCERT section covers this?

Section 8.2 (Area under Simple Curves) of Chapter 8, Application of Integrals, in NCERT Mathematics Part II for Class 12.

Do I need to draw the graph?

Yes. A neat sketch with the region shaded is expected and often carries a mark. It also tells you where to split the integral.