MathsCBSE Class 12 · Application of Integrals

Area Involving a Parabola: y² = 4ax and x² = 4ay by Integration (Class 12)

Updated 15 October 2026 · Fact-checked

For the parabola y² = 4ax, the upper arm is y = 2√(ax), so the area between it, the x-axis and x = c is ∫₀ᶜ 2√(ax) dx; the whole region cut off by the line x = c is twice that. For x² = 4ay use horizontal strips, x = 2√(ay), or vertical strips with y = x²/(4a).

Understand Area Involving a Parabola

The standard parabolas in the syllabus are $y^{2} = 4 a x$ (opening right), $y^{2} = - 4 a x$ (left), $x^{2} = 4 a y$ (up) and $x^{2} = - 4 a y$ (down), all with vertex at the origin. Each is symmetric about its axis, which halves the work.

**Parabola opening right, $y^{2} = 4 a x$ .** It is symmetric about the x-axis; the upper arm is $y = 2 a x$ . The region bounded by the parabola and the line $x = c$ (the 'mouth' of the parabola cut off by a vertical chord) has area
$A = 2 \int_{0}^{c} 2 a x d x = 4 a \cdot \frac{2}{3} c^{3/2} = \frac{8}{3} a c^{3/2}$ .
Only the first-quadrant part is $\int_{0}^{c} 2 a x d x$ . Read the question carefully: 'in the first quadrant' halves the area.

**Parabola opening up, $x^{2} = 4 a y$ .** It is symmetric about the y-axis. Two natural set-ups:

the region under the curve $y = \frac{x ^{2}}{4 a}$ from $x = p$ to $x = q$ : $\int_{p}^{q} \frac{x ^{2}}{4 a} d x$ (vertical strips);
the region between the curve and a horizontal line $y = c$ : horizontal strips of length $2 x = 4 a y$ , so $A = \int_{0}^{c} 2 4 a y d y$ (or twice the first-quadrant part).

Choosing strips. Use the variable that gives a single formula across the region. A region against the y-axis (bounded by $y = c$ and $y = d$ ) usually needs horizontal strips. Integrals of $x$ or $y$ use the power rule: $\int x d x = \frac{2}{3} x^{3/2}$ .

Archimedes' check. The region between a parabola and a chord perpendicular to its axis is $\frac{2}{3}$ of the enclosing rectangle. For $y^{2} = 4 x$ cut by $x = 4$ : rectangle $4 \times 8 = 32$ , region $\frac{2}{3} \times 32 = \frac{64}{3}$ . A fast way to check answers.

Exam note: a parabola cut by a slanting line. The separate NCERT section on a curve and a line was dropped, but CBSE sample papers still set such regions (for example a parabola and a line through the vertex). Solve the two equations together for the intersection points, then integrate (upper $-$ lower) between them. Keep to one standard parabola and one straight line.

Areas with parabolas appear as MCQs and as parts of 3- to 5-mark questions (NCERT Section 8.2).

Area Involving a Parabola: formulas to remember

Upper arm of y² = 4ax: $y = 2 a x$; Symmetric about the x-axis.
Region cut off by x = c: $A = 2 \int_{0}^{c} 2 a x d x = \frac{8}{3} a c^{3/2}$; Half of this lies in the first quadrant.
Parabola opening up: $x^{2} = 4 a y \Rightarrow y = \frac{x ^{2}}{4 a}, x = \pm 2 a y$; Use horizontal strips for regions bounded by $y = c$ .
Power rule for roots: $\int x d x = \frac{2}{3} x^{3/2} + C$; Most parabola areas reduce to this.
Archimedes' two-thirds: $Parabolic segment = \frac{2}{3} \times enclosing rectangle$; For a chord perpendicular to the axis.
Curve and line (exam note): $A = \int_{x_{1}}^{x_{2}} (y_{upper} - y_{lower}) d x$; $x_{1}, x_{2}$ from solving the two equations together.

How to solve Area Involving a Parabola questions

Use this routine for any area question with a standard parabola.

1Identify the parabola's direction and axis of symmetry; sketch it with any bounding lines and shade the region.
2Find intersection points of the parabola with the given lines (solve simultaneously).
3Choose vertical or horizontal strips so that each strip runs between the same two boundaries.
4Use symmetry (state it) to integrate over one half and double if appropriate.
5Integrate with the power rule, substitute limits, and check with Archimedes' $\frac{2}{3}$ rule where it applies.
6Write the answer in square units.

Area Involving a Parabola 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 root curve

Suppose a garden bed along a wall has its curved edge along $y = x$ (in metres), measured from the corner, and the bed runs from $x = 0$ to $x = 9$ along the wall (the x-axis).

Find the area of the bed using integration.

Try it first, then show the solutionSolution

Area $= \int_{0}^{9} x d x = [\frac{2}{3} x^{3/2}]_{0}^{9}$ .
$= \frac{2}{3} \times 27 = 18$ m².
Check: the enclosing rectangle is $9 \times 3 = 27$ and the bed is $\frac{2}{3}$ of it.

Answer: 18 m²

Memory hook: Under a half-parabola: two-thirds of the rectangle.

Area under an upward parabola

Imagine a skateboard ramp whose side profile is the region under $y = \frac{x ^{2}}{16}$ (in metres) from $x = 0$ to $x = 8$ .

Find the area of the side panel needed to cover the ramp's side.

Try it first, then show the solutionSolution

Area $= \int_{0}^{8} \frac{x ^{2}}{16} d x = [\frac{x ^{3}}{48}]_{0}^{8}$ .
$= \frac{512}{48} = \frac{32}{3} \approx 10.67$ m².
Check: the rectangle is $8 \times 4 = 32$ m², and the region under the curve is $\frac{1}{3}$ of it.

Answer: $\frac{32}{3} \approx 10.7$ m²

Memory hook: Under $y = k x^{2}$ from 0 is one-third of the rectangle; above it is two-thirds.

Tricks to remember Area Involving a Parabola

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

Two-thirds, one-third
A parabola from its vertex splits its rectangle into $\frac{2}{3}$ (the inside of the bowl) and $\frac{1}{3}$ (between curve and axis).
Use it when: Quick checks of every parabola area.
Double for symmetry
$y^{2} = 4 a x$ has arms above and below the x-axis: double the first-quadrant integral unless the question says 'first quadrant'.
Use it when: Regions cut off by $x = c$ .
Flat lines, flat strips
If the region is bounded by horizontal lines $y = c$ , $y = d$ , use horizontal strips.
Use it when: Parabolas like $x^{2} = 4 a y$ .
Meet first, integrate second
For a parabola and a line, solve them together before writing any integral.
Use it when: Exam-note regions.

Area Involving a Parabola: solved examples

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

Example 1: Area under y = x² · 2 marks · Very short answerBoard level

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

Try it first, then show the solutionSolution

Area $= \int_{1}^{3} x^{2} d x$ . (1 mark)
$= [\frac{x ^{3}}{3}]_{1}^{3} = 9 - \frac{1}{3} = \frac{26}{3}$ square units. (1 mark)

Answer: $\frac{26}{3}$ square units

Example 2: Parabola cut by a vertical line · 3 marks · Short answerBoard level

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

Try it first, then show the solutionSolution

Sketch: the parabola opens right; $x = 4$ meets it at $(4, \pm 4)$ . The region is symmetric about the x-axis. (1 mark)
Area $= 2 \int_{0}^{4} 2 x d x = 4 [\frac{2}{3} x^{3/2}]_{0}^{4}$ . (1 mark)
$= 4 \times \frac{2}{3} \times 8 = \frac{64}{3}$ square units (check: $\frac{2}{3} \times 4 \times 8$ ). (1 mark)

Answer: $\frac{64}{3}$ square units

Example 3: Horizontal strips for an upward parabola · 3 marks · Short answerHarder board level

Find the area of the region in the first quadrant bounded by $x^{2} = 4 y$ , the y-axis and the lines $y = 1$ and $y = 4$ .

Try it first, then show the solutionSolution

In the first quadrant, $x = 2 y$ ; the region runs from the y-axis to the curve for $1 \leq y \leq 4$ . (1 mark)
Area $= \int_{1}^{4} 2 y d y = [\frac{4}{3} y^{3/2}]_{1}^{4}$ . (1 mark)
$= \frac{4}{3} (8 - 1) = \frac{28}{3}$ square units. (1 mark)

Answer: $\frac{28}{3}$ square units

Example 4: Exam note: a parabola and a line through its vertex · 5 marks · Long answerHarder board level

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

Try it first, then show the solutionSolution

Intersections: $x^{2} = 4 x \Rightarrow x = 0$ or $4$ ; points $(0, 0)$ and $(4, 4)$ . (1 mark)
Sketch: between $x = 0$ and $x = 4$ the parabola's upper arm $y = 2 x$ lies above the line $y = x$ . (1 mark)
Area $= \int_{0}^{4} (2 x - x) d x$ . (1 mark)
$= [\frac{4}{3} x^{3/2} - \frac{x ^{2}}{2}]_{0}^{4} = \frac{32}{3} - 8$ . (1 mark)
$= \frac{8}{3}$ square units. (1 mark)

Answer: $\frac{8}{3}$ square units

Example 5: Exam note: an upward parabola cut by a slanting line · 5 marks · Long answerHOTS / competency

Find the area of the region bounded by $y = x^{2}$ and the line $y = 2 x + 3$ .

Try it first, then show the solutionSolution

Intersections: $x^{2} = 2 x + 3 \Rightarrow x^{2} - 2 x - 3 = 0 \Rightarrow x = - 1$ or $3$ . (1 mark)
Between them the line is above the parabola (at $x = 0$ : $3 > 0$ ). (1 mark)
Area $= \int_{- 1}^{3} (2 x + 3 - x^{2}) d x = [x^{2} + 3 x - \frac{x ^{3}}{3}]_{- 1}^{3}$ . (1 mark)
$= (9 + 9 - 9) - (1 - 3 + \frac{1}{3}) = 9 + \frac{5}{3}$ . (1 mark)
$= \frac{32}{3}$ square units. (1 mark)

Answer: $\frac{32}{3}$ square units

Example 6: Case study: a parabolic arch · 4 marks · Case-basedHOTS / competency

Suppose the opening of a garden arch is the region between the road (x-axis) and the curve $y = 4 - \frac{x ^{2}}{4}$ , in metres.
(i) Where does the arch meet the road? (1 mark)
(ii) Find the area of the opening. (2 marks)
(iii) What fraction of the enclosing rectangle does the opening fill? (1 mark)

Try it first, then show the solutionSolution

(i) $4 - \frac{x ^{2}}{4} = 0 \Rightarrow x = \pm 4$ : at $(\pm 4, 0)$ . (1 mark)
(ii) Area $= \int_{- 4}^{4} (4 - \frac{x ^{2}}{4}) d x = 2 [4 x - \frac{x ^{3}}{12}]_{0}^{4}$ (even function). (1 mark)
$= 2 (16 - \frac{16}{3}) = \frac{64}{3} \approx 21.3$ m². (1 mark)
(iii) Rectangle $8 \times 4 = 32$ m²; fraction $= \frac{64/3}{32} = \frac{2}{3}$ . (1 mark)

Answer: (i) $x = \pm 4$ (ii) $\frac{64}{3}$ m² (iii) $\frac{2}{3}$

Where students lose marks in Area Involving a Parabola

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

Mistake: Using only one arm of y² = 4ax
The region cut off by $x = c$ includes the part below the x-axis.
Wrong working
Area bounded by $y^{2} = 4 x$ and $x = 4$ : $\int_{0}^{4} 2 x d x = \frac{32}{3}$ .
Right working
$2 \int_{0}^{4} 2 x d x = \frac{64}{3}$ .
Do this: Check the sketch: if the region straddles the axis, double.
Usually costs: Half the value.
Mistake: Integrating √x wrongly
$\int x^{1/2} d x = \frac{x ^{3/2}}{3/2} = \frac{2}{3} x^{3/2}$ .
Wrong working
$\int_{0}^{4} 2 x d x = [x^{3/2}]_{0}^{4} = 8$ .
Right working
$= [\frac{4}{3} x^{3/2}]_{0}^{4} = \frac{32}{3}$ .
Do this: Add 1 to the power and divide by the new power.
Usually costs: The value.
Mistake: Wrong order of upper and lower curves
Subtracting the upper curve from the lower one gives a negative result.
Wrong working
$\int_{0}^{4} (x - 2 x) d x = - \frac{8}{3}$ , written as the area.
Right working
$\int_{0}^{4} (2 x - x) d x = \frac{8}{3}$ .
Do this: Test a point between the intersections to see which is on top.
Usually costs: Half to one mark.
Mistake: Skipping the intersection step
Limits for a parabola and a line come only from solving them together.
Wrong working
Area between $y^{2} = 4 x$ and $y = x$ taken from 0 to 1.
Right working
Solve: $x = 0$ or $4$ ; limits 0 to 4.
Do this: Always show the intersection working.
Usually costs: Most of the marks.

Exam tips

Parabola areas appeared as 1-mark MCQs (2024 board paper, 2025 sample paper) and in curve-and-line regions in the 2026 and 2027 sample papers; sketch every time.
State the symmetry you use, for example 'the region is symmetric about the x-axis'.
Check answers with Archimedes' $\frac{2}{3}$ rule when the chord is perpendicular to the axis.
For curve-and-line regions, the intersection points earn a mark on their own.
Keep answers as fractions such as $\frac{64}{3}$ unless decimals are asked.

Practice questions on Area Involving a Parabola

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 $y = x^{2}$ , the x-axis and the lines $x = 0$ , $x = 3$ is:
(a) 3
(b) 9
(c) 27
(d) 6
Hint
$\int_{0}^{3} x^{2} d x$ .
Answer
(b) 9 square units.
Q2 · 2 marks · Very short answerBoard level
Find the area of the region in the first quadrant bounded by $y^{2} = 9 x$ , the x-axis and the line $x = 4$ .
Hint
Upper arm $y = 3 x$ .
Answer
$\int_{0}^{4} 3 x d x = 3 \cdot \frac{2}{3} \cdot 8 = 16$ square units.
Q3 · 3 marks · Short answerBoard level
Find the area of the region bounded by $x^{2} = y$ and the line $y = 4$ .
Hint
Horizontal strips of length $2 y$ .
Answer
$\int_{0}^{4} 2 y d y = \frac{4}{3} \cdot 8 = \frac{32}{3}$ square units.
Q4 · 3 marks · Short answerHarder board level
Find the area enclosed between the curve $y = x^{2} - 4$ and the x-axis.
Hint
The curve is below the axis between its x-intercepts.
Answer
Intercepts $x = \pm 2$ . $\int_{- 2}^{2} (x^{2} - 4) d x = \frac{16}{3} - 16 = - \frac{32}{3}$ , so the area is $\frac{32}{3}$ square units.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The area of the region bounded by $y^{2} = 4 x$ and the line $x = 1$ is $\frac{8}{3}$ square units.
Reason (R): The area equals $2 \int_{0}^{1} 2 x d x$ .
(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
$\int_{0}^{1} 2 x d x = \frac{4}{3}$ .
Answer
(a). $2 \times \frac{4}{3} = \frac{8}{3}$ ; R is the correct set-up for A.
Q6 · 5 marks · Long answerHarder board level
Find the area of the region bounded by $y = x^{2}$ and the line $y = x + 2$ .
Hint
They meet at $x = - 1$ and $x = 2$ .
Answer
$\int_{- 1}^{2} (x + 2 - x^{2}) d x = [\frac{x ^{2}}{2} + 2 x - \frac{x ^{3}}{3}]_{- 1}^{2} = (2 + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3}) = \frac{10}{3} + \frac{7}{6} = \frac{9}{2}$ square units.
Q7 · 4 marks · Short answerHOTS / competency
The square with vertices $(0, 0)$ , $(4, 0)$ , $(4, 4)$ , $(0, 4)$ is divided by the parabola $y^{2} = 4 x$ into two parts. Find the area of each part and verify that they add up to the area of the square.
Hint
One part is between the parabola and the y-axis (horizontal strips); the other is below the parabola.
Answer
Part between the parabola and the y-axis: $\int_{0}^{4} \frac{y ^{2}}{4} d y = \frac{16}{3}$ . Part below the parabola: $\int_{0}^{4} 2 x d x = \frac{32}{3}$ . Sum $= 16$ , the area of the square.

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.Upper arm of $y^{2} = 4 a x$ ?Show answer
$y = 2 a x$ .
2.Area bounded by $y^{2} = 4 a x$ and $x = a$ ?Show answer
$\frac{8}{3} a^{2}$ .
3.What is $\int x d x$ ?Show answer
$\frac{2}{3} x^{3/2} + C$ .
4.Area under $y = x^{2}$ from 0 to a?Show answer
$\frac{a ^{3}}{3}$ .
5.Which strips suit $x^{2} = 4 a y$ with lines $y = c$ ?Show answer
Horizontal strips.
6.Area between $y^{2} = 4 x$ and $y = x$ ?Show answer
$\frac{8}{3}$ square units.

Area Involving a Parabola: frequently asked questions

How do I find the area bounded by a parabola and a vertical line?

Write the upper arm of the parabola as a function, for example y = 2√(ax) for y² = 4ax, integrate it from the vertex to the line, and double it if the region is symmetric about the x-axis.

Is the area between a parabola and a slanting line in the syllabus?

NCERT dropped the separate section on regions between a curve and a line, but CBSE sample papers still include simple cases with a standard parabola and one line. Find the intersection points and integrate upper minus lower.

When should I use horizontal strips?

When the region is bounded by horizontal lines such as y = 1 and y = 4, or lies against the y-axis. For x² = 4ay, horizontal strips have length 2√(4ay) across the whole parabola.

Which NCERT section covers parabola areas?

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

Are areas between two parabolas asked?

No. Areas between two curves (NCERT 8.3) are not in the 2026-27 syllabus.