JEE MainMathematics

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Area Enclosed by Two Intersecting Parabolas

Q: How do we know which curve is on top in the interval [1, 4]?

Pick any test point inside (1, 4), say x = 2. For the first curve, y = 4(2) - 2^2 = 4. For the second curve, y = (2 - 4)^2 / 3 = 4/3. Since 4 > 4/3, the curve y = 4x - x^2 lies above 3y = (x - 4)^2 on (1, 4).

The area of the region enclosed by the parabola y = 4x - x^2 and 3y = (x - 4)^2 is equal to

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A32/9B4C6D14/3

Step-by-step solutionView

Correct answer

The area enclosed by the two parabolas is 6 square units, found by integrating the difference between the upper and lower curves between their points of intersection at $x = 1$ and $x = 4$ .

Option analysis

Why each option works or fails

A · 32/9

Integrating with an incorrect denominator or missing the factor of $3$ on the quadratic coefficient leads to an incorrect fractional area. Ensure you correctly express $y = \frac{1}{3} (x - 4)^{2}$ before subtracting it from the upper curve $y = 4 x - x^{2}$ .

B · 4

Incorrectly simplifying the quadratic difference $(4 x - x^{2}) - \frac{1}{3} (x^{2} - 8 x + 16)$ leads to wrong coefficients in the integrand. Combine like terms carefully: $- x^{2} - \frac{1}{3} x^{2} = - \frac{4}{3} x^{2}$ , and $4 x + \frac{8}{3} x = \frac{20}{3} x$ .

C · 6

This is the correct calculation of the enclosed area. Correctly find the intersection points $x = 1$ and $x = 4$ , then evaluate $\int_{1}^{4} ((4 x - x^{2}) - \frac{1}{3} (x - 4)^{2}) d x = 6$ .

D · 14/3

Using incorrect integration limits (such as $[0, 4]$ instead of the actual intersection bounds $[1, 4]$ ). Always solve for the intersection points by equating $4 x - x^{2} = \frac{1}{3} (x - 4)^{2}$ to find the exact integration interval.

Reviewed route

Solution

StepWorking

01Given

Two parabolic curves: $y = 4 x - x^{2}$ (downward-opening parabola) and $y = \frac{(x - 4)^{2}}{3}$ (upward-opening parabola).

02Goal

Find the area of the bounded region between the two parabolas.

03Approach

Find the x-coordinates of the intersection points by equating $y_{1} = y_{2}$ . Then integrate $(y_{upper} - y_{lower}) d x$ between these limits.

04Execute

Equating the equations of the curves: $4 x - x^{2} = \frac{(x - 4)^{2}}{3}$ $3 (4 x - x^{2}) = x^{2} - 8 x + 16$ $12 x - 3 x^{2} = x^{2} - 8 x + 16$ $4 x^{2} - 20 x + 16 = 0 ⟹ x^{2} - 5 x + 4 = 0 ⟹ (x - 1) (x - 4) = 0$ So, the limits of integration are $x = 1$ to $x = 4$ . In this interval $[1, 4]$ , $y = 4 x - x^{2} \geq \frac{(x - 4)^{2}}{3}$ . $Area = \int_{1}^{4} (4 x - x^{2} - \frac{(x - 4)^{2}}{3}) d x$ $Area = {[2 x^{2} - \frac{x^{3}}{3} - \frac{(x - 4)^{3}}{9}]}_{1}^{4}$ Evaluating at the upper limit $x = 4$ : $2 (16) - \frac{64}{3} - 0 = 32 - \frac{64}{3} = \frac{32}{3}$ Evaluating at the lower limit $x = 1$ : $2 (1) - \frac{1}{3} - \frac{(1 - 4)^{3}}{9} = 2 - \frac{1}{3} - \frac{- 27}{9} = 2 - \frac{1}{3} + 3 = 5 - \frac{1}{3} = \frac{14}{3}$ $Area = \frac{32}{3} - \frac{14}{3} = \frac{18}{3} = 6$

✓Verify

Using Archimedes' formula for parabolic segment area: $Area = \frac{∣ a ∣}{6} (x_{2} - x_{1})^{3}$ . Here, the difference equation is $\frac{4}{3} (x^{2} - 5 x + 4) = 0$ , so $∣ a ∣ = \frac{4}{3}$ . Thus, $Area = \frac{4 / 3}{6} (4 - 1)^{3} = \frac{4}{18} \times 27 = \frac{2}{9} \times 27 = 6$ . Matches exactly.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

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How do we know which curve is on top in the interval

[1, 4]

Pick any test point inside $(1, 4)$ , say $x = 2$ . For the first curve, $y = 4 (2) - 2^{2} = 4$ . For the second curve, $y = (2 - 4)^{2} / 3 = 4 / 3$ . Since $4 > 4 / 3$ , the curve $y = 4 x - x^{2}$ lies above $3 y = (x - 4)^{2}$ on $(1, 4)$ .

Answer

The area enclosed by the two parabolas is 6 square units, found by integrating the difference between the upper and lower curves between their points of intersection at $x = 1$ and $x = 4$ .

Why each option works or fails

A: 32/9 - Integrating with an incorrect denominator or missing the factor of $3$ on the quadratic coefficient leads to an incorrect fractional area. Ensure you correctly express $y = \frac{1}{3} (x - 4)^{2}$ before subtracting it from the upper curve $y = 4 x - x^{2}$ .
B: 4 - Incorrectly simplifying the quadratic difference $(4 x - x^{2}) - \frac{1}{3} (x^{2} - 8 x + 16)$ leads to wrong coefficients in the integrand. Combine like terms carefully: $- x^{2} - \frac{1}{3} x^{2} = - \frac{4}{3} x^{2}$ , and $4 x + \frac{8}{3} x = \frac{20}{3} x$ .
C · correct: 6 - This is the correct calculation of the enclosed area. Correctly find the intersection points $x = 1$ and $x = 4$ , then evaluate $\int_{1}^{4} ((4 x - x^{2}) - \frac{1}{3} (x - 4)^{2}) d x = 6$ .
D: 14/3 - Using incorrect integration limits (such as $[0, 4]$ instead of the actual intersection bounds $[1, 4]$ ). Always solve for the intersection points by equating $4 x - x^{2} = \frac{1}{3} (x - 4)^{2}$ to find the exact integration interval.

Step-by-step solution

Given: Two parabolic curves: $y = 4 x - x^{2}$ (downward-opening parabola) and $y = \frac{(x - 4)^{2}}{3}$ (upward-opening parabola).
Goal: Find the area of the bounded region between the two parabolas.
Approach: Find the x-coordinates of the intersection points by equating $y_{1} = y_{2}$ . Then integrate $(y_{upper} - y_{lower}) d x$ between these limits.
Execute: Equating the equations of the curves: $4 x - x^{2} = \frac{(x - 4)^{2}}{3}$ $3 (4 x - x^{2}) = x^{2} - 8 x + 16$ $12 x - 3 x^{2} = x^{2} - 8 x + 16$ $4 x^{2} - 20 x + 16 = 0 ⟹ x^{2} - 5 x + 4 = 0 ⟹ (x - 1) (x - 4) = 0$ So, the limits of integration are $x = 1$ to $x = 4$ . In this interval $[1, 4]$ , $y = 4 x - x^{2} \geq \frac{(x - 4)^{2}}{3}$ . $Area = \int_{1}^{4} (4 x - x^{2} - \frac{(x - 4)^{2}}{3}) d x$ $Area = {[2 x^{2} - \frac{x^{3}}{3} - \frac{(x - 4)^{3}}{9}]}_{1}^{4}$ Evaluating at the upper limit $x = 4$ : $2 (16) - \frac{64}{3} - 0 = 32 - \frac{64}{3} = \frac{32}{3}$ Evaluating at the lower limit $x = 1$ : $2 (1) - \frac{1}{3} - \frac{(1 - 4)^{3}}{9} = 2 - \frac{1}{3} - \frac{- 27}{9} = 2 - \frac{1}{3} + 3 = 5 - \frac{1}{3} = \frac{14}{3}$ $Area = \frac{32}{3} - \frac{14}{3} = \frac{18}{3} = 6$
Verify: Using Archimedes' formula for parabolic segment area: $Area = \frac{∣ a ∣}{6} (x_{2} - x_{1})^{3}$ . Here, the difference equation is $\frac{4}{3} (x^{2} - 5 x + 4) = 0$ , so $∣ a ∣ = \frac{4}{3}$ . Thus, $Area = \frac{4 / 3}{6} (4 - 1)^{3} = \frac{4}{18} \times 27 = \frac{2}{9} \times 27 = 6$ . Matches exactly.

Shortcut:

When to use it: Whenever finding the area bounded between two intersecting parabolas of the form $y = a x^{2} + b x + c$ .

Given: Curves $y_{1} = - x^{2} + 4 x$ and $y_{2} = \frac{1}{3} x^{2} - \frac{8}{3} x + \frac{16}{3}$ .

Goal: Find bounded area using the direct quadratic difference formula $Area = \frac{∣ a ∣}{6} (β - α)^{3}$ .

Approach: Form $y_{1} - y_{2} = - \frac{4}{3} x^{2} + \frac{20}{3} x - \frac{16}{3} = - \frac{4}{3} (x^{2} - 5 x + 4) = - \frac{4}{3} (x - 1) (x - 4)$ . Identify $∣ a ∣ = \frac{4}{3}$ , $α = 1$ , $β = 4$ .

Execute: $Area = \frac{∣ a ∣}{6} (β - α)^{3} = \frac{4 / 3}{6} (4 - 1)^{3} = \frac{4}{18} \times 27 = 6$

Verify: Result 6 is positive and physically sensible for two intersecting parabolas spanning width 3 and height order $\sim 2.5$ .

Area Enclosed by Two Intersecting Parabolas

What feels right?

The area enclosed by the two parabolas is 6 square units, found by integrating the difference between the upper and lower curves between their points of intersection at $x = 1$ and $x = 4$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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