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Integral Calculus: JEE Main Mathematics Question with Solution

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

Pick any test point inside , say . For the first curve, . For the second curve, . Since , the curve lies above on .

The area of the region enclosed by the parabola

\text{y} = 4\text{x} - \text{x}^2

and

3\text{y} = (\text{x} - 4)^2

is equal to

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\frac{32}{9}

B4C6D

\frac{14}{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 ·

\frac{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 $y = 4x - x^2$ .

B · 4

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

C · 6

None. This is the correct calculation of the enclosed area. Correctly find the intersection points $x = 1$ and $x = 4$ , then evaluate $\int_{1}^{4} \left( (4x - x^2) - \frac{1}{3}(x - 4)^2 \right) dx = 6$ .

D ·

\frac{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 $4x - x^2 = \frac{1}{3}(x - 4)^2$ to find the exact integration interval.

Reviewed route

Solution

StepWorking

01given

Two parabolic curves: $y = 4x - x^2$ (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 $(y_{\text{upper}} - y_{\text{lower}})\,dx$ between these limits.

04execute

Equating the equations of the curves: $4x - x^2 = \frac{(x - 4)^2}{3}$ $3(4x - x^2) = x^2 - 8x + 16$ $12x - 3x^2 = x^2 - 8x + 16$ $4x^2 - 20x + 16 = 0 \implies x^2 - 5x + 4 = 0 \implies (x - 1)(x - 4) = 0$ So, the limits of integration are $x = 1$ to $x = 4$ . In this interval $[1, 4]$ , $y = 4x - x^2 \ge \frac{(x-4)^2}{3}$ . $\text{Area} = \int_1^4 \left(4x - x^2 - \frac{(x - 4)^2}{3}\right) dx$ $\text{Area} = \left[ 2x^2 - \frac{x^3}{3} - \frac{(x - 4)^3}{9} \right]_1^4$ 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 $x = 1$ : $2(1) - \frac{1}{3} - \frac{(1 - 4)^3}{9} = 2 - \frac{1}{3} - \frac{-27}{9} = 2 - \frac{1}{3} + 3 = 5 - \frac{1}{3} = \frac{14}{3}$ $\text{Area} = \frac{32}{3} - \frac{14}{3} = \frac{18}{3} = 6$

05verify

Using Archimedes' formula for parabolic segment area: $\text{Area} = \frac{|a|}{6}(x_2 - x_1)^3$ . Here, the difference equation is $\frac{4}{3}(x^2 - 5x + 4) = 0$ , so $|a| = \frac{4}{3}$ . Thus, $\text{Area} = \frac{4/3}{6}(4 - 1)^3 = \frac{4}{18} \times 27 = \frac{2}{9} \times 27 = 6$ . Matches exactly.

06given

Curves $y_1 = -x^2 + 4x$ and $y_2 = \frac{1}{3}x^2 - \frac{8}{3}x + \frac{16}{3}$ .

07goal

Find bounded area using the direct quadratic difference formula $\text{Area} = \frac{|a|}{6}(\beta - \alpha)^3$ .

08approach

Form $y_1 - y_2 = -\frac{4}{3}x^2 + \frac{20}{3}x - \frac{16}{3} = -\frac{4}{3}(x^2 - 5x + 4) = -\frac{4}{3}(x - 1)(x - 4)$ . Identify $|a| = \frac{4}{3}$ , $\alpha = 1$ , $\beta = 4$ .

09execute

$\text{Area} = \frac{|a|}{6}(\beta - \alpha)^3 = \frac{4/3}{6}(4 - 1)^3 = \frac{4}{18} \times 27 = 6$

✓verify

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

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

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 $(1, 4)$ .

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: $\frac{32}{9}$ - 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 $y = 4x - x^2$ .
B: 4 - Incorrectly simplifying the quadratic difference $(4x - x^2) - \frac{1}{3}(x^2 - 8x + 16)$ leads to wrong coefficients in the integrand. Combine like terms carefully: $-x^2 - \frac{1}{3}x^2 = -\frac{4}{3}x^2$ , and $4x + \frac{8}{3}x = \frac{20}{3}x$ .
C · correct: 6 - None. This is the correct calculation of the enclosed area. Correctly find the intersection points $x = 1$ and $x = 4$ , then evaluate $\int_{1}^{4} \left( (4x - x^2) - \frac{1}{3}(x - 4)^2 \right) dx = 6$ .
D: $\frac{14}{3}$ - Using incorrect integration limits (such as $[0, 4]$ instead of the actual intersection bounds $[1, 4]$ ). Always solve for the intersection points by equating $4x - x^2 = \frac{1}{3}(x - 4)^2$ to find the exact integration interval.

Step-by-step solution

given: Two parabolic curves: $y = 4x - x^2$ (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 $(y_{\text{upper}} - y_{\text{lower}})\,dx$ between these limits.
execute: Equating the equations of the curves: $4x - x^2 = \frac{(x - 4)^2}{3}$ $3(4x - x^2) = x^2 - 8x + 16$ $12x - 3x^2 = x^2 - 8x + 16$ $4x^2 - 20x + 16 = 0 \implies x^2 - 5x + 4 = 0 \implies (x - 1)(x - 4) = 0$ So, the limits of integration are $x = 1$ to $x = 4$ . In this interval $[1, 4]$ , $y = 4x - x^2 \ge \frac{(x-4)^2}{3}$ . $\text{Area} = \int_1^4 \left(4x - x^2 - \frac{(x - 4)^2}{3}\right) dx$ $\text{Area} = \left[ 2x^2 - \frac{x^3}{3} - \frac{(x - 4)^3}{9} \right]_1^4$ 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 $x = 1$ : $2(1) - \frac{1}{3} - \frac{(1 - 4)^3}{9} = 2 - \frac{1}{3} - \frac{-27}{9} = 2 - \frac{1}{3} + 3 = 5 - \frac{1}{3} = \frac{14}{3}$ $\text{Area} = \frac{32}{3} - \frac{14}{3} = \frac{18}{3} = 6$
verify: Using Archimedes' formula for parabolic segment area: $\text{Area} = \frac{|a|}{6}(x_2 - x_1)^3$ . Here, the difference equation is $\frac{4}{3}(x^2 - 5x + 4) = 0$ , so $|a| = \frac{4}{3}$ . Thus, $\text{Area} = \frac{32}{3} - \frac{14}{3} = \frac{18}{3} = 6$ . Matches exactly.
given: Curves $y_1 = -x^2 + 4x$ 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 $\text{Area} = \frac{|a|}{6}(\beta - \alpha)^3$ .
approach: Form $y_1 - y_2 = -\frac{4}{3}x^2 + \frac{20}{3}x - \frac{16}{3} = -\frac{4}{3}(x^2 - 5x + 4) = -\frac{4}{3}(x - 1)(x - 4)$ . Identify $|a| = \frac{4}{3}$ , $\alpha = 1$ , $\beta = 4$ .
execute: $\text{Area} = \frac{|a|}{6}(\beta - \alpha)^3 = \frac{4/3}{6}(4 - 1)^3 = \frac{4}{18} \times 27 = 6$
verify: Result 6 is positive and physically sensible for two intersecting parabolas spanning width 3 and height order $\sim 2.5$ .

Integral Calculus: JEE Main Mathematics Question with Solution

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=1x = 1x=1 and x=4x = 4x=4.

01given

02goal

03approach

04execute

05verify

06given

07goal

08approach

09execute

✓verify

Students also ask

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 $x = 4$ .