JEE MainMathematics

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Area of a bounded region defined by linear and parabolic inequalities

The area of the region {(x, y): 0 ≤ y ≤ 6 - x, y^2 ≥ 4x - 3, x ≥ 0} is:

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A8B9C12D15

Step-by-step solutionView

Correct answer

Integrating with respect to $y$ from $0$ to $3$ gives an area of $9$ .

Option analysis

Why each option works or fails

A · 8

Integrating with respect to $y$ correctly over $[0, 3]$ but miscalculating the evaluation of $[6 y - y^{2} / 2 - y^{3} / 12]_{0}^{3}$ as $8$ instead of $9$ . Carefully compute each term: $6 (3) - (3^{2}) / 2 - (3^{3}) / 12 = 18 - 4.5 - 2.25 = 11.25$ , and account for the remaining region or boundary evaluations precisely.

B · 9

This is the correct option. Set up the region with horizontal slices: for $y \in [0, 3]$ , $x$ ranges from $0$ to $(y^{2} + 3) / 4$ , and for $y \in [3, 6]$ , check the intersection with $x = 6 - y$ . The inequality $y^{2} \geq 4 x - 3$ means $x \leq (y^{2} + 3) / 4$ , which intersects $x = 6 - y$ at $y = 3$ . Evaluating $\int_{0}^{3} ((6 - y) - \frac{y^{2} + 3}{4}) d y$ or splitting appropriately yields an area of $9$ .

C · 12

Believing the upper limit of integration for the difference of functions is $y = 4$ or neglecting the parabolic subtraction over part of the interval. Solve the intersection equation $6 - y = \frac{y^{2} + 3}{4}$ to find the valid non-negative intersection point, which gives $y^{2} + 4 y - 21 = 0 ⟹ y = 3$ .

D · 15

Integrating only the linear function $6 - x$ from $x = 0$ to $x = 3$ or dropping the constraint $y^{2} \geq 4 x - 3$ . Include the parabolic boundary $x \leq \frac{y^{2} + 3}{4}$ ; omitting it overstates the area.

Reviewed route

Solution

StepWorking

01Given

The bounded region is defined by: 1. $0 \leq y \leq 6 - x ⟹ x \leq 6 - y$ and $y \geq 0$ 2. $y^{2} \geq 4 x - 3 ⟹ x \leq \frac{y^{2} + 3}{4}$ 3. $x \geq 0$

02Goal

Find the total area of the enclosed region in the first quadrant.

03Approach

Find the intersection points of $x = 6 - y$ and $x = \frac{y^{2} + 3}{4}$ . The bounds for $y$ run from $y = 0$ up to the intersection point. Integrate horizontal strips $Δ x = x_{right} - x_{left}$ with respect to $y$ . Here, the line is on the right and the parabola is on the left in the region of interest: $x_{right} = 6 - y$ and $x_{left} = \frac{y^{2} + 3}{4}$ .

04Execute

Set $6 - y = \frac{y^{2} + 3}{4} ⟹ 24 - 4 y = y^{2} + 3 ⟹ y^{2} + 4 y - 21 = 0 ⟹ (y + 7) (y - 3) = 0$ . Since $y \geq 0$ , the upper limit is $y = 3$ . The area is: $A = \int_{0}^{3} ((6 - y) - \frac{y^{2} + 3}{4}) d y = \int_{0}^{3} (\frac{21}{4} - y - \frac{y^{2}}{4}) d y$ $A = {[\frac{21}{4} y - \frac{y^{2}}{2} - \frac{y^{3}}{12}]}_{0}^{3} = \frac{63}{4} - \frac{9}{2} - \frac{27}{12} = \frac{63}{4} - \frac{18}{4} - \frac{9}{4} = \frac{36}{4} = 9$

✓Verify

Check at endpoints: for $y = 0$ , $x \in [3 / 4, 6]$ , width is $5.25$ . At $y = 3$ , $x = 3$ , width is $0$ . The average width is approximately $(5.25 + 0) / 2 = 2.625$ , multiplied by height $3$ gives roughly $7.875$ , close to $9$ (accounting for the concave parabolic left boundary which bulges out the area).

Hints that build this answer step by step

The region is bounded by $x \geq 0$ , $y \geq 0$ , $x + y \leq 6$ , and $x \leq \frac{y^{2} + 3}{4}$ . Where does the boundary curve $x = \frac{y^{2} + 3}{4}$ intersect the line $x = 6 - y$ in the first quadrant?
At $y = 3$ (where $x = 3$ )
Notice that for any $y \geq 0$ , $\frac{y^{2} + 3}{4} \geq \frac{3}{4} > 0$ . For $y \in [0, 3]$ , $\frac{y^{2} + 3}{4} \leq 6 - y$ , and for $y > 3$ , $\frac{y^{2} + 3}{4} > 6 - y$ . How is the range of $x$ bounded for each $y$ in the region?
For $y \in [0, 3]$ , $0 \leq x \leq \frac{y^{2} + 3}{4}$ ; for $y \in [3, 6]$ , $0 \leq x \leq 6 - y$
What is the total area evaluated by $\int_{0}^{3} \frac{y^{2} + 3}{4} d y + \int_{3}^{6} (6 - y) d y$ ?
$9$

✓ 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: 19 September 2026

Quick checks

Students also ask

Why integrate with respect to y instead of x?

Integrating with respect to x requires splitting the region at $x = 3 / 4$ and $x = 3$ , leading to multiple integrals involving square roots.

Answer

Integrating with respect to $y$ from $0$ to $3$ gives an area of $9$ .

Why each option works or fails

A: 8 - Integrating with respect to $y$ correctly over $[0, 3]$ but miscalculating the evaluation of $[6 y - y^{2} / 2 - y^{3} / 12]_{0}^{3}$ as $8$ instead of $9$ . Carefully compute each term: $6 (3) - (3^{2}) / 2 - (3^{3}) / 12 = 18 - 4.5 - 2.25 = 11.25$ , and account for the remaining region or boundary evaluations precisely.
B · correct: 9 - This is the correct option. Set up the region with horizontal slices: for $y \in [0, 3]$ , $x$ ranges from $0$ to $(y^{2} + 3) / 4$ , and for $y \in [3, 6]$ , check the intersection with $x = 6 - y$ . The inequality $y^{2} \geq 4 x - 3$ means $x \leq (y^{2} + 3) / 4$ , which intersects $y$ at $y = 3$ . Evaluating $\int_{0}^{3} ((6 - y) - \frac{y^{2} + 3}{4}) d y$ or splitting appropriately yields an area of $9$ .
C: 12 - Believing the upper limit of integration for the difference of functions is $y = 4$ or neglecting the parabolic subtraction over part of the interval. Solve the intersection equation $6 - y = \frac{y^{2} + 3}{4}$ to find the valid non-negative intersection point, which gives $y^{2} + 4 y - 21 = 0 ⟹ y = 3$ .
D: 15 - Integrating only the linear function $6 - x$ from $x = 0$ to $x = 3$ or dropping the constraint $y^{2} \geq 4 x - 3$ . Include the parabolic boundary $x \leq \frac{y^{2} + 3}{4}$ ; omitting it overstates the area.

Step-by-step solution

Given: The bounded region is defined by: 1. $0 \leq y \leq 6 - x ⟹ x \leq 6 - y$ and $y \geq 0$ 2. $y^{2} \geq 4 x - 3 ⟹ x \leq \frac{y^{2} + 3}{4}$ 3. $x \geq 0$
Goal: Find the total area of the enclosed region in the first quadrant.
Approach: Find the intersection points of $x = 6 - y$ and $x = \frac{y^{2} + 3}{4}$ . The bounds for $y$ run from $y = 0$ up to the intersection point. Integrate horizontal strips $Δ x = x_{right} - x_{left}$ with respect to $y$ . Here, the line is on the right and the parabola is on the left in the region of interest: $x_{right} = 6 - y$ and $x_{left} = \frac{y^{2} + 3}{4}$ .
Execute: Set $6 - y = \frac{y^{2} + 3}{4} ⟹ 24 - 4 y = y^{2} + 3 ⟹ y^{2} + 4 y - 21 = 0 ⟹ (y + 7) (y - 3) = 0$ . Since $y \geq 0$ , the upper limit is $y = 3$ . The area is: $A = \int_{0}^{3} ((6 - y) - \frac{y^{2} + 3}{4}) d y = \int_{0}^{3} (\frac{21}{4} - y - \frac{y^{2}}{4}) d y$ $A = {[\frac{21}{4} y - \frac{y^{2}}{2} - \frac{y^{3}}{12}]}_{0}^{3} = \frac{63}{4} - \frac{9}{2} - \frac{27}{12} = \frac{63}{4} - \frac{18}{4} - \frac{9}{4} = \frac{36}{4} = 9$
Verify: Check at endpoints: for $y = 0$ , $x \in [3 / 4, 6]$ , width is $5.25$ . At $y = 3$ , $x = 3$ , width is $0$ . The average width is approximately $(5.25 + 0) / 2 = 2.625$ , multiplied by height $3$ gives roughly $7.875$ , close to $9$ (accounting for the concave parabolic left boundary which bulges out the area).

Area of a bounded region defined by linear and parabolic inequalities

What feels right?

Integrating with respect to $y$ from $0$ to $3$ gives an area of $9$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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Integrating with respect to $y$ from $0$ to $3$ gives an area of $9$ .