JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Area of a Region Bounded by a Parabola and a Line

Q: Why do we take y = +√(8x) instead of both ±√(8x)?

Because y ≥ √(2)x and x ≥ 1 imply y > 0, so the region lies strictly in the first quadrant where y = +2√(2)√(x).

The area of the region S={(x, y): y^2 ≤ 8 x, y ≥ √(2) x, x ≥ 1} is

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A(13 √(2))/6B(11 √(2))/6C(5 √(2))/6D(19 √(2))/6

Step-by-step solutionView

Correct answer

The area of the region $S$ is $\frac{11 \sqrt{2}}{6}$ , obtained by integrating the difference between the upper curve $y = \sqrt{8 x}$ and the lower line $y = \sqrt{2} x$ from $x = 1$ to $x = 4$ .

Option analysis

Why each option works or fails

A · (13 √(2))/6

Incorrectly evaluating the lower limit contribution when integrating $\sqrt{8 x} - \sqrt{2} x$ , possibly by dropping the factor of $2 / 3$ or miscalculating $\sqrt{8} \cdot \frac{2}{3} (1)^{3 / 2}$ . Ensure that $\int 2 \sqrt{2} \sqrt{x} d x = \frac{4 \sqrt{2}}{3} x^{3 / 2}$ is correctly evaluated at both limits $x = 4$ and $x = 1$ .

B · (11 √(2))/6

This is the correct value of the definite integral. Correctly compute the upper bound $x = 4$ from the intersection of $y^{2} = 8 x$ and $y = \sqrt{2} x$ , and integrate $\int_{1}^{4} (2 \sqrt{2} x^{1 / 2} - \sqrt{2} x) d x = \frac{11 \sqrt{2}}{6}$ .

C · (5 √(2))/6

Subtracting the bounds incorrectly or evaluating the integral from $x = 2$ to $x = 4$ instead of $x = 1$ to $x = 4$ . Check the condition $x \geq 1$ given in the stem to set the lower integration limit accurately at $x = 1$ .

D · (19 √(2))/6

Adding the terms instead of subtracting the lower boundary, or miscalculating the term at $x = 4$ as $32 \sqrt{2} / 3 - 8 \sqrt{2}$ . Subtract the lower curve $y = \sqrt{2} x$ from the upper curve $y = 2 \sqrt{2} \sqrt{x}$ , keeping careful track of signs at each boundary.

Reviewed route

Solution

StepWorking

01Given

Region $S = {(x, y) : y^{2} \leq 8 x, y \geq \sqrt{2} x, x \geq 1}$ .

02Goal

Find the total area of the bounded region $S$ .

03Approach

Find intersection points of the upper boundary curve $y = \sqrt{8 x} = 2 \sqrt{2} \sqrt{x}$ and the lower boundary line $y = \sqrt{2} x$ . Since $x \geq 1$ , integrate the vertical difference $(y_{upper} - y_{lower})$ from $x = 1$ to the intersection point $x = 4$ .

04Execute

Solve for intersection: $2 x^{2} = 8 x ⟹ x (x - 4) = 0$ , giving $x = 0$ and $x = 4$ . For $x \in [1, 4]$ , $y_{upper} = 2 \sqrt{2} \sqrt{x} \geq \sqrt{2} x = y_{lower}$ . Integrate: $Area = \int_{1}^{4} (2 \sqrt{2} x^{1 / 2} - \sqrt{2} x) d x = {[\frac{4 \sqrt{2}}{3} x^{3 / 2} - \frac{\sqrt{2}}{2} x^{2}]}_{1}^{4} = \frac{4 \sqrt{2}}{3} (8 - 1) - \frac{\sqrt{2}}{2} (16 - 1) = \frac{28 \sqrt{2}}{3} - \frac{15 \sqrt{2}}{2} = \frac{11 \sqrt{2}}{6}$ .

✓Verify

Check at midpoint $x = 2$ : $y_{upper} = 4$ , $y_{lower} = 2 \sqrt{2} \approx 2.828$ , difference $\approx 1.172$ . Interval width is $3$ . Area estimate $\approx 3 \times 0.9 \approx 2.7$ . Exact value $\frac{11 \sqrt{2}}{6} \approx 2.59$ , which perfectly matches.

Hints that build this answer step by step

What are the integration limits for $x$ that define the region $S$ ?
$x \in [1, 4]$
Which integrand correctly represents the vertical height of a strip from the lower boundary to the upper boundary for $x \in [1, 4]$ ?
$2 \sqrt{2} \sqrt{x} - \sqrt{2} x$
What is the value of the definite integral $\int_{1}^{4} (2 \sqrt{2} x^{1 / 2} - \sqrt{2} x) d x$ ?
$\frac{11 \sqrt{2}}{6}$

✓ 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: 22 September 2026

Quick checks

Students also ask

Why do we take

y = + \sqrt{8 x}

instead of both

\pm \sqrt{8 x}

Because $y \geq \sqrt{2} x$ and $x \geq 1$ imply $y > 0$ , so the region lies strictly in the first quadrant where $y = + 2 \sqrt{2} \sqrt{x}$ .

Answer

The area of the region $S$ is $\frac{11 \sqrt{2}}{6}$ , obtained by integrating the difference between the upper curve $y = \sqrt{8 x}$ and the lower line $y = \sqrt{2} x$ from $x = 1$ to $x = 4$ .

Why each option works or fails

A: (13 √(2))/6 - Incorrectly evaluating the lower limit contribution when integrating $\sqrt{8 x} - \sqrt{2} x$ , possibly by dropping the factor of $2 / 3$ or miscalculating $\sqrt{8} \cdot \frac{2}{3} (1)^{3 / 2}$ . Ensure that $\int 2 \sqrt{2} \sqrt{x} d x = \frac{4 \sqrt{2}}{3} x^{3 / 2}$ is correctly evaluated at both limits $x = 4$ and $x = 1$ .
B · correct: (11 √(2))/6 - This is the correct value of the definite integral. Correctly compute the upper bound $x = 4$ from the intersection of $y^{2} = 8 x$ and $y = \sqrt{2} x$ , and integrate $\int_{1}^{4} (2 \sqrt{2} x^{1 / 2} - \sqrt{2} x) d x = \frac{11 \sqrt{2}}{6}$ .
C: (5 √(2))/6 - Subtracting the bounds incorrectly or evaluating the integral from $x = 2$ to $x = 4$ instead of $x = 1$ to $x = 4$ . Check the condition $x \geq 1$ given in the stem to set the lower integration limit accurately at $x = 1$ .
D: (19 √(2))/6 - Adding the terms instead of subtracting the lower boundary, or miscalculating the term at $x = 4$ as $32 \sqrt{2} / 3 - 8 \sqrt{2}$ . Subtract the lower curve $y = \sqrt{2} x$ from the upper curve $y = 2 \sqrt{2} \sqrt{x}$ , keeping careful track of signs at each boundary.

Step-by-step solution

Given: Region $S = {(x, y) : y^{2} \leq 8 x, y \geq \sqrt{2} x, x \geq 1}$ .
Goal: Find the total area of the bounded region $S$ .
Approach: Find intersection points of the upper boundary curve $y = \sqrt{8 x} = 2 \sqrt{2} \sqrt{x}$ and the lower boundary line $y = \sqrt{2} x$ . Since $x \geq 1$ , integrate the vertical difference $(y_{upper} - y_{lower})$ from $x = 1$ to the intersection point $x = 4$ .
Execute: Solve for intersection: $2 x^{2} = 8 x ⟹ x (x - 4) = 0$ , giving $x = 0$ and $x = 4$ . For $x \in [1, 4]$ , $y_{upper} = 2 \sqrt{2} \sqrt{x} \geq \sqrt{2} x = y_{lower}$ . Integrate: $Area = \int_{1}^{4} (2 \sqrt{2} x^{1 / 2} - \sqrt{2} x) d x = {[\frac{4 \sqrt{2}}{3} x^{3 / 2} - \frac{\sqrt{2}}{2} x^{2}]}_{1}^{4} = \frac{4 \sqrt{2}}{3} (8 - 1) - \frac{\sqrt{2}}{2} (16 - 1) = \frac{28 \sqrt{2}}{3} - \frac{15 \sqrt{2}}{2} = \frac{11 \sqrt{2}}{6}$ .
Verify: Check at midpoint $x = 2$ : $y_{upper} = 4$ , $y_{lower} = 2 \sqrt{2} \approx 2.828$ , difference $\approx 1.172$ . Interval width is $3$ . Area estimate $\approx 3 \times 0.9 \approx 2.7$ . Exact value $\frac{11 \sqrt{2}}{6} \approx 2.59$ , which perfectly matches.

Area of a Region Bounded by a Parabola and a Line

What feels right?

The area of the region $S$ is $\frac{11 \sqrt{2}}{6}$ , obtained by integrating the difference between the upper curve $y = \sqrt{8 x}$ and the lower line $y = \sqrt{2} x$ from $x = 1$ to $x = 4$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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04Execute

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