JEE MainMathematics

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Area Between Curves Defined by Absolute Values and Lower Bounds

The area of the region {(x, y): x^2 ≤ y ≤ |x^2 - 4|, y ≥ 1} is

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A3/4(4√(2) - 1)B4/3(4√(2) - 1)C4/3(4√(2) + 1)D3/4(4√(2) + 1)

Step-by-step solutionView

Correct answer

The area of the bounded region is $\frac{4}{3} (4 \sqrt{2} - 1)$ , found by integrating between the symmetric curves for $x \in [1, \sqrt{2}]$ and doubling the result.

Option analysis

Why each option works or fails

A · 3/4(4√(2) - 1)

Inverting the fractional multiplier during integration of polynomial terms (using $\frac{3}{4}$ instead of $\frac{4}{3}$ ). Integrating $x^{2}$ yields $\frac{x^{3}}{3}$ ; ensure the denominator 3 is correctly placed outside when factoring out constants.

B · 4/3(4√(2) - 1)

The boundaries are correctly identified as $y = x^{2}$ , $y = 4 - x^{2}$ , and $y = 1$ , leading to $2 \int_{1}^{\sqrt{2}} (4 - 2 x^{2}) d x = \frac{4}{3} (4 \sqrt{2} - 1)$ . This is the correct answer.

C · 4/3(4√(2) + 1)

A sign error when substituting the lower limit $x = 1$ , adding the lower limit term rather than subtracting it. Recall that $\int_{a}^{b} f (x) d x = F (b) - F (a)$ ; subtracting $F (1) = 4 (1) - \frac{2}{3} (1)^{3} = \frac{10}{3}$ produces a negative contribution.

D · 3/4(4√(2) + 1)

Combining both an inverted coefficient fraction and a sign error at the lower limit. Carefully track both the integration fraction $\frac{1}{3}$ and the subtraction rule of the fundamental theorem of calculus.

Reviewed route

Solution

StepWorking

01Given

Region bounded by $x^{2} \leq y \leq ∣ x^{2} - 4 ∣$ and $y \geq 1$ .

02Goal

Find the total area of the bounded region.

03Approach

The region is symmetric about the y-axis ( $x \to - x$ ). We can calculate the area for $x \geq 0$ and multiply by 2. Integrating with respect to $y$ is simpler because for $y \in [1, 2]$ , $x$ ranges from $0$ to $\sqrt{y}$ , and for $y \in [2, 4]$ , $x$ ranges from $0$ to $\sqrt{4 - y}$ (since $∣ x^{2} - 4 ∣ = 4 - x^{2} = y ⟹ x = \sqrt{4 - y}$ where $x^{2} \leq 2$ ).

04Execute

Set up the integral along $y$ : $Area = 2 [\int_{1}^{2} \sqrt{y} d y + \int_{2}^{4} \sqrt{4 - y} d y]$

05Execute

Evaluate $\int_{1}^{2} y^{1 / 2} d y = {[\frac{2}{3} y^{3 / 2}]}_{1}^{2} = \frac{2}{3} (2 \sqrt{2} - 1)$ . Evaluate $\int_{2}^{4} (4 - y)^{1 / 2} d y = {[- \frac{2}{3} (4 - y)^{3 / 2}]}_{2}^{4} = 0 - (- \frac{2}{3} (2)^{3 / 2}) = \frac{2}{3} (2 \sqrt{2}) = \frac{4 \sqrt{2}}{3}$ . Sum $= \frac{2}{3} (2 \sqrt{2} - 1 + 2 \sqrt{2}) = \frac{2}{3} (4 \sqrt{2} - 1)$ . Total Area $= 2 \times \frac{2}{3} (4 \sqrt{2} - 1) = \frac{4}{3} (4 \sqrt{2} - 1)$ .

✓Verify

Check at $y = 2$ : both curves give $x = \sqrt{2}$ , confirming continuity of the boundary. The computed area $\approx 6.21$ is positive and bounded inside a rectangle of width $2 \sqrt{2} \approx 2.83$ and height $3$ , whose area is $\approx 8.49$ .

Hints that build this answer step by step

What are the relevant upper and lower curves defining the region for $x > 0$ under the constraints $x^{2} \leq y \leq ∣ x^{2} - 4 ∣$ and $y \geq 1$ ?
Upper curve: $y = 4 - x^{2}$ ; lower boundary: $y = x^{2}$ for $x \in [1, \sqrt{2}]$ , with $y \geq 1$ setting the left bound to $x = 1$ .
Using the symmetry of the region about the $y$ -axis, which definite integral represents the total area?
$2 \int_{1}^{\sqrt{2}} ((4 - x^{2}) - x^{2}) d x = 2 \int_{1}^{\sqrt{2}} (4 - 2 x^{2}) d x$
What is the evaluated value of $2 \int_{1}^{\sqrt{2}} (4 - 2 x^{2}) d x$ ?
$\frac{4}{3} (4 \sqrt{2} - 1)$

✓ 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 finding the intersection of $x^{2}$ and $4 - x^{2}$ (at $x = \sqrt{2}$ ), as well as intersections with $y = 1$ , splitting the region into multiple x-strips with upper minus lower curves. Integrating along y gives simple single-term boundaries $x (y)$ in each interval.

Answer

The area of the bounded region is $\frac{4}{3} (4 \sqrt{2} - 1)$ , found by integrating between the symmetric curves for $x \in [1, \sqrt{2}]$ and doubling the result.

Why each option works or fails

A: 3/4(4√(2) - 1) - Inverting the fractional multiplier during integration of polynomial terms (using $\frac{3}{4}$ instead of $\frac{4}{3}$ ). Integrating $x^{2}$ yields $\frac{x^{3}}{3}$ ; ensure the denominator 3 is correctly placed outside when factoring out constants.
B · correct: 4/3(4√(2) - 1) - The boundaries are correctly identified as $y = x^{2}$ , $y = 4 - x^{2}$ , and $y = 1$ , leading to $2 \int_{1}^{\sqrt{2}} (4 - 2 x^{2}) d x = \frac{4}{3} (4 \sqrt{2} - 1)$ . This is the correct answer.
C: 4/3(4√(2) + 1) - A sign error when substituting the lower limit $x = 1$ , adding the lower limit term rather than subtracting it. Recall that $\int_{a}^{b} f (x) d x = F (b) - F (a)$ ; subtracting $F (1) = 4 (1) - \frac{2}{3} (1)^{3} = \frac{10}{3}$ produces a negative contribution.
D: 3/4(4√(2) + 1) - Combining both an inverted coefficient fraction and a sign error at the lower limit. Carefully track both the integration fraction $\frac{1}{3}$ and the subtraction rule of the fundamental theorem of calculus.

Step-by-step solution

Given: Region bounded by $x^{2} \leq y \leq ∣ x^{2} - 4 ∣$ and $y \geq 1$ .
Goal: Find the total area of the bounded region.
Approach: The region is symmetric about the y-axis ( $x \to - x$ ). We can calculate the area for $x \geq 0$ and multiply by 2. Integrating with respect to $y$ is simpler because for $y \in [1, 2]$ , $x$ ranges from $0$ to $\sqrt{y}$ , and for $y \in [2, 4]$ , $x$ ranges from $0$ to $\sqrt{4 - y}$ (since $∣ x^{2} - 4 ∣ = 4 - x^{2} = y ⟹ x = \sqrt{4 - y}$ where $x^{2} \leq 2$ ).
Execute: Set up the integral along $y$ : $Area = 2 [\int_{1}^{2} \sqrt{y} d y + \int_{2}^{4} \sqrt{4 - y} d y]$
Execute: Evaluate $\int_{1}^{2} y^{1 / 2} d y = {[\frac{2}{3} y^{3 / 2}]}_{1}^{2} = \frac{2}{3} (2 \sqrt{2} - 1)$ . Evaluate $\int_{2}^{4} (4 - y)^{1 / 2} d y = {[- \frac{2}{3} (4 - y)^{3 / 2}]}_{2}^{4} = 0 - (- \frac{2}{3} (2)^{3 / 2}) = \frac{2}{3} (2 \sqrt{2}) = \frac{4 \sqrt{2}}{3}$ . Sum $= \frac{2}{3} (2 \sqrt{2} - 1 + 2 \sqrt{2}) = \frac{2}{3} (4 \sqrt{2} - 1)$ . Total Area $= 2 \times \frac{2}{3} (4 \sqrt{2} - 1) = \frac{4}{3} (4 \sqrt{2} - 1)$ .
Verify: Check at $y = 2$ : both curves give $x = \sqrt{2}$ , confirming continuity of the boundary. The computed area $\approx 6.21$ is positive and bounded inside a rectangle of width $2 \sqrt{2} \approx 2.83$ and height $3$ , whose area is $\approx 8.49$ .

Area Between Curves Defined by Absolute Values and Lower Bounds

What feels right?

The area of the bounded region is $\frac{4}{3} (4 \sqrt{2} - 1)$ , found by integrating between the symmetric curves for $x \in [1, \sqrt{2}]$ and doubling the result.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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