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Integral Calculus: JEE Main Mathematics Question with Solution

Q: Why do we split the integral at x = 1?

Because when , so the horizontal line is lower than the parabola on this interval, making the active upper boundary.

The area of the region

\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^{2} \leq \mathrm{y} \leq 8-\mathrm{x}^{2}, \mathrm{y} \leq 7\right\} \text { is }

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21

18

24

20

Step-by-step solutionView

Correct answer

The area bounded by $y \ge x^2$ , $y \le 8-x^2$ , and $y \le 7$ is $20$ .

Option analysis

Why each option works or fails

A ·

21

Integrating without accounting for the correct boundaries, or miscalculating the area of the parabolic cap removed by $y \le 7$ . Ensure the cap region $8-x^2 > 7$ is integrated between $x = -1$ and $x = 1$ , giving an area reduction of $\frac{4}{3}$ , rather than $1$ or another approximate value.

B ·

18

Making an arithmetic error when evaluating the fractional definite integrals $\int (8-2x^2)\,dx$ . Carefully compute $\left[8x - \frac{2x^3}{3}\right]_{-2}^{2} = \frac{64}{3}$ , then subtract the excess area $\frac{4}{3}$ to obtain $\frac{60}{3} = 20$ .

C ·

24

Neglecting the condition $y \le 7$ entirely and computing the full area between $y = x^2$ and $y = 8-x^2$ , or making an addition error on the total bounds. Check all given inequalities; $y \le 7$ cuts off the top section of the region above $y = 7$ .

D ·

20

This is the correct option. Correctly compute the total area between the parabolas $\frac{64}{3}$ and subtract the top segment above $y = 7$ , which is $\int_{-1}^1 (8-x^2 - 7)\,dx = \frac{4}{3}$ , yielding $20$ .

Reviewed route

Solution

StepWorking

01given

The region is defined by $x^2 \le y \le 8-x^2$ and $y \le 7$ .

02goal

Find the total area of this closed region.

03approach

The curves $y = x^2$ and $y = 8-x^2$ intersect where $x^2 = 8-x^2 \implies x^2 = 4 \implies x = \pm 2$ . At $x = \pm 1$ , the upper curve $y = 8-x^2$ reaches $y = 7$ . Thus, for $|x| \le 1$ , the upper boundary is capped at $y = 7$ . For $1 \le |x| \le 2$ , the upper boundary is $y = 8-x^2$ . The lower boundary everywhere is $y = x^2$ . Due to symmetry about the y-axis, we calculate the area for $x \ge 0$ and multiply by 2.

04execute

Area $= 2 \left[ \int_{0}^{1} (7 - x^2) \, dx + \int_{1}^{2} ((8 - x^2) - x^2) \, dx \right] = 2 \left[ \left(7(1) - \frac{1}{3}\right) + \int_{1}^{2} (8 - 2x^2) \, dx \right] = 2 \left[ \frac{20}{3} + \left( 8(2-1) - \frac{2}{3}(8 - 1) \right) \right] = 2 \left[ \frac{20}{3} + 8 - \frac{14}{3} \right] = 2 \left[ \frac{6}{3} + 8 \right] = 2 [2 + 8] = 20$ .

05verify

Total area between $y = 8-x^2$ and $y = x^2$ without the cap $y \le 7$ is $\int_{-2}^{2} (8 - 2x^2) \, dx = 2 \left[8(2) - \frac{2}{3}(8)\right] = 2 \left[16 - \frac{16}{3}\right] = \frac{64}{3} \approx 21.33$ . The region cut off at the top is bounded by $y = 8-x^2$ and $y = 7$ , spanning $x \in [-1, 1]$ , which has area $\int_{-1}^{1} ((8-x^2) - 7) \, dx = \int_{-1}^{1} (1-x^2) \, dx = 2(1 - 1/3) = \frac{4}{3}$ . Subtracting gives $\frac{64}{3} - \frac{4}{3} = \frac{60}{3} = 20$ . Both methods yield 20.

06given

Region between parabolas $y = x^2$ and $y = 8-x^2$ subject to $y \le 7$ .

07goal

Compute the required area by: $\text{Area}_{\text{total parabolas}} - \text{Area}_{\text{cap above } y=7}$ .

08approach

Area between parabolas: $A_1 = \int_{-2}^2 (8 - 2x^2) \, dx$ . Area of the cap where $y > 7$ : $A_2 = \int_{-1}^1 ((8 - x^2) - 7) \, dx = \int_{-1}^1 (1 - x^2) \, dx$ . Desired Area $= A_1 - A_2$ .

09execute

$A_1 = 2\left[8(2) - \frac{2}{3}(8)\right] = \frac{64}{3}$ . $A_2 = 2\left[1 - \frac{1}{3}\right] = \frac{4}{3}$ . Area $= \frac{64}{3} - \frac{4}{3} = \frac{60}{3} = 20$ .

✓verify

Value is an exact integer 20, matching option (3).

✓ 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

Why do we split the integral at x = 1?

Because $8 - x^2 > 7$ when $|x| < 1$ , so the horizontal line $y = 7$ is lower than the parabola $8 - x^2$ on this interval, making $y = 7$ the active upper boundary.

Answer

The area bounded by $y \ge x^2$ , $y \le 8-x^2$ , and $y \le 7$ is $20$ .

Why each option works or fails

A: $21$ - Integrating without accounting for the correct boundaries, or miscalculating the area of the parabolic cap removed by $y \le 7$ . Ensure the cap region $8-x^2 > 7$ is integrated between $x = -1$ and $x = 1$ , giving an area reduction of $\frac{4}{3}$ , rather than $1$ or another approximate value.
B: $18$ - Making an arithmetic error when evaluating the fractional definite integrals $\int (8-2x^2)\,dx$ . Carefully compute $\left[8x - \frac{2x^3}{3}\right]_{-2}^{2} = \frac{64}{3}$ , then subtract the excess area $\frac{4}{3}$ to obtain $\frac{60}{3} = 20$ .
C: $24$ - Neglecting the condition $y \le 7$ entirely and computing the full area between $y = x^2$ and $y = 8-x^2$ , or making an addition error on the total bounds. Check all given inequalities; $y \le 7$ cuts off the top section of the region above $y = 7$ .
D · correct: $20$ - This is the correct option. Correctly compute the total area between the parabolas $\frac{64}{3}$ and subtract the top segment above $y = 7$ , which is $\int_{-1}^1 (8-x^2 - 7)\,dx = \frac{4}{3}$ , yielding $20$ .

Step-by-step solution

given: The region is defined by $x^2 \le y \le 8-x^2$ and $y \le 7$ .
goal: Find the total area of this closed region.
approach: The curves $y = x^2$ and $y = 8-x^2$ intersect where $x^2 = 8-x^2 \implies x^2 = 4 \implies x = \pm 2$ . At $x = \pm 1$ , the upper curve $y = 8-x^2$ reaches $y = 7$ . Thus, for $|x| \le 1$ , the upper boundary is capped at $y = 7$ . For $1 \le |x| \le 2$ , the upper boundary is $y = 8-x^2$ . The lower boundary everywhere is $y = x^2$ . Due to symmetry about the y-axis, we calculate the area for $x \ge 0$ and multiply by 2.
execute: Area $= 2 \left[ \int_{0}^{1} (7 - x^2) \, dx + \int_{1}^{2} ((8 - x^2) - x^2) \, dx \right] = 2 \left[ \left(7(1) - \frac{1}{3}\right) + \int_{1}^{2} (8 - 2x^2) \, dx \right] = 2 \left[ \frac{20}{3} + \left( 8(2-1) - \frac{2}{3}(8 - 1) \right) \right] = 2 \left[ \frac{20}{3} + 8 - \frac{14}{3} \right] = 2 \left[ \frac{6}{3} + 8 \right] = 2 [2 + 8] = 20$ .
verify: Total area between $y = 8-x^2$ and $y = x^2$ without the cap $y \le 7$ is $\int_{-2}^{2} (8 - 2x^2) \, dx = 2 \left[8(2) - \frac{2}{3}(8)\right] = 2 \left[16 - \frac{16}{3}\right] = \frac{64}{3} \approx 21.33$ . The region cut off at the top is bounded by $y = 8-x^2$ and $y = 7$ , spanning $x \in [-1, 1]$ , which has area $\int_{-1}^{1} ((8-x^2) - 7) \, dx = \int_{-1}^{1} (1-x^2) \, dx = 2(1 - 1/3) = \frac{4}{3}$ . Subtracting gives $\frac{64}{3} - \frac{4}{3} = \frac{60}{3} = 20$ . Both methods yield 20.
given: Region between parabolas $y = x^2$ and $y = 8-x^2$ subject to $y \le 7$ .
goal: Compute the required area by: $\text{Area}_{\text{total parabolas}} - \text{Area}_{\text{cap above } y=7}$ .
approach: Area between parabolas: $A_1 = \int_{-2}^2 (8 - 2x^2) \, dx$ . Area of the cap where $y > 7$ : $A_2 = \int_{-1}^1 ((8 - x^2) - 7) \, dx = \int_{-1}^1 (1 - x^2) \, dx$ . Desired Area $= A_1 - A_2$ .
execute: $A_1 = 2\left[8(2) - \frac{2}{3}(8)\right] = \frac{64}{3}$ . $A_2 = 2\left[1 - \frac{1}{3}\right] = \frac{4}{3}$ . Area $= \frac{64}{3} - \frac{4}{3} = \frac{60}{3} = 20$ .
verify: Value is an exact integer 20, matching option (3).

Integral Calculus: JEE Main Mathematics Question with Solution

The area bounded by y≥x2y \ge x^2y≥x2, y≤8−x2y \le 8-x^2y≤8−x2, and y≤7y \le 7y≤7 is 202020.

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✓verify

Students also ask

The area bounded by $y \ge x^2$ , $y \le 8-x^2$ , and $y \le 7$ is $20$ .