JEE MainMathematics

Reviewed by official_key

+4 marks−1 if incorrectNumericalpyq

Integral Calculus: JEE Main Mathematics Question with Solution

Q: Why integrate with respect to y instead of x?

Integrating with respect to x requires solving for y as , splitting the region into two pieces at , which is algebraically more cumbersome.

If the area of the region bounded by the curves

y^{2}-2 y=-x, x+y=0

A

, then

8 \mathrm{~A}

is equal to

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Type the value - units or words beside it are fine.

Step-by-step solutionView

Correct answer

The bounded area is $A = \frac{1}{6}$ , so $8A = \frac{4}{3}$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Curves: $y^2 - 2y = -x$ (parabola opening leftwards) and $x + y = 0$ (straight line). Area of bounded region is $A$ .

02goal

Find the value of $8A$ , where $A$ is the area enclosed between the parabola and the line.

03approach

Express $x$ in terms of $y$ for both curves: $x_1 = 2y - y^2$ and $x_2 = -y$ . Find their points of intersection in $y$ , then compute $A = \int_{y_1}^{y_2} (x_{\text{right}} - x_{\text{left}}) \, dy$ . Finally, multiply by 8.

04execute

Equating $x$ : $2y - y^2 = -y \implies y^2 - 3y = 0 \implies y(y - 3) = 0$ . Thus, $y_1 = 0$ and $y_2 = 3$ . For $y \in [0, 3]$ , $2y - y^2 \ge -y$ .

05execute

Evaluate the integral: $A = \int_{0}^{3} ((2y - y^2) - (-y)) \, dy = \int_{0}^{3} (3y - y^2) \, dy = \left[ \frac{3y^2}{2} - \frac{y^3}{3} \right]_0^3 = \frac{27}{2} - 9 = \frac{9}{2}$ .

06execute

Calculate $8A = 8 \times \frac{9}{2} = 36$ .

✓verify

Using Archimedes' formula for parabolic segment: $\text{Area} = \frac{1}{6} |a| (y_2 - y_1)^3$ . Here, the quadratic in $y$ is $-(y^2 - 3y)$ , so $|a| = 1$ and root difference is $3 - 0 = 3$ . Area $= \frac{1}{6} (1) (3)^3 = \frac{27}{6} = \frac{9}{2}$ . Then $8A = 36$ . Matches perfectly.

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why integrate with respect to y instead of x?

Integrating with respect to x requires solving for y as $1 \pm \sqrt{1-x}$ , splitting the region into two pieces at $x=0$ , which is algebraically more cumbersome.

Answer

The bounded area is $A = \frac{1}{6}$ , so $8A = \frac{4}{3}$ .

Why each option works or fails

Step-by-step solution

given: Curves: $y^2 - 2y = -x$ (parabola opening leftwards) and $x + y = 0$ (straight line). Area of bounded region is $A$ .
goal: Find the value of $8A$ , where $A$ is the area enclosed between the parabola and the line.
approach: Express $x$ in terms of $y$ for both curves: $x_1 = 2y - y^2$ and $x_2 = -y$ . Find their points of intersection in $y$ , then compute $A = \int_{y_1}^{y_2} (x_{\text{right}} - x_{\text{left}}) \, dy$ . Finally, multiply by 8.
execute: Equating $x$ : $2y - y^2 = -y \implies y^2 - 3y = 0 \implies y(y - 3) = 0$ . Thus, $y_1 = 0$ and $y_2 = 3$ . For $y \in [0, 3]$ , $2y - y^2 \ge -y$ .
execute: Evaluate the integral: $A = \int_{0}^{3} ((2y - y^2) - (-y)) \, dy = \int_{0}^{3} (3y - y^2) \, dy = \left[ \frac{3y^2}{2} - \frac{y^3}{3} \right]_0^3 = \frac{27}{2} - 9 = \frac{9}{2}$ .
execute: Calculate $8A = 8 \times \frac{9}{2} = 36$ .
verify: Using Archimedes' formula for parabolic segment: $\text{Area} = \frac{1}{6} |a| (y_2 - y_1)^3$ . Here, the quadratic in $y$ is $-(y^2 - 3y)$ , so $|a| = 1$ and root difference is $3 - 0 = 3$ . Area $= \frac{1}{6} (1) (3)^3 = \frac{27}{6} = \frac{9}{2}$ . Then $8A = 36$ . Matches perfectly.

Shortcut:

When to use it: Whenever finding the area bounded between a parabola and a secant line.

quick_kill: Intersection roots of $y^2 - 2y = -y$ are $y = 0, 3$ . The area of any parabolic segment bounded by a line is $\frac{|a|}{6}(\Delta y)^3 = \frac{1}{6}(3 - 0)^3 = \frac{27}{6} = \frac{9}{2}$ . Thus, $8A = 8 \times \frac{9}{2} = 36$ .

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

The bounded area is $A = \frac{1}{6}$ , so $8A = \frac{4}{3}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

We think you should solve this next

The value of $\int_{0}^{20\pi} (\sin^4 x + \cos^4 x)dx$ is equal to…

Let…

If the area of the region…

Students also ask

Integral Calculus: JEE Main Mathematics Question with Solution

The bounded area is A=16A = \frac{1}{6}A=61​, so 8A=438A = \frac{4}{3}8A=34​.

01given

02goal

03approach

04execute

05execute

06execute

✓verify

The value of ∫020π(sin⁡4x+cos⁡4x)dx\int_{0}^{20\pi} (\sin^4 x + \cos^4 x)dx∫020π​(sin4x+cos4x)dx is equal to…

Let…

If the area of the region…

Students also ask

The bounded area is $A = \frac{1}{6}$ , so $8A = \frac{4}{3}$ .

The value of $\int_{0}^{20\pi} (\sin^4 x + \cos^4 x)dx$ is equal to…