JEE MainMathematics

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+4 marks−1 if incorrectNumericalPrevious-year question

Area Bounded by an Absolute-Value Curve and the X-Axis

Let A be the area bounded by the curve y = x|x-3|, the x-axis and the ordinates x = -1 and x = 2. Then 12A is equal to

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Correct answer

The bounded area is A = 61/6, so 12A equals 122.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

The curve is $y = x ∣ x - 3 ∣$ , bounded by the $x$ -axis from $x = - 1$ to $x = 2$ . We need to find $12 A$ , where $A$ is the total enclosed area.

02Goal

Find $12 A = 12 \int_{- 1}^{2} ∣ y ∣ d x$ .

03Approach

For $x \in [- 1, 2]$ , $x - 3 < 0$ , so $∣ x - 3 ∣ = - (x - 3) = 3 - x$ . Thus $y = x (3 - x) = 3 x - x^{2}$ . Over $[- 1, 0]$ , $y \leq 0$ , so $∣ y ∣ = - (3 x - x^{2}) = x^{2} - 3 x$ . Over $[0, 2]$ , $y \geq 0$ , so $∣ y ∣ = 3 x - x^{2}$ . Split the integral at $x = 0$ .

04Execute

Compute the first integral $A_{1} = \int_{- 1}^{0} (x^{2} - 3 x) d x = {[\frac{x^{3}}{3} - \frac{3 x^{2}}{2}]}_{- 1}^{0} = 0 - (- \frac{1}{3} - \frac{3}{2}) = \frac{11}{6}$ .

05Execute

Compute the second integral $A_{2} = \int_{0}^{2} (3 x - x^{2}) d x = {[\frac{3 x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{2} = (\frac{3 (4)}{2} - \frac{8}{3}) - 0 = 6 - \frac{8}{3} = \frac{10}{3} = \frac{20}{6}$ .

06Execute

Total area $A = A_{1} + A_{2} = \frac{11}{6} + \frac{20}{6} = \frac{31}{6}$ . Therefore, $12 A = 12 \times \frac{31}{6} = 62$ .

✓Verify

Check: Direct integral without absolute values $\int_{- 1}^{2} (3 x - x^{2}) d x = \frac{3 (3)}{2} - \frac{9}{3} = \frac{9}{2} - 3 = \frac{3}{2} = \frac{9}{6}$ . Total area must exceed net signed integral: $\frac{31}{6} > \frac{9}{6}$ , with difference $2 A_{1} = 2 (11 / 6) = 22 / 6$ , which is correct since $\frac{31}{6} - \frac{22}{6} = \frac{9}{6}$ .

Hints that build this answer step by step

How does the function f(x) = x|x - 3| simplify on the interval [-1, 2]?
Since x - 3 < 0 for all x in [-1, 2], |x - 3| = -(x - 3), giving f(x) = -x(x - 3) = 3x - x^2.
Where does f(x) = 3x - x^2 cross the x-axis on [-1, 2], and how should the area integral be split?
It crosses at x = 0; f(x) <= 0 on [-1, 0] and f(x) >= 0 on [0, 2], so A = ∫_{-1}^0 -(3x - x^2) dx + ∫_0^2 (3x - x^2) dx.
What is the value of the integral ∫_{-1}^0 (x^2 - 3x) dx?
11/6
What is the value of the integral ∫_0^2 (3x - x^2) dx?
10/3
Combining the two regions, what is the value of 12A?
62

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

Quick checks

Students also ask

Why do we split the integral at x = 0 when the modulus was at x = 3?

The curve crosses the x-axis at x = 0 (since y = x(3-x) = 0). Area bounded by the x-axis requires taking |y|, so regions where y < 0 must be negated to contribute positive area.

Answer

The bounded area is A = 61/6, so 12A equals 122.

Why each option works or fails

Step-by-step solution

Given: The curve is $y = x ∣ x - 3 ∣$ , bounded by the $x$ -axis from $x = - 1$ to $x = 2$ . We need to find $12 A$ , where $A$ is the total enclosed area.
Goal: Find $12 A = 12 \int_{- 1}^{2} ∣ y ∣ d x$ .
Approach: For $x \in [- 1, 2]$ , $x - 3 < 0$ , so $∣ x - 3 ∣ = - (x - 3) = 3 - x$ . Thus $y = x (3 - x) = 3 x - x^{2}$ . Over $[- 1, 0]$ , $y \leq 0$ , so $∣ y ∣ = - (3 x - x^{2}) = x^{2} - 3 x$ . Over $[0, 2]$ , $y \geq 0$ , so $∣ y ∣ = 3 x - x^{2}$ . Split the integral at $x = 0$ .
Execute: Compute the first integral $A_{1} = \int_{- 1}^{0} (x^{2} - 3 x) d x = {[\frac{x^{3}}{3} - \frac{3 x^{2}}{2}]}_{- 1}^{0} = 0 - (- \frac{1}{3} - \frac{3}{2}) = \frac{11}{6}$ .
Execute: Compute the second integral $A_{2} = \int_{0}^{2} (3 x - x^{2}) d x = {[\frac{3 x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{2} = (\frac{3 (4)}{2} - \frac{8}{3}) - 0 = 6 - \frac{8}{3} = \frac{10}{3} = \frac{20}{6}$ .
Execute: Total area $A = A_{1} + A_{2} = \frac{11}{6} + \frac{20}{6} = \frac{31}{6}$ . Therefore, $12 A = 12 \times \frac{31}{6} = 62$ .
Verify: Check: Direct integral without absolute values $\int_{- 1}^{2} (3 x - x^{2}) d x = \frac{3 (3)}{2} - \frac{9}{3} = \frac{9}{2} - 3 = \frac{3}{2} = \frac{9}{6}$ . Total area must exceed net signed integral: $\frac{31}{6} > \frac{9}{6}$ , with difference $2 A_{1} = 2 (11 / 6) = 22 / 6$ , which is correct since $\frac{31}{6} - \frac{22}{6} = \frac{9}{6}$ .

Area Bounded by an Absolute-Value Curve and the X-Axis

What feels right?

The bounded area is A = 61/6, so 12A equals 122.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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