Integral Calculus: JEE Main Mathematics Question with Solution

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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Step-by-step solutionView

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 $12A$ , where $A$ is the total enclosed area.

02goal

Find $12A = 12 \int_{-1}^2 |y|\,dx$ .

03approach

For $x \in [-1, 2]$ , $x - 3 < 0$ , so $|x-3| = -(x-3) = 3-x$ . Thus $y = x(3-x) = 3x - x^2$ . Over $[-1, 0]$ , $y \le 0$ , so $|y| = -(3x-x^2) = x^2-3x$ . Over $[0, 2]$ , $y \ge 0$ , so $|y| = 3x-x^2$ . Split the integral at $x = 0$ .

04execute

Compute the first integral $A_1 = \int_{-1}^0 (x^2 - 3x)\,dx = \left[\frac{x^3}{3} - \frac{3x^2}{2}\right]_{-1}^0 = 0 - \left(-\frac{1}{3} - \frac{3}{2}\right) = \frac{11}{6}$ .

05execute

Compute the second integral $A_2 = \int_0^2 (3x - x^2)\,dx = \left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_0^2 = \left(\frac{3(4)}{2} - \frac{8}{3}\right) - 0 = 6 - \frac{8}{3} = \frac{10}{3} = \frac{20}{6}$ .

06execute

Total area $A = A_1 + A_2 = \frac{11}{6} + \frac{20}{6} = \frac{31}{6}$ . Therefore, $12A = 12 \times \frac{31}{6} = 62$ .

✓verify

Check: Direct integral without absolute values $\int_{-1}^2 (3x-x^2)dx = \frac{3(3)}{2} - \frac{9}{3} = \frac{9}{2} - 3 = \frac{3}{2} = \frac{9}{6}$ . Total area must exceed net signed integral: $\frac{31}{6} > \frac{9}{6}$ , with difference $2A_1 = 2(11/6) = 22/6$ , which is correct since $\frac{31}{6} - \frac{22}{6} = \frac{9}{6}$ .

✓ 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: 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 $12A$ , where $A$ is the total enclosed area.
goal: Find $12A = 12 \int_{-1}^2 |y|\,dx$ .
approach: For $x \in [-1, 2]$ , $x - 3 < 0$ , so $|x-3| = -(x-3) = 3-x$ . Thus $y = x(3-x) = 3x - x^2$ . Over $[-1, 0]$ , $y \le 0$ , so $|y| = -(3x-x^2) = x^2-3x$ . Over $[0, 2]$ , $y \ge 0$ , so $|y| = 3x-x^2$ . Split the integral at $x = 0$ .
execute: Compute the first integral $A_1 = \int_{-1}^0 (x^2 - 3x)\,dx = \left[\frac{x^3}{3} - \frac{3x^2}{2}\right]_{-1}^0 = 0 - \left(-\frac{1}{3} - \frac{3}{2}\right) = \frac{11}{6}$ .
execute: Compute the second integral $A_2 = \int_0^2 (3x - x^2)\,dx = \left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_0^2 = \left(\frac{3(4)}{2} - \frac{8}{3}\right) - 0 = 6 - \frac{8}{3} = \frac{10}{3} = \frac{20}{6}$ .
execute: Total area $A = A_1 + A_2 = \frac{11}{6} + \frac{20}{6} = \frac{31}{6}$ . Therefore, $12A = 12 \times \frac{31}{6} = 62$ .
verify: Check: Direct integral without absolute values $\int_{-1}^2 (3x-x^2)dx = \frac{3(3)}{2} - \frac{9}{3} = \frac{9}{2} - 3 = \frac{3}{2} = \frac{9}{6}$ . Total area must exceed net signed integral: $\frac{31}{6} > \frac{9}{6}$ , with difference $2A_1 = 2(11/6) = 22/6$ , which is correct since $\frac{31}{6} - \frac{22}{6} = \frac{9}{6}$ .