Let A be the area bounded by the curve y = x|x - 3| the x-axis, a… — Solution | JEE Mathematics

NVAMediumJEE 2023Applications of Integrals (Area)

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:

Answer

Correct answer:62

Standard Method

Given: The curve is $y = x|x - 3|$ , and the required area is bounded by the x-axis between $x = -1$ and $x = 2$ .

Find: The value of $12A$ .

Computing the area:

A = \int_{-1}^{0} (x^2 - 3x) \, dx + \int_{0}^{2} (3x - x^2) \, dx

A = \left[ \frac{x^3}{3} - \frac{3x^2}{2} \right]_{-1}^{0} + \left[ \frac{3x^2}{2} - \frac{x^3}{3} \right]_{0}^{2}

A = \frac{11}{6} + \frac{10}{3}

A = \frac{31}{6}

\Rightarrow 12A = 62

Therefore, the value of $12A$ is $62$ .

Using a single expression for $|x-3|$ on the whole interval is incorrect. The modulus must be handled piecewise; here $|x-3| = 3-x$ on the relevant interval. Rewrite the integrand carefully before integrating.
Ignoring that area with the x-axis must be taken as positive is incorrect. The curve changes sign at $x=0$ , so split the integral at $x=0$ and add positive area contributions.
Multiplying by $12$ before simplifying the fractional area often leads to algebraic errors. First compute $A$ correctly, then evaluate $12A$ at the end.

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