Let be the area bounded by the curve , the -axis and the ordinates and . Then is equal to:

62 Explanation: Solution To find the area , we need to integrate the function from to . Step 1: Define the function without absolute value The term changes its behavior based on whether is greater than or less than . If , then . If , then . Since our interval of integration is , is always less than . Therefore, in this range: Step 2: Set up the integral for Area The area is given by: Let's check if the curve crosses the -axis between and . . The roots are and . Since lies within our interval , the function changes sign at . For : is negative. For : is positive. So, the total area is: Step 3: Evaluate the integrals Applying the power rule : First part: Second part: Total Area : Step 4: Find

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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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Explanation

Solution

To find the area $A$ , we need to integrate the function $y = x|x - 3|$ from $x = -1$ to $x = 2$ .

Step 1: Define the function without absolute value

The term $|x - 3|$ changes its behavior based on whether $x$ is greater than or less than $3$ .

If $x < 3$ , then $|x - 3| = -(x - 3) = 3 - x$ .

If $x \geq 3$ , then $|x - 3| = x - 3$ .

Since our interval of integration is $[-1, 2]$ , $x$ is always less than $3$ . Therefore, in this range:

$y = x(3 - x) = 3x - x^2$

Step 2: Set up the integral for Area $A$

The area $A$ is given by:

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

Let's check if the curve crosses the $x$ -axis between $-1$ and $2$ .

$3x - x^2 = 0 \implies x(3 - x) = 0$ . The roots are $x = 0$ and $x = 3$ .

Since $x=0$ lies within our interval $[-1, 2]$ , the function changes sign at $0$ .

For $x \in [-1, 0]$ : $y$ is negative.

For $x \in [0, 2]$ : $y$ is positive.

So, the total area $A$ is:

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

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

Step 3: Evaluate the integrals

Applying the power rule $\int x^n \, dx = \frac{x^{n+1}}{n+1}$ :

First part:

$\left[ \frac{x^3}{3} - \frac{3x^2}{2} \right]_{-1}^{0} = (0 - 0) - \left( \frac{(-1)^3}{3} - \frac{3(-1)^2}{2} \right)$

$= - \left( -\frac{1}{3} - \frac{3}{2} \right) = - \left( -\frac{2+9}{6} \right) = \frac{11}{6}$

Second part:

$\left[ \frac{3x^2}{2} - \frac{x^3}{3} \right]_{0}^{2} = \left( \frac{3(2)^2}{2} - \frac{(2)^3}{3} \right) - (0)$

$= \left( \frac{12}{2} - \frac{8}{3} \right) = 6 - \frac{8}{3} = \frac{18-8}{3} = \frac{10}{3}$

Total Area $A$ :

$A = \frac{11}{6} + \frac{10}{3} = \frac{11}{6} + \frac{20}{6} = \frac{31}{6}$

Step 4: Find $12A$

$12A = 12 \times \frac{31}{6}$

$12A = 2 \times 31 = 62$

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The area of the region enclosed by the curve

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