Question:

Step 1: Define the function and determine its behavior. The given curve is $y = x|x-3|$. We need to analyze the absolute value function. $$ |x-3| = \begin{cases} x-3 & \text{if } x-3 \ge 0 \implies x \ge 3 \\ -(x-3) & \text{if } x-3 < 0 \implies x < 3 \end{cases} $$ Therefore, the function $y$ can be written as: $$ y = \begin{cases} x(x-3) = x^2 - 3x & \text{if } x \ge 3 \\ x(-(x-3)) = -x^2 + 3x & \text{if } x < 3 \end{cases} $$ Step 2: Identify the relevant part of the function for the given interval. The area is bounded by the ordinates $x=-1$ and $x=2$. This interval $[-1, 2]$ lies entirely within the region where $x < 3$. Thus, for the interval of integration, the function is $y = -x^2 + 3x$. Step 3: Determine the sign of the function within the integration interval. The area $A$ is given by $\int_{-1}^{2} |y| dx$. We need to ensure that the integrand is positive. Let's analyze the sign of $y = -x^2 + 3x = x(3-x)$ for $x \in [-1, 2]$. - If $x \in [-1, 0)$, $x$ is negative and $(3-x)$ is positive, so $y$ is negative. - If $x = 0$, $y = 0$. - If $x \in (0, 2]$, $x$ is positive and $(3-x)$ is positive, so $y$ is positive. Since $y$ changes sign in the interval $[-1, 2]$, we must split the integral into two parts to calculate the total area $A = \int_{-1}^{2} |y| dx$. Step 4: Set up the integral for the area. Based on the sign analysis, we have: - For $x \in [-1, 0)$, $y < 0$, so $|y| = -( -x^2 + 3x ) = x^2 - 3x$. - For $x \in [0, 2]$, $y \ge 0$, so $|y| = -x^2 + 3x$. The total area $A$ is given by: $$ A = \int_{-1}^{0} (x^2 - 3x) dx + \int_{0}^{2} (-x^2 + 3x) dx $$ Step 5: Evaluate the integrals. First, evaluate the integral from $x=-1$ to $x=0$: $$ \int_{-1}^{0} (x^2 - 3x) dx = \left[ \frac{x^3}{3} - \frac{3x^2}{2} \right]_{-1}^{0} $$ $$ = \left( \frac{0^3}{3} - \frac{3(0)^2}{2} \right) - \left( \frac{(-1)^3}{3} - \frac{3(-1)^2}{2} \right) $$ $$ = 0 - \left( -\frac{1}{3} - \frac{3}{2} \right) = - \left( -\frac{2}{6} - \frac{9}{6} \right) = - \left( -\frac{11}{6} \right) = \frac{11}{6} $$ Next, evaluate the integral from $x=0$ to $x=2$: $$ \int_{0}^{2} (-x^2 + 3x) dx = \left[ -\frac{x^3}{3} + \frac{3x^2}{2} \right]_{0}^{2} $$ $$ = \left( -\frac{2^3}{3} + \frac{3(2)^2}{2} \right) - \left( -\frac{0^3}{3} + \frac{3(0)^2}{2} \right) $$ $$ = \left( -\frac{8}{3} + \frac{12}{2} \right) - 0 = \left( -\frac{8}{3} + 6 \right) = \left( -\frac{8}{3} + \frac{18}{3} \right) = \frac{10}{3} $$ Now, sum the areas to find $A$: $$ A = \frac{11}{6} + \frac{10}{3} = \frac{11}{6} + \frac{20}{6} = \frac{31}{6} $$ Step 6: Calculate $12A$. The problem asks for the value of $12A$. $$ 12A = 12 \times \frac{31}{6} = 2 \times 31 = 62 $$ The final answer is $\boxed{62}$.