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 _____.

$$A = \int_{-1}^{2} |3x - x^2| \, dx$$

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

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

$$\Rightarrow A = \left( 0 - \left[ \frac{(-1)^3}{3} - \frac{3(-1)^2}{2} \right] \right) + \left( \left[ \frac{3(2)^2}{2} - \frac{(2)^3}{3} \right] - 0 \right)$$

$$\Rightarrow A = -\left( -\frac{1}{3} - \frac{3}{2} \right) + \left( 6 - \frac{8}{3} \right)$$

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

$$\Rightarrow 12A = 12 \times \frac{31}{6} = 2 \times 31 = 62$$