Let A be the area bounded by the curve y = x|x - 3| , the x -axis...

Plotting the $y = x\left|x - 3\right|$ and $x -$axis between $x = - 1$ and $x = 2$ we get,

Now area of the above region is given by $A = \left|{∫}_{- 1}^{0}\left(3x - {x}^{2}\right)dx\right| + {∫}_{0}^{2}\left(3x - {x}^{2}\right)dx$

$⇒ A = \left|{\left[3 \times \dfrac{{x}^{2}}{2} - \dfrac{{x}^{3}}{3}\right]}_{- 1}^{0}\right| + {\left[3 \times \dfrac{{x}^{2}}{2} - \dfrac{{x}^{3}}{3}\right]}_{0}^{2}$

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

Hence, the value of $12A = 62$