Mathematics - Area Under Curves Question with Solution | TestHub

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 $A=\left|{\int }_{-1}^0\left(3x-x^2\right)dx\right|+{\int }_0^2\left(3x-x^2\right)dx$

$\Rightarrow A=\left|{\left[3\times \frac{x^2}{2}-\frac{x^3}{3}\right]}_{-1}^0\right|+{\left[3\times \frac{x^2}{2}-\frac{x^3}{3}\right]}_0^2$

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

Hence, the value of $12A=62$