Mathematics - Area Under Curves Question with Solution | TestHub

$\begin{array}{ll}& \text{ For X-axis, }\\ & y=0\\ \therefore & x(x-2)(x+1)=0\\ & \Rightarrow x=0\text{ or }x=2\text{ or }x=-1\end{array}$
$\begin{array}{rl}& \text{ Required area }={\int }_{-1}^{0}y\mathrm{d}x+\left|{\int }_0^{2}y\mathrm{d}x\right|\\ & ={\int }_{-1}^{0}x(x-2)(x+1)\mathrm{d}x+\left|{\int }_0^{2}x(x-2)(x+1)\mathrm{d}x\right|\\ & ={\int }_{-1}^0\left(x^3-x^2-2x\right)\mathrm{d}x+\left|{\int }_0^2\left(x^3-x^2-2x\right)\mathrm{d}x\right|\\ & ={\left[\frac{x^4}{4}-\frac{x^3}{3}-x^2\right]}_{-1}^0+\left|{\left[\frac{x^4}{4}-\frac{x^3}{3}-x^2\right]}_0^2\right|\\ & =\frac{5}{12}+\left|-\frac{8}{3}\right|\\ & =\frac{37}{12}\text{ sq. units }\end{array}$