The area bounded by the curve $\mathrm{y}=\mathrm{x}(\mathrm{x}-1{)}^2$ , Y-axis and the line y = 2 is ___ sq.units

The given curve is   $y=x{\left(x-1\right)}^2$ and the line $y=2$

Substitute $y=2$ in the above equation 

$$\begin{gathered} x\left(x^2-2x+1\right)=2 \\ {x}^3-2x^2+x-2=0 \end{gathered}$$

Observing the above cubic equation, $x=2$ is a solution

Hence, $x^3-2x^2+x-2=\left(x-2\right)\left(x^2+1\right)$

The area of the required region is as below

The area of the shaded region is the region bouned by the curve $y=x{\left(x-1\right)}^2$ and the lines $x=0,y=0,x=2,y=2$

Hence the area of the shaded region is ${\int }_0^{2}x{\left(x-1\right)}^{2}dx$

$\begin{array}{rcl}{\int }_0^{2}2-x{\left(x-1\right)}^{2}dx& =& {\int }_0^{2}2-x^3+2x^2-xdx\\ & =& {\left[2x-\frac{x^4}{4}+2\frac{x^3}{3}-\frac{x^2}{2}\right]}_0^2\\ & =& 4-\frac{16}{4}+\frac{16}{3}-2\\ & =& \frac{16}{3}-2\\ & =& \frac{10}{3}\\ & =& 3.33\end{array}$