Mathematics - Area Under Curves Question with Solution | TestHub

Point of intersection of curve $y=2x-x^2$ and $x$ axis, is $0=2\mathrm{x}-{\mathrm{x}}^2\Rightarrow \mathrm{x}(\mathrm{x}-2)=0\Rightarrow \mathrm{x}=0,2$ When $\mathrm{x}=0,\mathrm{y}=0$ and when $\mathrm{x}=2,\mathrm{y}=0$ Refer figure Required area is shaded $\begin{array}{rl}& A={\int }_0^2\left(2x-x^2\right)dx\\ & =2{\left[\frac{x^2}{2}\right]}_0^2-{\left[\frac{x^2}{3}\right]}_0^2\\ & =(4)-\left(\frac{8}{3}\right)=\frac{4}{3}\end{array}$