The area of the region bounded by the curve $y=2 x-x^2$ and $\mathrm{X}$-axis is
The area of the region bounded by the curve $y=2 x-x^2$ and $\mathrm{X}$-axis is
$\frac{2}{3}$ sq. units
$\frac{4}{3}$ sq. units
$\frac{5}{3}$ sq. units
$\frac{8}{3}$ sq. units
Solution
Point of intersection of curve $y=2 x-x^2$ 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
$\begin{aligned}
& A=\int_0^2\left(2 x-x^2\right) d x \\
& =2\left[\frac{x^2}{2}\right]_0^2-\left[\frac{x^2}{3}\right]_0^2 \\
& =(4)-\left(\frac{8}{3}\right)=\frac{4}{3}
\end{aligned}$
