Mathematics - Area Under Curves Question with Solution | TestHub

The given curve is $y=2x-x^2$.

To find the points of intersection with the X-axis, set $y=0$:

$2x-x^2=0$

$x(2-x)=0\Rightarrow x=0,x=2$

The required area is the integral of the curve from $x=0$ to $x=2$:

$A={\int }_0^2(2x-x^2)dx$

$A={\left[x^2-\frac{x^3}{3}\right]}_0^2$

$A=\left(2^2-\frac{2^3}{3}\right)-0$

$A=4-\frac{8}{3}=\frac{4}{3}$ sq. units.

Answer: $\frac{4}{3}$