The area bounded by the curve y=2x-x^2 and the straight line Y = -x i…

The correct answer is Option (3) → $\frac{9}{2}$

$y=2x-x^2$

$y=1-(x-1)^2$

Curve 1 = $(y-1)=-(x-1)^2$

Curve 2 = $y=-x$

$y=-x$

so $-x=2x-x^2$

so $x^2=3x$

$x=0,3$ points of intersection

Required area = $\int\limits_0^3\left|(2x-x^2)-(-x)\right|dx$

$=\int\limits_0^33x^2-x^2dx=\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^3=\frac{9}{2}$ sq. unit