The area (in sq. units) bounded by the curve | vedclass

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(A) To find the area bounded by the curve $y=2x-x^2$ and the line $y=-x$, we first find their points of intersection by setting $2x-x^2 = -x$.$2x-x^2+

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$\frac{9}{2}$

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(A) To find the area bounded by the curve $y=2x-x^2$ and the line $y=-x$, we first find their points of intersection by setting $2x-x^2 = -x$.$2x-x^2+x = 0$$3x-x^2 = 0$$x(3-x) = 0$So, the points of intersection are $x=0$ and $x=3$.In the interval $[0, 3]$, the curve $y=2x-x^2$ lies above the line $y=-x$.The area is given by the integral:$	ext{Area} = \int_0^3 ((2x-x^2) - (-x)) dx$$= \int_0^3 (3x-x^2) dx$$= \left[ \frac{3x^2}{2} - \frac{x^3}{3} ight]_0^3$$= \left( \frac{3(3)^2}{2} - \frac{(3)^3}{3} ight) - (0 - 0)$$= \frac{27}{2} - \frac{27}{3}$$= \frac{27}{2} - 9$$= \frac{27-18}{2} = \frac{9}{2} 	ext{ sq. units.}$

The area (in sq. units) bounded by the curve $y=2x-x^2$ and the line $y=-x$ is

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