The area of the region bounded by y=2x-x^{2} | vedclass

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(B) The given curve is $y=2x-x^{2}$.To find the points where the curve intersects the $x$-axis,we set $y=0$:$2x-x^{2}=0 \implies x(2-x)=0$,which gives

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$\frac{4}{3} 	ext{ sq unit}$

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(B) The given curve is $y=2x-x^{2}$.To find the points where the curve intersects the $x$-axis,we set $y=0$:$2x-x^{2}=0 \implies x(2-x)=0$,which gives $x=0$ and $x=2$.Thus,the curve intersects the $x$-axis at $(0,0)$ and $(2,0)$.The required area is given by the integral of $y$ with respect to $x$ from $x=0$ to $x=2$:$	ext{Area} = \int_{0}^{2} (2x-x^{2}) dx$$= \left[ x^{2} - \frac{x^{3}}{3} ight]_{0}^{2}$$= \left( 2^{2} - \frac{2^{3}}{3} ight) - (0 - 0)$$= 4 - \frac{8}{3}$$= \frac{12-8}{3} = \frac{4}{3} 	ext{ sq unit}$.

The area of the region bounded by $y=2x-x^{2}$ and the $x$-axis is

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