The area of the region bounded by the curve y | vedclass

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(B) To find the area bounded by the curve $y=2x-x^2$ and the $X$-axis,we first find the intersection points by setting $y=0$: $0 = 2x - x^2$ $x(2-x) =

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$\frac{4}{3}$ sq. units

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(B) To find the area bounded by the curve $y=2x-x^2$ and the $X$-axis,we first find the intersection points by setting $y=0$: $0 = 2x - x^2$ $x(2-x) = 0$ So,$x=0$ and $x=2$. The required area $A$ is given by the integral of the curve from $x=0$ to $x=2$: $A = \int_{0}^{2} (2x - x^2) dx$ $A = \left[ \frac{2x^2}{2} - \frac{x^3}{3} \right]_{0}^{2}$ $A = \left[ x^2 - \frac{x^3}{3} \right]_{0}^{2}$ $A = (2^2 - \frac{2^3}{3}) - (0^2 - \frac{0^3}{3})$ $A = (4 - \frac{8}{3}) - 0$ $A = \frac{12-8}{3} = \frac{4}{3}$ sq. units.

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

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