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(D) The area $A$ bounded by the curves $y=2x-x^2$, $y=0$, and $x=1$ is given by:$A = \int_{0}^{1} (2x-x^2) dx = [x^2 - \frac{x^3}{3}]_{0}^{1} = 1 - \f

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$y=\frac{2}{3}x$

Vedclass · Answer

(D) The area $A$ bounded by the curves $y=2x-x^2$, $y=0$, and $x=1$ is given by:$A = \int_{0}^{1} (2x-x^2) dx = [x^2 - \frac{x^3}{3}]_{0}^{1} = 1 - \frac{1}{3} = \frac{2}{3}$ sq units.Let the line $y=mx$ divide this area into two equal halves.The area of the region bounded by $y=2x-x^2$ and $y=mx$ from $x=0$ to the intersection point $x_0$ is $\frac{A}{2} = \frac{1}{3}$.First, find the intersection: $2x-x^2 = mx \implies x(2-x-m) = 0$. Since $x 
eq 0$, $x = 2-m$.However, the problem implies the line passes through the region bounded by $x=1$. The area of the triangle formed by the line $y=mx$ and the vertical line $x=1$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 1 	imes m = \frac{m}{2}$.Equating this to half the total area: $\frac{m}{2} = \frac{1}{3} \implies m = \frac{2}{3}$.Thus, the equation of the line is $y=\frac{2}{3}x$.

The straight line through the origin which divides the area formed by the curves $y=2x-x^2$, $y=0$, and $x=1$ into two equal halves is

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