Consider a region R={(x, y) in mathbb{R}^{2}: | vedclass

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(D) The region $R$ is bounded by the parabola $y=x^{2}$ and the line $y=2x$. The intersection points are found by $x^{2}=2x$,which gives $x=0$ and $x=

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$3 \alpha^{2}-8 \alpha^{3 / 2}+8=0$

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(D) The region $R$ is bounded by the parabola $y=x^{2}$ and the line $y=2x$. The intersection points are found by $x^{2}=2x$,which gives $x=0$ and $x=2$. Thus,the points are $(0,0)$ and $(2,4)$.The total area $A$ of region $R$ is given by:$A = \int_{0}^{2} (2x - x^{2}) dx = [x^{2} - \frac{x^{3}}{3}]_{0}^{2} = 4 - \frac{8}{3} = \frac{4}{3}$.Alternatively,integrating with respect to $y$,the region is bounded by $x = \sqrt{y}$ (right) and $x = y/2$ (left) for $y \in [0, 4]$:$A = \int_{0}^{4} (\sqrt{y} - \frac{y}{2}) dy = [\frac{2}{3}y^{3/2} - \frac{y^{2}}{4}]_{0}^{4} = \frac{2}{3}(8) - \frac{16}{4} = \frac{16}{3} - 4 = \frac{4}{3}$.The line $y=\alpha$ divides the area into two equal parts. The area of the lower part (from $y=0$ to $y=\alpha$) is half of the total area:$\int_{0}^{\alpha} (\sqrt{y} - \frac{y}{2}) dy = \frac{1}{2} 	imes \frac{4}{3} = \frac{2}{3}$.Evaluating the integral:$[\frac{2}{3}y^{3/2} - \frac{y^{2}}{4}]_{0}^{\alpha} = \frac{2}{3} \Rightarrow \frac{2}{3}\alpha^{3/2} - \frac{\alpha^{2}}{4} = \frac{2}{3}$.Multiplying by $12$ to clear the denominators:$8\alpha^{3/2} - 3\alpha^{2} = 8 \Rightarrow 3\alpha^{2} - 8\alpha^{3/2} + 8 = 0$.

Consider a region $R=\{(x, y) \in \mathbb{R}^{2}: x^{2} \leq y \leq 2 x\}$. If a line $y=\alpha$ divides the area of region $R$ into two equal parts,then which of the following is true?

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