If the line x=α divides the area of region R=x,y∈R2:x3≤y≤x,0≤x≤1into two equal parts, then

α 4 + 4 α 2 − 1 = 0 , 0 < α ≤ 1 2

∫ 0 1 x − x 3 d x = 2 ∫ 0 α x − x 3 d x 1 4 = 2 α 2 2 − α 4 4 2 α 4 − 4 α 2 + 1 = 0 ⇒ α 2 = 4 − 16 − 8 4 α 2 = 1 − 1 2 ∵ α ∈ 1 2 , 1

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If the line $x = α$ divides the area of region $R = \{(x, y) \in R^{2} : x^{3} \leq y \leq x, 0 \leq x \leq 1\}$ into two equal parts, then

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$α^{4} + 4 α^{2} - 1 = 0$

$\frac{1}{2} < α < 1$

$0 < α \leq \frac{1}{2}$

$2 α^{4} - 4 α^{2} + 1 = 0$

answer is A, C.

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$\int_{0}^{1} (x - x^{3}) d x = 2 \int_{0}^{α} (x - x^{3}) d x$

$\frac{1}{4} = 2 (\frac{α^{2}}{2} - \frac{α^{4}}{4})$

$2 α^{4} - 4 α^{2} + 1 = 0$

$\Rightarrow α^{2} = \frac{4 - \sqrt{16 - 8}}{4}$

$α^{2} = 1 - \frac{1}{\sqrt{2}}$ $(∵ α \in (\frac{1}{2}, 1))$

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