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(B) Given $f(x) = \frac{x^3}{3} - x^2 + \frac{5x}{9} + \frac{17}{36}$.
First,calculate the total area of the red region $A_R = \int_0^1 f(x) dx = \le

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$B, C, D$

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(B) Given $f(x) = \frac{x^3}{3} - x^2 + \frac{5x}{9} + \frac{17}{36}$.
First,calculate the total area of the red region $A_R = \int_0^1 f(x) dx = \left[ \frac{x^4}{12} - \frac{x^3}{3} + \frac{5x^2}{18} + \frac{17x}{36} ight]_0^1 = \frac{1}{12} - \frac{1}{3} + \frac{5}{18} + \frac{17}{36} = \frac{3 - 12 + 10 + 17}{36} = \frac{18}{36} = \frac{1}{2}$.
Since the total area of the square $S$ is $1 	imes 1 = 1$,the area of the green region $A_G = 1 - A_R = 1 - \frac{1}{2} = \frac{1}{2}$.
Let $A_R^-(h)$ and $A_G^-(h)$ be the areas of the red and green regions below the line $L_h$,respectively. Let $A_R^+(h)$ and $A_G^+(h)$ be the areas above $L_h$.
For $(B)$: We want $A_R^+(h) = A_R^-(h)$,which implies $A_R^-(h) = \frac{1}{2} A_R = \frac{1}{4}$. Since $f(x) \ge \frac{13}{36} > \frac{1}{4}$,for $h = \frac{1}{4}$,$A_R^-(h) = \int_0^1 \frac{1}{4} dx = \frac{1}{4}$. Thus,$(B)$ is true.
For $(C)$: Let $g(h) = A_G^+(h) - A_R^-(h)$. At $h = \frac{13}{36}$,$A_R^-(h) = 0$ and $A_G^+(h) = \frac{1}{2}$,so $g(h) = \frac{1}{2}$. At $h = \frac{181}{324}$,$A_R^-(h) = \frac{1}{2}$ and $A_G^+(h) = 0$,so $g(h) = -\frac{1}{2}$. By the Intermediate Value Theorem,there exists $h$ such that $g(h) = 0$,i.e.,$A_G^+(h) = A_R^-(h)$. Thus,$(C)$ is true.
For $(D)$: Let $k(h) = A_R^+(h) - A_G^-(h)$. Since $A_R^+(h) + A_R^-(h) = A_R = 1/2$ and $A_G^+(h) + A_G^-(h) = A_G = 1/2$,we have $A_R^+(h) = 1/2 - A_R^-(h)$ and $A_G^-(h) = 1/2 - A_G^+(h)$. Then $k(h) = (1/2 - A_R^-(h)) - (1/2 - A_G^+(h)) = A_G^+(h) - A_R^-(h) = g(h)$. Since $g(h)$ takes values from $1/2$ to $-1/2$,$k(h)$ also takes the value $0$. Thus,$(D)$ is true.
Therefore,options $(B), (C),$ and $(D)$ are correct.

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