Let A be the area of the region {(x, y): y ge | vedclass

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(B) The region is bounded by $y = x^2$,$y = (1-x)^2$,and $y = 2x(1-x)$.First,find the intersection points:$x^2 = 2x(1-x) \Rightarrow x^2 = 2x - 2x^2 \

Vedclass · Accepted Answer

$25$

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(B) The region is bounded by $y = x^2$,$y = (1-x)^2$,and $y = 2x(1-x)$.First,find the intersection points:$x^2 = 2x(1-x) \Rightarrow x^2 = 2x - 2x^2 \Rightarrow 3x^2 - 2x = 0 \Rightarrow x(3x-2) = 0$. So $x = 0$ or $x = 2/3$.$(1-x)^2 = 2x(1-x) \Rightarrow 1-2x+x^2 = 2x-2x^2 \Rightarrow 3x^2-4x+1 = 0 \Rightarrow (3x-1)(x-1) = 0$. So $x = 1/3$ or $x = 1$.$x^2 = (1-x)^2 \Rightarrow x^2 = 1-2x+x^2 \Rightarrow 2x = 1 \Rightarrow x = 1/2$.The region is symmetric about $x = 1/2$. The area $A$ is given by:$A = 2 \int_{1/3}^{1/2} (2x(1-x) - (1-x)^2) dx$$A = 2 \int_{1/3}^{1/2} (2x - 2x^2 - (1 - 2x + x^2)) dx$$A = 2 \int_{1/3}^{1/2} (-3x^2 + 4x - 1) dx$$A = 2 [-x^3 + 2x^2 - x]_{1/3}^{1/2}$$A = 2 [(-1/8 + 2/4 - 1/2) - (-1/27 + 2/9 - 1/3)]$$A = 2 [(-1/8) - (-1/27 + 6/27 - 9/27)] = 2 [-1/8 - (-4/27)] = 2 [4/27 - 1/8]$$A = 2 [(32 - 27) / 216] = 2 [5 / 216] = 5 / 108$Therefore,$540A = 540 	imes (5 / 108) = 5 	imes 5 = 25$.

Let $A$ be the area of the region $\{(x, y): y \geq x^2, y \geq(1-x)^2, y \leq 2x(1-x)\}$. Then $540A$ is equal to

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