Consider the functions f, g: R rightarrow R d | vedclass

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(D) To find the area $\alpha$,we first find the intersection points of $f(x)$ and $g(x)$ for $x \geq 0$. For $x \in [0, 3/4]$,$g(x) = 2(1 - 4x/3) = 2

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$6$

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(D) To find the area $\alpha$,we first find the intersection points of $f(x)$ and $g(x)$ for $x \geq 0$. For $x \in [0, 3/4]$,$g(x) = 2(1 - 4x/3) = 2 - 8x/3$.Setting $f(x) = g(x)$:$x^2 + \frac{5}{12} = 2 - \frac{8x}{3}$$x^2 + \frac{8x}{3} + \frac{5}{12} - 2 = 0$$x^2 + \frac{8x}{3} - \frac{19}{12} = 0$$12x^2 + 32x - 19 = 0$$12x^2 + 38x - 6x - 19 = 0$$2x(6x + 19) - 1(6x + 19) = 0$$(6x + 19)(2x - 1) = 0$Since $x \geq 0$,we have $x = 1/2$.The area $\alpha$ is symmetric about the $y$-axis,so $\alpha = 2 \int_0^{3/4} \min\{f(x), g(x)\} dx$.For $x \in [0, 1/2]$,$f(x) \leq g(x)$,and for $x \in [1/2, 3/4]$,$g(x) \leq f(x)$.$\alpha = 2 \left[ \int_0^{1/2} (x^2 + \frac{5}{12}) dx + \int_{1/2}^{3/4} (2 - \frac{8x}{3}) dx ight]$$\alpha = 2 \left[ \left( \frac{x^3}{3} + \frac{5x}{12} ight)_0^{1/2} + \left( 2x - \frac{4x^2}{3} ight)_{1/2}^{3/4} ight]$$\alpha = 2 \left[ (\frac{1}{24} + \frac{5}{24}) + ( (2(\frac{3}{4}) - \frac{4}{3}(\frac{9}{16})) - (2(\frac{1}{2}) - \frac{4}{3}(\frac{1}{4})) ) ight]$$\alpha = 2 \left[ \frac{6}{24} + ( (\frac{3}{2} - \frac{3}{4}) - (1 - \frac{1}{3}) ) ight]$$\alpha = 2 \left[ \frac{1}{4} + ( \frac{3}{4} - \frac{2}{3} ) ight] = 2 \left[ \frac{1}{4} + \frac{1}{12} ight] = 2 \left[ \frac{3+1}{12} ight] = 2 	imes \frac{4}{12} = \frac{2}{3}$.Therefore,$9\alpha = 9 	imes \frac{2}{3} = 6$.

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