Let f(alpha) denote the area of the region in | vedclass

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(C) For $\alpha = 0$, the function becomes $y = |0 \cdot x - 5| - |1 - 0 \cdot x| + 0 \cdot x^2 = |-5| - |1| + 0 = 5 - 1 = 4$.Thus, $f(0)$ is the area

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$7$

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(C) For $\alpha = 0$, the function becomes $y = |0 \cdot x - 5| - |1 - 0 \cdot x| + 0 \cdot x^2 = |-5| - |1| + 0 = 5 - 1 = 4$.Thus, $f(0)$ is the area bounded by $x=0, x=1, y^2=x$ and $y=4$. Since $y^2=x$ implies $y=\sqrt{x}$ in the first quadrant, the area is:$f(0) = \int_0^1 (4 - \sqrt{x}) dx = [4x - \frac{2}{3}x^{3/2}]_0^1 = 4 - \frac{2}{3} = \frac{10}{3}$.For $\alpha = 1$, the function becomes $y = |x - 5| - |1 - x| + x^2$. For $x \in (0, 1)$, $x-5  0$, so $|x-5| = 5-x$ and $|1-x| = 1-x$.Thus, $y = (5-x) - (1-x) + x^2 = 5 - x - 1 + x + x^2 = 4 + x^2$.The area $f(1)$ is bounded by $x=0, x=1, y^2=x$ and $y=4+x^2$:$f(1) = \int_0^1 ((4+x^2) - \sqrt{x}) dx = [4x + \frac{x^3}{3} - \frac{2}{3}x^{3/2}]_0^1 = 4 + \frac{1}{3} - \frac{2}{3} = 4 - \frac{1}{3} = \frac{11}{3}$.Finally, $f(0) + f(1) = \frac{10}{3} + \frac{11}{3} = \frac{21}{3} = 7$.

Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x=0, x=1, y^{2}=x$ and $y=|\alpha x-5|-|1-\alpha x|+\alpha x^{2}$. Then $f(0)+f(1)$ is equal to

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