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

at $ \alpha = 0 \Rightarrow f(0) $

$x = 0,; x = 1,; y^2 = x$

$y = |0x - 5| - |1 - 0x| + 0x^2$

$y = 5 - 1 = 4$

$A_1 = \int_0^1 (4 - \sqrt{x}) dx$

$ = 4x - \frac{2}{3}x^{3/2} \Big|_0^1 $

$ = 4 - \frac{2}{3} = \frac{10}{3}$

at $ \alpha = 1 \Rightarrow f(1) $

$x = 0,; x = 1,; y^2 = x$

$y = |x - 5| - |1 - x| + x^2$

$x \in (0,1)$

$y = 5 - x - (1 - x) + x^2$

$y = 4 + x^2$

$A_2 = \int_0^1 \left( (4 + x^2) - \sqrt{x} \right) dx$

$ = 4x + \frac{x^3}{3} - \frac{2}{3}x^{3/2} \Big|_0^1 $

$ = 4 + \frac{1}{3} - \frac{2}{3} = \frac{11}{3}$

$f(0) + f(1) = |A_1 + A_2| = \left| \frac{10}{3} + \frac{11}{3} \right| = \left| \frac{21}{3} \right| = 7$