Let A = \ (x, y): y ^ 2 ≤ 4x, y - 2x ≥ - 4\ . The area of the...

Given, ${y}^{2} = 4x . . . \left(1\right)$

$y = 2x - 4 ⇒ 2x = y + 4 . . . \left(2\right)$

Solving equations \left(1\right) \text{ and } \left(2\right)$, we get</p><p>$⇒ {y}^{2} = 2y + 8$</p><p>$⇒ {y}^{2} - 2y - 8 = 0$</p><p>$⇒ \left(y - 4\right)\left(y + 2\right) = 0$</p><p>$⇒ y = 4, - 2$</p><p>From equation \left(2\right)$,

If $y = 4$ then $x = 4$ and if $y = - 2$ then $x = 1$.

$A = \left[\mathop{∫}_{0}^{4}\left(\sqrt{4x}\right)dx - \dfrac{1}{2} \times 2 \times 4\right] + \left[\mathop{∫}_{0}^{1}\sqrt{4x}dx\right] + \dfrac{1}{2} \times 1 \times 2$

$= 2{\left[\dfrac{{x}^{3 / 2}}{\dfrac{3}{2}}\right]}_{0}^{4} - 4 + 2{\left[\dfrac{{x}^{3 / 2}}{\dfrac{3}{2}}\right]}_{0}^{1} + 1$

$= \dfrac{4}{3}\left[8\right] - 4 + \dfrac{4}{3}\left[1\right] + 1$

$= \dfrac{36}{3} - 3$

$= 9$