The area of the region S = x , y : y 2 ≤ 8 x , y ≥ 2 x , x ≥ 1 is

Plotting the region for the given inequalities $y^2\le 8x, y\ge \sqrt{2}x, x\le 1$, we get

Required area will be ${\int }_1^4\left(2\sqrt{2}\sqrt{x}-\sqrt{2}x\right)dx$

$=2\sqrt{2}{\left(\frac{x^{\frac{3}{2}}}{3/2}\right)}_1^4-\sqrt{2}{\left(\frac{x^2}{2}\right)}_1^4$

$=\frac{4\sqrt{2}}{3}\left(8-1\right)-\frac{\sqrt{2}}{3}\left(16-1\right)$

$=\frac{28\sqrt{2}}{3}-\frac{15\sqrt{2}}{2}=\frac{11\sqrt{2}}{6}$