If the area of the region S = \ ( x, y ) : 2y - y ^ 2 ≤ x ^ 2 ≤ 2y,...

Given,

$S = \left\{( x, y ) : 2y - {y}^{2} \le {x}^{2} \le 2y, x \ge y\right\}$,

Now plotting the diagram, we get

Now from above diagram, given region are,

$2y - {y}^{2} \le {x}^{2}$, ${x}^{2} \le 2y$ and $x \ge y$

Now the intersecting point of ${x}^{2} + {y}^{2} - 2y = 0 \text{ and } x = y$ will be \left(1, 1\right)$and${x}^{2} = 2y \text{ and } x = y$will be \left(2, 2\right)$

So, required area will be,

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

$⇒ A = {\left(x - \dfrac{x}{2}\sqrt{1 - {x}^{2}} - \dfrac{1}{2} \cdot \sin^{- 1}x - \dfrac{{x}^{3}}{6}\right)}_{0}^{1} + {\left(\dfrac{{x}^{2}}{2} - \dfrac{{x}^{3}}{6}\right)}_{1}^{2}$

$⇒ A = \dfrac{7}{6} - \dfrac{\pi}{4} = \dfrac{n + 2}{n + 1} - \dfrac{\pi}{x - 1}$

So, on comparing we get, $n = 5$