If the area of the region S = ( x , y ) : 2 y - y 2 ≤ x 2 ≤ 2 y , x ≥ y is equal to n + 2…

Given,

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

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 $x^2+y^2-2y=0 \& x=y$ will be $\left(1,1\right)$ and $x^2=2y \& x=y$ will be $\left(2,2\right)$

So, required area will be,

$A={\int }_0^1\left(1+\sqrt{1-x^2}-\frac{x^2}{2}\right)+{\int }_1^2\left(x-\frac{x^2}{2}\right)dx$

$\Rightarrow A={\left(x-\frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}·{\sin }^{-1}x-\frac{x^3}{6}\right)}_0^1+{\left(\frac{x^2}{2}-\frac{x^3}{6}\right)}_1^2$

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

So, on comparing we get, $n=5$