If the area of the region \ (x, y): | x ^ 2 - 2| ≤ y ≤ x\ is A ,...

Given,

$\left|{x}^{2} - 2\right| \le y \text{ and } y \le x$

Now on finding the intersection point we get, \left(x, y\right) ≡ \left(1, 1\right) \text{ and } \left(2, 2\right)$</p><p>Now plotting the curve of the above function we get,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/2f60ecd7-918c-4d46-bb3d-88e6e1bfae54.png" style="width: 250px; height: 232px;" /></p><p>Now from above diagram required bounded area will be,</p><p>$A = \left[{∫}_{1}^{\sqrt{2}}\left(x - \left(2 - {x}^{2}\right)\right)dx + {∫}_{\sqrt{2}}^{2}\left(x - \left({x}^{2} - 2\right)\right)dx\right]$</p><p>$⇒ A = {\left[\dfrac{{x}^{2}}{2} - 2x + \dfrac{{x}^{3}}{3}\right]}_{1}^{\sqrt{2}} + {\left[\dfrac{{x}^{2}}{2} - \dfrac{{x}^{3}}{3} + 2x\right]}_{\sqrt{2}}^{2}$</p><p>$⇒ A = \left[1 - 2\sqrt{2} + \dfrac{2\sqrt{2}}{3}\right] - \left[\dfrac{1}{2} - 2 + \dfrac{1}{3}\right] + \left[2 - \dfrac{8}{3} + 4\right] - \left[1 - \dfrac{2\sqrt{2}}{3} + 2\sqrt{2}\right]$</p><p>$⇒ A = \dfrac{9}{2} - \dfrac{8\sqrt{2}}{3}$</p><p>$⇒ 6A = 27 - 16\sqrt{2}$</p><p>$⇒ 6A + 16\sqrt{2} = 27