If the area of the region { ( x , y ) : | x 2 - 2 | y x } is A, then 6 A + 16 2 is equal to _______ . [2023]

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Solution

Q.
If the area of the region ${(x, y) : | x^{2} - 2 | \leq y \leq x}$ is A, then $6 A + 16 \sqrt{2}$ is equal to _______ . [2023]

Ans.
(27)

$We have,$

$Region = {(x, y) : | x^{2} - 2 | \leq y \leq x}$

The graph of the region is as shown in figure.

$Required area, A = \int_{1}^{\sqrt{2}} [x - {- (x^{2} - 2)}] d x + \int_{\sqrt{2}}^{2} {x - (x^{2} - 2)} d x$

$= \int_{1}^{\sqrt{2}} (x^{2} + x - 2) d x + \int_{\sqrt{2}}^{2} (- x^{2} + x + 2) d x$

$= {(\frac{x^{3}}{3} + \frac{x^{2}}{2} - 2 x)}_{1}^{\sqrt{2}} + {(\frac{- x^{3}}{3} + \frac{x^{2}}{2} + 2 x)}_{\sqrt{2}}^{2}$

$= (\frac{2 \sqrt{2}}{3} + 1 - 2 \sqrt{2}) - (\frac{1}{3} + \frac{1}{2} - 2)$ $+ (\frac{- 8}{3} + 2 + 4) - (\frac{- 2 \sqrt{2}}{3} + 1 + 2 \sqrt{2})$

$= \frac{27 - 16 \sqrt{2}}{6}$ $∴$ $6 A + 16 \sqrt{2} = 27 - 16 \sqrt{2} + 16 \sqrt{2} = 27$

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