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QJEEM23Made37: JEE Main 2023 Mathematics - Class 12 JEEM23Made37 Solution: If the area of the region is , then is equal to __________.

If the area of the region {(x,y):∣x2−2∣≤y≤x}\left\{(x, \mathrm{y}):\left|x^{2}-2\right| \leq y \leq x\right\} is A\mathrm{A} , then 6A+1626 \mathrm{A}+16 \sqrt{2} is equal to __________.

Step-by-Step Verified Solution

 We have, {(x,y):∣x2−2∣≤y≤x}\text { We have, }\left\{(x, y):\left|x^2-2\right| \leq y \leq x\right\}

JEE Main 2023 (Online) 10th April Evening Shift Mathematics - Area Under The Curves Question 38 English Explanation

On solving, ∣x2−2∣=y\left|x^2-2\right|=y and y=xy=x , we get (1,1)(1,1) and (2,2)(2,2)

amp;A=∫12(x−(2−x2))dx+∫22(x−(x2−2))dxamp;A=[x22−2x+x33]12+[x22−x33+2x]22\begin{aligned}& A=\int\limits_1^{\sqrt{2}}\left(x-\left(2-x^2\right)\right) d x+\int\limits_{\sqrt{2}}^2\left(x-\left(x^2-2\right)\right) d x \\\\& A=\left[\frac{x^2}{2}-2 x+\frac{x^3}{3}\right]_1^{\sqrt{2}}+\left[\frac{x^2}{2}-\frac{x^3}{3}+2 x\right]_{\sqrt{2}}^2\end{aligned}

A=(1−22+223)−(12−2+13)+amp;(2−83+4)amp;−(1−223+22)\begin{aligned}A=\left(1-2 \sqrt{2}+\frac{2 \sqrt{2}}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)+ & \left(2-\frac{8}{3}+4\right) \\& -\left(1-\frac{2 \sqrt{2}}{3}+2 \sqrt{2}\right)\end{aligned}

amp;A=−42+423+76+103amp;A=−823+92⇒A=−162+276amp;6A=−162+27amp;6A+162=27\begin{aligned}& A=-4 \sqrt{2}+\frac{4 \sqrt{2}}{3}+\frac{7}{6}+\frac{10}{3} \\\\& A=\frac{-8 \sqrt{2}}{3}+\frac{9}{2} \Rightarrow A=\frac{-16 \sqrt{2}+27}{6} \\\\& 6 A=-16 \sqrt{2}+27 \\\\& 6 A+16 \sqrt{2}=27\end{aligned}

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