JEE MAIN - Mathematics (2025 - 2nd April Morning Shift - No. 23)
If the area of the region $\left\{(x, y):\left|4-x^2\right| \leq y \leq x^2, y \leq 4, x \geq 0\right\}$ is $\left(\frac{80 \sqrt{2}}{\alpha}-\beta\right), \alpha, \beta \in \mathbf{N}$, then $\alpha+\beta$ is equal to _________.
Answer
22
Explanation
$$\begin{aligned} &\text { Area = }\\ &\int_{\sqrt{2}}^2\left(x^2-\left(4-x^2\right)\right) d x+(2 \sqrt{2}-2) \times 4-\int_2^{2 \sqrt{2}}\left(x^2-4\right) d x \end{aligned}$$
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$$\begin{aligned} & =\left[\frac{2 x^3}{3}-4 x\right]_{\sqrt{2}}^2+8 \sqrt{2}-8-\left[\frac{x^3}{3}-4 x\right]_2^{2 \sqrt{2}} \\ & =\frac{40 \sqrt{2}}{3}-16 \\ & \Rightarrow \alpha=6, \beta=16 \Rightarrow \alpha+\beta=22 \end{aligned}$$
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