JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Area Under Curves

For $x\ge 0$ the region is $|4-x^2|\le y\le x^2$ together with $y\le 4$. The lower curve $y=|4-x^2|$ is $4-x^2$ for $0\le x\le 2$ and $x^2-4$ for $x>2$. The requirement $|4-x^2|\le x^2$ forces $x\ge \sqrt{2}$. For $\sqrt{2}\le x\le 2$ the strip runs from $y=4-x^2$ to $y=x^2$; for $2\le x\le 2\sqrt{2}$ it runs from $y=x^2-4$ to the cap $y=4$, beyond which the lower bound exceeds $4$. Hence



The first integral is $\frac{8\sqrt{2}-8}{3}$ and the second $\frac{32\sqrt{2}-40}{3}$, so



Comparing with $\frac{80\sqrt{2}}{\alpha }-\beta$ gives $\alpha =6$ and $\beta =16$, so $\alpha +\beta =22$.