JEE Main 2025 Apr 2 Shift 1, Mathematics Q23: Area as Definite Integral

$A={\int }_0^4\sqrt{4+y}dy-{\int }_0^2\sqrt{4-y}dy-{\int }_2^4\sqrt{y}dy$

$={\left[\frac{2}{3}(4+y{)}^{\frac{3}{2}}\right]}_0^4-{\left[-\frac{2}{3}(4-y{)}^{\frac{3}{2}}\right]}_0^2-{\left[\frac{2}{3}{y}^{\frac{3}{2}}\right]}_2^4$

$=\frac{32\sqrt{2}-16}{3}-\frac{16-4\sqrt{2}}{3}-\frac{16-4\sqrt{2}}{3}$

$=\frac{40\sqrt{2}}{3}-16$

Comparing with $\frac{80\sqrt{2}}{\alpha }-\beta$,

$\alpha =6,\beta =16$

$\therefore \boxed{\alpha +\beta =22}$