JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Area Under Curves

With $1\le y\le x^2$ and $x>0$, the region sits above $y=1$ and below $y=x^2$ while $x^2\le \frac{27}{x}$, that is up to $x=3$; beyond that the hyperbola $xy=27$ caps it, down to $y=1$ at $x=27$. Hence the area is ${\int }_1^3(x^2-1)dx+{\int }_3^{27}\left(\frac{27}{x}-1\right)dx$. The first integral is ${\left[\frac{x^3}{3}-x\right]}_1^3=\frac{20}{3}$, and the second is $[27\ln x-x{]}_3^{27}=27\ln 9-24=54\ln 3-24$. Adding these, the area of the region equals $54{\log }_{e}3-\frac{52}{3}$.