JEE Main 2026 Apr 5 Shift 1, Mathematics Q16: Area as Definite Integral

The curves $y=x^2$ and $y=1$ meet at $x=1$; the curves $y=x^2$ and $xy=27$ meet where $x^3=27$, i.e. $x=3$; and $y=1$ meets $xy=27$ at $x=27$.

For $1\le x\le 3$, the region runs from $y=1$ up to $y=x^2$. For $3\le x\le 27$, it runs from $y=1$ up to $y=27/x$.

$\text{Area}={\int }_1^3(x^2-1)dx+{\int }_3^{27}\left(\frac{27}{x}-1\right)dx$

$={\left[\frac{x^3}{3}-x\right]}_1^3+{\left[27\ln x-x\right]}_3^{27}$

$=\left(9-3-\frac{1}{3}+1\right)+(27(\ln 27-\ln 3)-24)$

$=\frac{20}{3}+27\ln 9-24=54\ln 3+\frac{20-72}{3}=54\ln 3-\frac{52}{3}$.