JEE Main 2026 — Area Under Curves Question with Solution

The region $R$ is defined by $xy\le 8$, $1\le y\le x^2$, and $x\ge 0$.
First, find the intersection points of the curves.
Intersection of $y=1$ and $y=x^2$: $x^2=1⟹x=1$ (since $x\ge 0$).
Intersection of $y=x^2$ and $xy=8$: $x(x^2)=8⟹x^3=8⟹x=2$. At $x=2$, $y=4$.
Intersection of $xy=8$ and $y=1$: $x(1)=8⟹x=8$.
The region is bounded below by $y=1$ from $x=1$ to $x=8$.
The upper boundary consists of two parts:
From $x=1$ to $x=2$, the upper curve is $y=x^2$.
From $x=2$ to $x=8$, the upper curve is $y=\frac{8}{x}$.
The area $A$ is given by:
$A={\int }_1^2(x^2-1)dx+{\int }_2^8(\frac{8}{x}-1)dx$
$A=[\frac{x^3}{3}-x{]}_1^2+[8{\log }_{e}x-x{]}_2^8$
$A=(\frac{8}{3}-2)-(\frac{1}{3}-1)+(8{\log }_{e}8-8)-(8{\log }_{e}2-2)$
$A=(\frac{2}{3})-(-\frac{2}{3})+8(3{\log }_{e}2)-8-8{\log }_{e}2+2$
$A=\frac{4}{3}+24{\log }_{e}2-8{\log }_{e}2-6$
$A=16{\log }_{e}2+\frac{4}{3}-6$
$A=16{\log }_{e}2-\frac{14}{3}$
$A=\frac{2}{3}(24{\log }_{e}2-7)$