JEE Main 2026 — Area Under Curves Question with Solution

The given region is bounded by the curves $y=1$, $y=x^2$, and $xy=27$ in the first quadrant.

Let us find the points of intersection of these curves:
Intersection of $y=1$ and $y=x^2$ gives $x=1$.
Intersection of $y=x^2$ and $xy=27$ gives $x(x^2)=27\Rightarrow x^3=27\Rightarrow x=3$, so $y=9$.
Intersection of $y=1$ and $xy=27$ gives $x=27$.

We can find the area by integrating with respect to $y$ from $y=1$ to $y=9$. For a given $y$, the value of $x$ ranges from the parabola $x=\sqrt{y}$ to the hyperbola $x=\frac{27}{y}$.

The area $A$ is given by:
$A={\int }_1^9\left(\frac{27}{y}-\sqrt{y}\right)dy$

Integrating the terms:
$A={\left[27\ln y-\frac{2}{3}{y}^{3/2}\right]}_1^9$

Substituting the limits:
$A=\left(27\ln 9-\frac{2}{3}(9{)}^{3/2}\right)-\left(27\ln 1-\frac{2}{3}(1{)}^{3/2}\right)$

$A=\left(27\ln (3^2)-\frac{2}{3}(27)\right)-\left(0-\frac{2}{3}\right)$

$A=54\ln 3-18+\frac{2}{3}$

$A=54\ln 3-\frac{54}{3}+\frac{2}{3}$

$A=54\ln 3-\frac{52}{3}$

Answer: $54{\log }_{e}3-\frac{52}{3}$