JEE Main 2026 — Mathematics Calculus

The given region is bounded by the curves $y = 1$y=1, $y = x^2$y=x2, and $xy = 27$xy=27 in the first quadrant.

Let us find the points of intersection of these curves:

Intersection of $y = 1$y=1 and $y = x^2$y=x2 gives $x = 1$x=1.

Intersection of $y = x^2$y=x2 and $xy = 27$xy=27 gives $x(x^2) = 27 \Rightarrow x^3 = 27 \Rightarrow x = 3$x(x2)=27⇒x3=27⇒x=3, so $y = 9$y=9.

Intersection of $y = 1$y=1 and $xy = 27$xy=27 gives $x = 27$x=27.

We can find the area by integrating with respect to $y$y from $y = 1$y=1 to $y = 9$y=9. For a given $y$y, the value of $x$x ranges from the parabola $x = \sqrt{y}$x=y​ to the hyperbola $x = \dfrac{27}{y}$x=y27​.

The area $A$A is given by:

$$A = \int_{1}^{9} \left( \dfrac{27}{y} - \sqrt{y} \right) dy$$A=∫19​(y27​−y​)dy

Integrating the terms:

$$A = \left[ 27 \ln y - \dfrac{2}{3} y^{3/2} \right]_{1}^{9}$$A=[27lny−32​y3/2]19​

Substituting the limits:

$$A = \left( 27 \ln 9 - \dfrac{2}{3} (9)^{3/2} \right) - \left( 27 \ln 1 - \dfrac{2}{3} (1)^{3/2} \right)$$A=(27ln9−32​(9)3/2)−(27ln1−32​(1)3/2)

$$A = \left( 27 \ln(3^2) - \dfrac{2}{3} (27) \right) - \left( 0 - \dfrac{2}{3} \right)$$A=(27ln(32)−32​(27))−(0−32​)

$$A = 54 \ln 3 - 18 + \dfrac{2}{3}$$A=54ln3−18+32​

$$A = 54 \ln 3 - \dfrac{54}{3} + \dfrac{2}{3}$$A=54ln3−354​+32​

$$A = 54 \ln 3 - \dfrac{52}{3}$$A=54ln3−352​

Answer: $54\log_e 3 - \dfrac{52}{3}$54loge​3−352​