JEE Main 2026 — Mathematics Calculus

The region $R$R is defined by $xy \le 8$xy≤8, $1 \le y \le x^2$1≤y≤x2, and $x \ge 0$x≥0.

First, find the intersection points of the curves.

Intersection of $y = 1$y=1 and $y = x^2$y=x2: $x^2 = 1 \implies x = 1$x2=1⟹x=1 (since $x \ge 0$x≥0).

Intersection of $y = x^2$y=x2 and $xy = 8$xy=8: $x(x^2) = 8 \implies x^3 = 8 \implies x = 2$x(x2)=8⟹x3=8⟹x=2. At $x=2$x=2, $y=4$y=4.

Intersection of $xy = 8$xy=8 and $y = 1$y=1: $x(1) = 8 \implies x = 8$x(1)=8⟹x=8.

The region is bounded below by $y = 1$y=1 from $x = 1$x=1 to $x = 8$x=8.

The upper boundary consists of two parts:

From $x = 1$x=1 to $x = 2$x=2, the upper curve is $y = x^2$y=x2.

From $x = 2$x=2 to $x = 8$x=8, the upper curve is $y = \frac{8}{x}$y=x8​.

The area $A$A is given by:

$$A = \int_{1}^{2} (x^2 - 1) dx + \int_{2}^{8} (\frac{8}{x} - 1) dx$$A=∫12​(x2−1)dx+∫28​(x8​−1)dx

$$A = [\frac{x^3}{3} - x]_1^2 + [8 \log_e x - x]_2^8$$A=[3x3​−x]12​+[8loge​x−x]28​

$$A = (\frac{8}{3} - 2) - (\frac{1}{3} - 1) + (8 \log_e 8 - 8) - (8 \log_e 2 - 2)$$A=(38​−2)−(31​−1)+(8loge​8−8)−(8loge​2−2)

$$A = (\frac{2}{3}) - (-\frac{2}{3}) + 8(3 \log_e 2) - 8 - 8 \log_e 2 + 2$$A=(32​)−(−32​)+8(3loge​2)−8−8loge​2+2

$$A = \frac{4}{3} + 24 \log_e 2 - 8 \log_e 2 - 6$$A=34​+24loge​2−8loge​2−6

$$A = 16 \log_e 2 + \frac{4}{3} - 6$$A=16loge​2+34​−6

$$A = 16 \log_e 2 - \frac{14}{3}$$A=16loge​2−314​

$$A = \frac{2}{3} (24 \log_e 2 - 7)$$A=32​(24loge​2−7)