JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The region R is defined by xy 8, 1 y x^2, and x 0. First, find the intersection points of the curves. Intersection of y = 1 and y = x^2: x^2 = 1 x = 1 (since x 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 = 8 x . The area A is given by: A = _ 1 ^ 2 (x^2 - 1) dx + _ 2 ^ 8 ( 8 x - 1) dx A = [ x^3 3 - x]_1^2 + [8 _e x - x]_2^8 A = ( 8 3 - 2) - ( 1 3 - 1) + (8 _e 8 - 8) - (8 _e 2 - 2) A = ( 2 3 ) - (- 2 3 ) + 8(3 _e 2) - 8 - 8 _e 2 + 2 A = 4 3 + 24 _e 2 - 8 _e 2 - 6 A = 16 _e 2 + 4 3 - 6 A = 16 _e 2 - 14 3 A = 2 3 (24 _e 2 - 7)