JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given integral is I = _ /6 ^ /3 ( 4 - ^2 x ^4 x ) dx. We can rewrite the integrand in terms of x and x. Using the identity ^2 x = 1 + ^2 x = 1 + 1 ^2 x , we get: I = _ /6 ^ /3 (4 - 1 - 1 ^2 x ) ^4 x \, dx I = _ /6 ^ /3 (3 - 1 ^2 x ) (1 + ^2 x) ^2 x \, dx Substitute t = x, which gives dt = ^2 x \, dx. The limits of integration change as follows: When x = 6 , t = 1 3 . When x = 3 , t = 3 . The integral becomes: I = _ 1/ 3 ^ 3 (3 - 1 t^2 ) (1 + t^2) dt Expanding the integrand, we get: I = _ 1/ 3 ^ 3 (3 + 3t^2 - 1 t^2 - 1 ) dt I = _ 1/ 3 ^ 3 (3t^2 + 2 - t^ -2 ) dt Integrating with respect to t: I = [ t^3 + 2t + 1 t ]_ 1/ 3 ^ 3 Substitute the upper limit t = 3 : ( 3 )^3 + 2 3 + 1 3 = 3 3 + 2 3 + 3 3 = 16 3 3 = 48 3 3 Substitute the lower limit t = 1 3 : ( 1 3 )^3 + 2 ( 1 3 ) + 3 = 1 3 3 + 2 3 + 3 = 1 + 6 + 9 3 3 = 16 3 3 Subtracting the lower limit value from the upper limit value: I = 48 3 3 - 16 3 3 = 32 3 3 Answer: 32 3 3