JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given integral can be split at integer values of x because of the greatest integer function [x]. I = _0^3 ( e^x + e^ -x [x]! ) dx I = _0^1 e^x + e^ -x 0! dx + _1^2 e^x + e^ -x 1! dx + _2^3 e^x + e^ -x 2! dx Since 0! = 1, 1! = 1, and 2! = 2, we get: I = _0^2 (e^x + e^ -x ) dx + 1 2 _2^3 (e^x + e^ -x ) dx Integrating the terms: I = [ e^x - e^ -x ]_0^2 + 1 2 [ e^x - e^ -x ]_2^3 I = (e^2 - e^ -2 - (1 - 1)) + 1 2 (e^3 - e^ -3 - (e^2 - e^ -2 )) I = e^2 - e^ -2 + 1 2 e^3 - 1 2 e^ -3 - 1 2 e^2 + 1 2 e^ -2 I = 1 2 e^2 + 1 2 e^3 - 1 2 e^ -2 - 1 2 e^ -3 I = 1 2 ( e^2 + e^3 - 1 e^2 - 1 e^3 ) Answer: 1 2 (e^2 + e^3 - 1 e^2 - 1 e^3 )