JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let I = _0^ _e(x) x^2 + 4 \,dx Substitute x = 2 , which gives dx = 2 ^2 \,d . The limits of integration change from x = 0 = 0 to x = 2 . I = _0^ /2 _e(2 ) 4 ^2 + 4 (2 ^2 ) \,d I = _0^ /2 _e(2 ) 4 ^2 (2 ^2 ) \,d I = 1 2 _0^ /2 _e(2 ) \,d I = 1 2 _0^ /2 ( _e 2 + _e( )) \,d I = 1 2 _e 2 _0^ /2 1 \,d + 1 2 _0^ /2 _e( ) \,d Let I_1 = _0^ /2 _e( ) \,d Using the property _0^a f(x)\,dx = _0^a f(a-x)\,dx, we get: I_1 = _0^ /2 _e ( ( 2 - ) ) \,d = _0^ /2 _e( ) \,d I_1 = _0^ /2 _e ( 1 ) \,d = - _0^ /2 _e( ) \,d = -I_1 2I_1 = 0 I_1 = 0 Substituting I_1 = 0 back into the equation for I: I = 1 2 _e 2 [ ]_0^ /2 + 0 I = 1 2 _e 2 ( 2 ) = _e 2 4 Answer: _e(2) 4