If _ /6 ^ /3 1 1+ x + x dx = 2 , then the value of _ /6 ^ /3 x 1+ x + x dx = ............

Step-by-step solution

Let I = _ /6 ^ /3 x 1+ x + x dx Using the definite integral property _ a ^ b f(x)dx = _ a ^ b f(a+b-x)dx , we substitute x with 6 + 3 - x = 2 - x : I = _ /6 ^ /3 ( /2 - x) 1+ ( /2 - x) + ( /2 - x) dx I = _ /6 ^ /3 x 1+ x + x dx Adding the two expressions for I : 2I = _ /6 ^ /3 x + x 1+ x + x dx 2I = _ /6 ^ /3 1+ x + x - 1 1+ x + x dx 2I = _ /6 ^ /3 (1 - 1 1+ x + x )dx 2I = _ /6 ^ /3 1 dx - _ /6 ^ /3 1 1+ x + x dx Given that _ /6 ^ /3 1 1+ x + x dx = 2 , we substitute this value: 2I = [x]_ /6 ^ /3 - 2 2I = ( 3 - 6 )