Let f( ) = _ 12 ^ 5 12 d x 1 + ^ x - x for > 0 . The value of the integral ₁^ e^2 f( ) d is :

Step-by-step solution

First, evaluate f( ) . The sum of the limits is 12 + 5 12 = 6 12 = 2 . Using the property _ a ^ b g(x) d x = _ a ^ b g(a+b-x) d x , we replace x with 2 - x : f( ) = _ 12 ^ 5 12 d x 1 + ^ ( 2 -x) - ( 2 -x) f( ) = _ 12 ^ 5 12 d x 1 + ^ x - x = _ 12 ^ 5 12 d x 1 + ^ -( x - x) Multiplying numerator and denominator by ^ x - x : f( ) = _ 12 ^ 5 12 ^ x - x 1 + ^ x - x d x Adding this to the original integral for f( ) : 2f( ) = _ 12 ^ 5 12 1 + ^ x - x 1 + ^ x - x d x = _ 12 ^ 5 12 1 d x 2f( ) = 5 12 - 12 = 4 12 = 3 f( ) =