If f a + b + 1 - x = f x , for all x , where a and b are fixed positive real numbers, then 1 a + b ∫ a b x f x + f x + 1 d x is equal to

Step-by-step solution

I = 1 a + b ∫ a b x f x + f x + 1 d x             … 1 x → a + b - x I = 1 a + b ∫ a b a + b - x f a + b - x + a + b + 1 - x d x I = 1 a + b ∫ a b a a + b - x f x + 1 + f x d x             … 2 ∵ Keeping x → x + 1 in given relation 1 + 2 2 I = ∫ a b f x + 1 + f x d x 2 I = ∫ a b f x + 1 d x + ∫ a b f x d x ∫ a b f a + b + 1 - x d x + ∫ a b f x d x 2 I = 2 ∫ a b f x d x I = &#