integrate 0 to a f(x)/(f(x)+f(a-x)) dx

Question

integrate 0 to a f(x)/(f(x)+f(a-x)) dx, where u = a-x, use substituation to solve it

Answer 1

We need to place the condition that f(a-x) ≠ - f(x) on [0, a]

 

Then, using I[0, a] as the integral from 0 to a, I[0,a] f(x)/(f(x)+f(a-x)) dx = 

I[0,1/2 a] f(x)/(f(x)+f(a-x)) dx + I[1/2 a, a] f(x)/(f(x)+f(a-x)) dx = 

I[0,1/2 a] f(x)/(f(x)+f(a-x)) dx + I[1/2 a, a] 1 -  f(a-x)/(f(x)+f(a-x)) dx =  (substituting  x = a-x)

I[0,1/2 a] f(x)/(f(x)+f(a-x)) dx + I[0, 1/2a] 1 - f(x)/(f(x)+f(a-x)) dx = 

I[0, 1/2a] 1 dx = 

1/2a