If fa+b+1−x=fx, for all x, where a and b are fixed positive real numbers, then 1a+b∫abxfx+fx+1dxis equal to
∫a+1b+1fxdx
∫a+1b+1fx+1dx
∫a−1b−1fx+1dx
∫a−1b−1fxdx
Given f a + b + 1 − x = f x
it implies that fx+1=fa+b−x
I=1a+b∫abxfx+fx+1dx. .......1
⇒I=1a+b∫aba+b−xfx+1+fxdx. .....2
Adding (1) and (2)
2I=∫abfx+fx+1dx
⇒2I=∫abfa+b−xdx+∫abfx+1dx
⇒2I=2∫abfx+1dx⇒I=∫abfx+1dx
=∫a+1b+1fxdx