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

Given f  a  +  b  +  1  −  x  =  f  xit implies that  fx+1=fa+b−xI=1a+b∫abxfx+fx+1dx.                                      .......1⇒I=1a+b∫aba+b−xfx+1+fxdx.                         .....2Adding (1) and (2)  2I=∫abfx+fx+1dx⇒2I=∫abfa+b−xdx+∫abfx+1dx⇒2I=2∫abfx+1dx⇒I=∫abfx+1dx                                           =∫a+1b+1fxdx