Let f be a real-valued function satisfying f(x)+f(x+4)=f(x+2)+f(x+6)  then  ∫xx+8 f(t)dt is

Given that f(x)+f(x+4)=f(x+2)+f(x+6)----(1) Replacing x by x+2, we get f(x+2)+f(x+6)=f(x+4)+f(x+8)----(2) From equations ( 1 ) and ( 2 ), we get f(x)=f(x+8)----(3) or ∫xx+8 f(t)dt=∫08 f(t)dt= constant