Let f:R→R be a continuous function such that f(x)+f(x+1)=2, for all x∈R . If I1=∫08f(x)dx and I2=∫-13f(x)dx, then the value of I1+2I2 is equal to

f(x)+f(x+1)=2  and  f(x+1)+f(x+2)=2 ⇒f(x+2)=f(x)The function fx is periodic with periodicity 2I1=4∫02f(x)dx.   substitute x=t+1 in I2             I2=∫-22f(t+1)dt=2∫02f(x+1)dt                     I1+2I2=4∫02f(x)+f(x+1)dx =4×4=16