$\text{ Let }f:R\to R\text{ be a continuous function such that }f\left(x\right)+f(x+1)=2,\text{ for all }x\in R\text{ . If }$$I_1={\int }_0^{8}f\left(x\right)dx\text{ and }{I}_2={\int }_{-1}^{3}f\left(x\right)dx,\text{ then the value of }{I}_1+2I_2\text{ is equal to }$

$f\left(x\right)+f(x+1)=2 and f(x+1)+f(x+2)=2 \Rightarrow f(x+2)=f\left(x\right)$

$The function f\left(x\right) is periodic with periodicity 2$

$I_1=4{\int }_0^{2}f\left(x\right)dx. substitute x=t+1 in I_2$

$I_2={\int }_{-2}^{2}f(t+1)dt=2{\int }_0^{2}f(x+1)dt I_1+2I_2=4{\int }_0^{2}f\left(x\right)+f(x+1)dx =4\times 4=16$