Question

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

Answer 1

$f\left(x\right)+f(x+1)=2\cdots \left(1\right)x\to (x+1)f(x+1)+f(x+2)=2\cdots \left(ii\right)\text{by}\left(i\right)\&\left(ii\right)f\left(x\right)-f(x+2)=0f(x+2)=f\left(x\right)$
$f\left(x\right)$ is periodic with $T=2$
$I_1={\int }_0^{2\times 4}f\left(x\right)dx=4{\int }_0^{2}f\left(x\right)dx{I}_2={\int }_{-1}^{3}f\left(x\right)dx={\int }_0^{4}f(x+1)dx={\int }_0^4(2-f(x\left)\right)dx{I}_2=8-2{\int }_0^{2}f\left(x\right)dx{I}_1+2I_2=16$