Question:
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that $f(x)+f(x+1)=2$, for all $x \in \mathbb{R}$. If
$\mathrm{I}_{1}=\int_{0}^{8} f(\mathrm{x}) \mathrm{d} \mathrm{x}$ and $\mathrm{I}_{2}=\int_{-1}^{3} f(\mathrm{x}) \mathrm{d} \mathrm{x}$, then the value of $\mathrm{I}_{1}+2 \mathrm{I}_{2}$ is equal to______.
Solution:
$f(x)+f(x+1)=2$
$\Rightarrow f(x)$ is periodic with period $=2$
$\mathrm{I}_{1}=\int_{0}^{8} f(\mathrm{x}) \mathrm{d} \mathrm{x}=4 \int_{0}^{2} f(\mathrm{x}) \mathrm{d} \mathrm{x}$
$=4 \int_{0}^{1}(f(x)+f(1+x)) d x=8$
Similarly $\mathrm{I}_{2}=2 \times 2=4$
$\mathrm{I}_{1}+2 \mathrm{I}_{2}=16$