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(B) Given $f(x)+f(x+1)=2$. Integrating both sides from $0$ to $1$,we get $\int_{0}^{1} f(x) d x + \int_{0}^{1} f(x+1) d x = \int_{0}^{1} 2 d x$. Using

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$16$

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(B) Given $f(x)+f(x+1)=2$. Integrating both sides from $0$ to $1$,we get $\int_{0}^{1} f(x) d x + \int_{0}^{1} f(x+1) d x = \int_{0}^{1} 2 d x$. Using the substitution $t=x+1$ in the second integral,we have $\int_{0}^{1} f(x) d x + \int_{1}^{2} f(t) d t = 2$,which implies $\int_{0}^{2} f(x) d x = 2$. Since $f(x)+f(x+1)=2$,$f(x+2) = 2-f(x+1) = 2-(2-f(x)) = f(x)$,so $f(x)$ is periodic with period $T=2$. $I_{1} = \int_{0}^{8} f(x) d x = \frac{8}{2} \int_{0}^{2} f(x) d x = 4 	imes 2 = 8$. $I_{2} = \int_{-1}^{3} f(x) d x$. Since $f(x)$ has period $2$,$\int_{a}^{a+T} f(x) d x$ is independent of $a$. Thus,$I_{2} = \int_{0}^{4} f(x) d x = \frac{4}{2} \int_{0}^{2} f(x) d x = 2 	imes 2 = 4$. Therefore,$I_{1}+2 I_{2} = 8 + 2(4) = 8 + 8 = 16$.

Let $f: R \rightarrow R$ be a continuous function such that $f(x)+f(x+1)=2$ for all $x \in R$. If $I_{1}=\int_{0}^{8} f(x) d x$ and $I_{2}=\int_{-1}^{3} f(x) d x$,then the value of $I_{1}+2 I_{2}$ is equal to:

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