Let $$f : R \to R$$ be a continuous function such that $$f(x) + f(x+1) = 2$$ for all $$x \in R$$. If $$I_1 = \int_0^8 f(x)\,dx$$ and $$I_2 = \int_{-1}^3 f(x)\,dx$$, then the value of $$I_1 + 2I_2$$ is equal to ________.

Given $$f(x) + f(x+1) = 2$$ for all $$x \in \mathbb{R}$$. This means the function has period 2, since replacing $$x$$ by $$x+1$$: $$f(x+1) + f(x+2) = 2$$, and subtracting from the original: $$f(x) - f(x+2) = 0$$, so $$f(x+2) = f(x)$$.

For any interval of length 2, say $$[a, a+2]$$: $$\int_a^{a+2} f(x)\,dx = \int_a^{a+1} f(x)\,dx + \int_{a+1}^{a+2} f(x)\,dx$$. In the second integral, let $$x = t + 1$$: $$\int_a^{a+1} f(t+1)\,dt$$. So the total is $$\int_a^{a+1} [f(x) + f(x+1)]\,dx = \int_a^{a+1} 2\,dx = 2$$.

Now $$I_1 = \int_0^8 f(x)\,dx$$. Since the interval has length 8 = 4 periods of length 2: $$I_1 = 4 \times 2 = 8$$.

$$I_2 = \int_{-1}^3 f(x)\,dx$$. The interval has length 4 = 2 periods of length 2: $$I_2 = 2 \times 2 = 4$$.

Therefore $$I_1 + 2I_2 = 8 + 2(4) = 16$$.