Integrate

$\begin{aligned} \int_0^2 f(x) \ dx & = \int_0^1 f(x) \ dx + \color{#3D99F6} \int_1^2 f(x) \ dx & \small \color{#3D99F6} \text{Let }x = u+1 \implies dx = du \\ & = \int_0^1 f(x) \ dx + \color{#3D99F6} \int_0^1 f(u+1) \ du & \small \color{#3D99F6} \text{Since }f(x) + f(x+1) = 1 \\ & = \int_0^1 f(x) \ dx + \color{#3D99F6} \int_0^1 (1-f(u)) \ du \\ & = \int_0^1 du = \boxed 1 \end{aligned}$

If we treat this integral as the area under the curve between 0 and 2, then for each $x$ in [0,1], $x+1$ is in [1,2] . The sums of the heights of the curve at these two points is 1, from the problem statement. For example, $f(x)$ could be 0.2 and $f(x+1)$ could be 0.8. The average height at the 2 points is 0.5. Each of the $x$ heights can be replaced with 0.5 for every $x$ in [0,1]. This produces a rectangle on [0,2] with height 0.5. The area is $2 \times 0.5 = \boxed{1}$

a solution which would work for f(x) + f(x+1) = 1 would be f(x) = 0.5

Integrating 0.5 between 0 and 2 gets you the answer of 1.