设连续函数 $f(x)$ 满足 $f(x+2)-f(x)=x, \int_{0}^{2} f(x) d x=0$, 则 $\int_{1}^{3} f(x) d x=$__？
$\begin{aligned} & \int_1^3 f(x) d x=\int_1^2 f(x) d x+\int_2^3 f(x) d x \\ & =\int_1^2 f(x) d x+\int_0^1 f(t+2) d t(\text { 令 } x=t+2) \\ & =\int_1^2 f(x) d x+\int_0^1 f(x+2) d x \\ & =\int_1^2 f(x) d x+\int_0^1[f(x)+x] d x \\ & =\int_1^2 f(x) d x+\int_0^1 f(x) d x+\int_0^1 x d x \\ & =\int_0^2 f(x) d x+\int_0^1 x d x \\ & =\int_0^1 x d x \\ & =\frac{1}{2} \\ & \end{aligned}$
因此答案为：$\frac{1}{2}$