设连续函数 $f(x)$ 满足 $f(x+2)-f(x)=x, \int_0^2 f(x) d x=0$, 则 $\int_1^3 f(x) d x=$
【答案】 $\frac{1}{2}$

【解析】 $\int_1^3 f(x) d x=\int_1^2 f(x) d x+\int_2^3 f(x) d x$
$$
\begin{aligned}
& =\int_1^2 f(x) d x+\int_0^1 f(t+2) d t(\text { 令 } x=t+2) \\
& =\int_1^2 f(x) d x+\int_0^1 f(x+2) d x \\
& =\int_1^2 f(x) d x+\int_0^1[f(x)+x] d x \\
& =\int_1^2 f(x) d x+\int_0^1 f(x) d x+\int_0^1 x d x \\
& =\int_0^2 f(x) d x+\int_0^1 x d x \\
& =\int_0^1 x d x
\end{aligned}
$$
$=\dfrac{1}{2}$
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