2023年考研数二第15题解析：积分区间的拆分与合并

一、题目

设连续函数 $f(x)$ 满足: $f(x+2)-f(x)=x, \int_{0}^{2} f(x) \mathrm{~ d} x=0$, 则 $\int_{1}^{3} f(x) \mathrm{~ d} x=?$

难度评级：

二、解析

首先：

$$
\begin{aligned}
\int_{1}^{3} f(x) \mathrm{~ d} x = \\
& \int_{1}^{0} f(x) \mathrm{~ d} x+\int_{0}^{2} f(x) \mathrm{~ d} x+\int_{2}^{3} f(x) \mathrm{~ d} x = \\ \\
& \int_{1}^{0} f(x) \mathrm{~ d} x + \textcolor{orangered}{\int_{2}^{3} f(x) \mathrm{~ d} x }
\end{aligned}
$$

若令 $\bar{x} + 2=x$, 则由 $x \in (2,3)$ 可得：

$$
\bar{x} \in (0,1)
$$

接着，若令 $\bar{x} = x$, 则：

$$
x \in (0, 1)
$$

因此：

$$
\begin{aligned}
\int_{1}^{3} f(x) \mathrm{~ d} x = \\
& \int_{1}^{0} f(x) \mathrm{~ d} x + \textcolor{orangered}{\int_{2}^{3} f(x) \mathrm{~ d} x } = \\ \\
& \int_{1}^{0} f(x) \mathrm{~ d} x + \textcolor{orangered}{\int_{2-2}^{3-2} f(\bar{x} + 2) \mathrm{~ d} (\bar{x} + 2) } = \\ \\
& \int_{1}^{0} f(x) \mathrm{~ d} x + \textcolor{springgreen}{\int_{0}^{1} f(x+2) \mathrm{~ d} x} = \\ \\
& \int_{0}^{1} f(x+2) \mathrm{~ d} x-\int_{0}^{1} f(x) \mathrm{~ d} x = \\ \\
& \int_{0}^{1}[f(x+2)-f(x)] \mathrm{~ d} x = \\ \\
& \int_{0}^{1} x \mathrm{~ d} x = \left.\frac{1}{2} x^{2}\right|_{0} ^{1}=\frac{1}{2}
\end{aligned}
$$

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