2026年考研数二第15题解析：积分平均值的计算

一、题目

二、解析

解法 1

根据积分平均值的计算公式，得：

$$
\frac{1}{2-0} \times \int_{0}^{2} \ln(2 + x) \mathrm{~d} x = \frac{1}{2} \int_{0}^{2} \ln(2 + x) \mathrm{~d} x
$$

接着，令 $t = \ln(2 + x)$, 可知：

$$
\begin{aligned}
& x = \mathrm{e}^{t} -2 \\
& \mathrm{d} x = \mathrm{e}^{t} \mathrm{~d} t \\
& x = 0 \textcolor{lightgreen}{ \leadsto } t = \ln 2 \\
& x = 2 \textcolor{lightgreen}{ \leadsto } t = \ln 4 = \ln 2^{2} = 2 \ln 2
\end{aligned}
$$

于是：

$$
\frac{1}{2} \int_{0}^{2} \ln(2 + x) \mathrm{~d} x = \frac{1}{2} \int_{\ln 2}^{2\ln 2} t\mathrm{e}^{t} \mathrm{~d} t
$$

接着，由《$\mathrm{e}^{x}$ 的求解技巧》这篇讲义，可得：

$$
\begin{aligned}
\frac{1}{2} \int_{\ln 2}^{2\ln 2} t\mathrm{e}^{t} \mathrm{~d} t & = \frac{1}{2} (t \mathrm{e}^{t} – \mathrm{e}^{t}) \Bigg|_{\ln 2}^{2 \ln 2} \\ \\
& = \frac{1}{2} \mathrm{e}^{t} (t – 1) \Bigg|_{\ln 2}^{2 \ln 2} \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{e^{\ln 2} = 2, \ \mathrm{e}^{2 \ln 2} = \left( \mathrm{e}^{\ln 2} \right)^{2}} \\ \\
& = \frac{1}{2} \left[ \mathrm{e}^{2 \ln 2} (2 \ln 2 – 1) – \mathrm{e}^{\ln 2} (\ln 2 – 1) \right] \\ \\
& = \frac{1}{2} \left[ 4 (2 \ln 2 – 1) – 2 (\ln 2 – 1) \right] \\ \\
& = \frac{1}{2} \left( 8 \ln 2 – 4 – 2 \ln 2 + 2 \right) \\ \\
& = \frac{1}{2} (6 \ln 2 – 2) \\ \\
& = 3 \ln 2 – 1
\end{aligned}
$$

解法 2

函数 $f(x) = \ln(2+x)$ 在区间 $[0,2]$ 上的平均值为：

$$
\begin{aligned}
\frac{1}{2} \int_{0}^{2} f(x) \mathrm{~d} x & = \frac{1}{2} \int_{0}^{2} \ln(x + 2) \mathrm{~d} x \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{t = x + 2} \\ \\
& = \frac{1}{2} \int_{2}^{4} \ln t \mathrm{~d} t \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{\left( t \ln t \right) ^{\prime} = \ln t + 1} \\ \\
& = \frac{1}{2} t \ln t \Bigg|_{2}^{4} – \frac{1}{2} \int_{2}^{4} 1 \mathrm{~d} t \\ \\
& = \frac{1}{2} (4 \ln 4 – 2 \ln 2) – \frac{1}{2} \cdot 2 \\ \\
& = \frac{1}{2} (4 \cdot 2 \ln 2 – 2 \ln 2) – \frac{1}{2} \cdot 2 \\ \\
& = \frac{1}{2} (6 \ln 2 -2) \\ \\
& = 3 \ln 2 – 1
\end{aligned}
$$

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