本题见于考研数学李林老师的微信公众号“李林考研数学”

问题：设\(\displaystyle f\left( x \right)\)在\(\displaystyle\left[ a,b \right]\)上可导，且\(\displaystyle f'\left( x \right) \ne 0\)。

(1)证明：至少存在一点\(\displaystyle\xi \in \left( a,b \right)\)，使得

\[\int_a^b{f\left( x \right) \text{d}x}=f\left( b \right) \left( \xi -a \right) +f\left( a \right) \left( b-\xi \right) \]

(2)对(1)中的\(\displaystyle\xi\)，求

\[\lim_{b\rightarrow a^+} \frac{\xi -a}{b-a} \]

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过程如下：

(1)由于\(\displaystyle f'\left( x \right) \ne 0\)，不妨设\(\displaystyle f'\left( x \right) >0\)

采用零点定理证明\(\displaystyle\xi\)的存在性，令

\[F\left( x \right) =f\left( b \right) \left( x-a \right) +f\left( a \right) \left( b-x \right) -\int_a^b{f\left( x \right) \text{d}x} \]

则有

\[\begin{equation*} F\left( a \right) =f\left( a \right) \left( b-a \right) -\int_a^b{f\left( x \right) \text{d}x}=\int_a^b{\left[ f\left( a \right) -f\left( x \right) \right] \text{d}x}<0 \\ F\left( b \right) =f\left( b \right) \left( b-a \right) -\int_a^b{f\left( x \right) \text{d}x}=\int_a^b{\left[ f\left( b \right) -f\left( x \right) \right] \text{d}x}>0 \end{equation*} \]

函数\(\displaystyle F\left( x \right)\)满足零点定理的条件，由定理内容可知至少存在一点\(\displaystyle\xi \in \left( a,b \right)\)使得\(\displaystyle F\left( \xi \right) =0\)，即

\[\int_a^b{f\left( x \right) \text{d}x}=f\left( b \right) \left( \xi -a \right) +f\left( a \right) \left( b-\xi \right) \]

(2)基于问题(1)得到的结论，不难发现可将\(\displaystyle\xi\)写成用\(\displaystyle a\)与\(\displaystyle b\)构成的表达式，得到

\[\xi =\frac{\int_a^b{f\left( x \right) \text{d}x}+af\left( b \right) -bf\left( a \right)}{f\left( b \right) -f\left( a \right)} \]

将其代入至极限式中，则有

\[\begin{align*} \lim_{b\rightarrow a^+} \frac{\xi -a}{b-a}&=\lim_{b\rightarrow a^+} \frac{\frac{\int_a^b{f\left( x \right) \text{d}x}+af\left( b \right) -bf\left( a \right)}{f\left( b \right) -f\left( a \right)}-a}{b-a} \\ &=\lim_{b\rightarrow a^+} \frac{\int_a^b{f\left( x \right) \text{d}x}+af\left( b \right) -bf\left( a \right) -a\left[ f\left( b \right) -f\left( a \right) \right]}{\left( b-a \right) \left[ f\left( b \right) -f\left( a \right) \right]} \\ &=\lim_{b\rightarrow a^+} \frac{\int_a^b{f\left( x \right) \text{d}x}-bf\left( a \right) +af\left( a \right)}{\left( b-a \right) ^2}\cdot \frac{b-a}{f\left( b \right) -f\left( a \right)} \\ &=\frac{1}{f'\left( a \right)}\lim_{b\rightarrow a^+} \frac{\int_a^b{f\left( x \right) \text{d}x}-bf\left( a \right) +af\left( a \right)}{\left( b-a \right) ^2} \\ &=\frac{1}{f'\left( a \right)}\lim_{b\rightarrow a^+} \frac{f\left( b \right) -f\left( a \right)}{2\left( b-a \right)} \\ &=\frac{1}{2f'\left( a \right)}\times f'\left( a \right) \\ &=\frac{1}{2} \end{align*} \]

其中用到了洛必达法则及导数的定义。

对于上述过程中，如果不想洛必达，也可进行\(\text{Taylor}\)展开，如下所示：

\[\begin{align*} \int_a^b{f\left( x \right) \text{d}x}&=\int_a^a{f\left( x \right) \text{d}x}+\left( b-a \right) f\left( a \right) +\frac{\left( b-a \right) ^2}{2!}f'\left( \eta \right) \\ &=\left( b-a \right) f\left( a \right) +\frac{\left( b-a \right) ^2}{2!}f'\left( \eta \right) \end{align*} \]

其中\(\displaystyle a<\eta <b\)，对于过程中的一块极限，则可以如下进行处理：

\[\begin{align*} \lim_{b\rightarrow a^+} \frac{\int_a^b{f\left( x \right) \text{d}x}-bf\left( a \right) +af\left( a \right)}{\left( b-a \right) ^2}&=\lim_{b\rightarrow a^+} \frac{\left( b-a \right) f\left( a \right) +\frac{\left( b-a \right) ^2}{2!}f'\left( \eta \right) -bf\left( a \right) +af\left( a \right)}{\left( b-a \right) ^2} \\ &=\frac{1}{2}\lim_{b\rightarrow a^+} f'\left( \eta \right) \\ &=\frac{1}{2}f'\left( a \right) \end{align*} \]