本題見於考研數學李林老師的微信公眾號“李林考研數學”
問題：設\(\displaystyle f\left( x \right)\)在\(\displaystyle\left[ a,b \right]\)上可導，且\(\displaystyle f'\left( x \right) \ne 0\)。
(1)證明：至少存在一點\(\displaystyle\xi \in \left( a,b \right)\)，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)對(1)中的\(\displaystyle\xi\)，求
\[\lim_{b\rightarrow a^+} \frac{\xi -a}{b-a} \]
過程如下：
(1)由於\(\displaystyle f'\left( x \right) \ne 0\)，不妨設\(\displaystyle f'\left( x \right) >0\)
采用零點定理證明\(\displaystyle\xi\)的存在性，令
\[F\left( x \right) =f\left( b \right) \left( x-a \right) +f\left( a \right) \left( b-x \right) -\int_a^b{f\left( x \right) \text{d}x} \]
則有
\[\begin{equation*} F\left( a \right) =f\left( a \right) \left( b-a \right) -\int_a^b{f\left( x \right) \text{d}x}=\int_a^b{\left[ f\left( a \right) -f\left( x \right) \right] \text{d}x}<0 \\ F\left( b \right) =f\left( b \right) \left( b-a \right) -\int_a^b{f\left( x \right) \text{d}x}=\int_a^b{\left[ f\left( b \right) -f\left( x \right) \right] \text{d}x}>0 \end{equation*} \]
函數\(\displaystyle F\left( x \right)\)滿足零點定理的條件，由定理內容可知至少存在一點\(\displaystyle\xi \in \left( a,b \right)\)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)基於問題(1)得到的結論，不難發現可將\(\displaystyle\xi\)寫成用\(\displaystyle a\)與\(\displaystyle b\)構成的表達式，得到
\[\xi =\frac{\int_a^b{f\left( x \right) \text{d}x}+af\left( b \right) -bf\left( a \right)}{f\left( b \right) -f\left( a \right)} \]
將其代入至極限式中，則有
\[\begin{align*} \lim_{b\rightarrow a^+} \frac{\xi -a}{b-a}&=\lim_{b\rightarrow a^+} \frac{\frac{\int_a^b{f\left( x \right) \text{d}x}+af\left( b \right) -bf\left( a \right)}{f\left( b \right) -f\left( a \right)}-a}{b-a} \\ &=\lim_{b\rightarrow a^+} \frac{\int_a^b{f\left( x \right) \text{d}x}+af\left( b \right) -bf\left( a \right) -a\left[ f\left( b \right) -f\left( a \right) \right]}{\left( b-a \right) \left[ f\left( b \right) -f\left( a \right) \right]} \\ &=\lim_{b\rightarrow a^+} \frac{\int_a^b{f\left( x \right) \text{d}x}-bf\left( a \right) +af\left( a \right)}{\left( b-a \right) ^2}\cdot \frac{b-a}{f\left( b \right) -f\left( a \right)} \\ &=\frac{1}{f'\left( a \right)}\lim_{b\rightarrow a^+} \frac{\int_a^b{f\left( x \right) \text{d}x}-bf\left( a \right) +af\left( a \right)}{\left( b-a \right) ^2} \\ &=\frac{1}{f'\left( a \right)}\lim_{b\rightarrow a^+} \frac{f\left( b \right) -f\left( a \right)}{2\left( b-a \right)} \\ &=\frac{1}{2f'\left( a \right)}\times f'\left( a \right) \\ &=\frac{1}{2} \end{align*} \]
其中用到了洛必達法則及導數的定義。
對於上述過程中，如果不想洛必達，也可進行\(\text{Taylor}\)展開，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\(\displaystyle a<\eta <b\)，對於過程中的一塊極限，則可以如下進行處理：
\[\begin{align*} \lim_{b\rightarrow a^+} \frac{\int_a^b{f\left( x \right) \text{d}x}-bf\left( a \right) +af\left( a \right)}{\left( b-a \right) ^2}&=\lim_{b\rightarrow a^+} \frac{\left( b-a \right) f\left( a \right) +\frac{\left( b-a \right) ^2}{2!}f'\left( \eta \right) -bf\left( a \right) +af\left( a \right)}{\left( b-a \right) ^2} \\ &=\frac{1}{2}\lim_{b\rightarrow a^+} f'\left( \eta \right) \\ &=\frac{1}{2}f'\left( a \right) \end{align*} \]