# 第二积分中值定理的一个非常规证明An Unconventional Proof of the Second Integral Mean Value Theorem

DOI: 10.12677/AAM.2020.911227, PDF, HTML, XML, 下载: 61  浏览: 114

Abstract: The second integral mean value theorem is an independent theorem from the first integral mean value theorem, which belongs to the integral mean value theorem. It can be used to prove the Riemann integral criterion for Dirichlet Abel anomaly. In mathematics competition and postgraduate entrance examination, there are often problems related to the second integral mean value theorem. In this paper, we prove the second mean value theorem of integrals by using the Weierstrass approximation theorem and Bernstein polynomials.

$f\left(x\right)$ 是闭区间 $\left[a,b\right]$ 上的连续函数，则对任意给定的 $\epsilon >0$，存在多项式 $P\left(x\right)$，使得 $|P\left(x\right)-f\left(x\right)|<\epsilon$ 对于一切 $x\in \left[a,b\right]$ 成立。

$f:\left[0,1\right]\to R$，称 ${B}_{n}\left(f;x\right)=\underset{i=0}{\overset{n}{\sum }}f\left(\frac{i}{n}\right)\left(\begin{array}{l}n\hfill \\ i\hfill \end{array}\right){x}^{i}{\left(1-x\right)}^{n-i}$ 为f的n次Bernstein多项式。

(1) ${B}_{n}\left(f;0\right)=f\left(0\right),{B}_{n}\left(f;1\right)=f\left(1\right)$

(2) 如果 $f\ge 0$ 那么 ${B}_{n}\left(f\right)\ge 0$，如果 $f\le 0$，那么 ${B}_{n}\left(f\right)\le 0$

(3) 如果f递增(减)，那么 ${B}_{n}\left(f\right)$ 也递增(减)；

(4) 如果f是上凸(下凸)函数，那么 ${B}_{n}\left(f\right)$ 也是上凸(下凸)函数。

$f\left(x\right)$$\left[a,b\right]$ 上可积， $g\left(x\right)$$\left[a,b\right]$ 上单调,则存在 $\xi \in \left[a,b\right]$，使得

${\int }_{a}^{b}f\left(x\right)g\left(x\right)\text{d}x=g\left(a\right){\int }_{a}^{\xi }f\left(x\right)\text{d}x+g\left(b\right){\int }_{\xi }^{b}f\left(x\right)\text{d}x$

$\begin{array}{c}{\int }_{a}^{b}f\left(x\right)g\left(x\right)\text{d}x={F\left(x\right)g\left(x\right)|}_{a}^{b}-{\int }_{a}^{b}F\left(x\right){g}^{\prime }\left(x\right)\text{d}x\\ =g\left(b\right){\int }_{a}^{b}f\left(x\right)\text{d}x-{\int }_{a}^{b}F\left(x\right){g}^{\prime }\left(x\right)\text{d}x\\ =g\left(b\right){\int }_{a}^{b}f\left(x\right)\text{d}x-F\left(\xi \right){\int }_{a}^{b}{g}^{\prime }\left(x\right)\text{d}x\\ =g\left(b\right){\int }_{a}^{b}f\left(x\right)\text{d}x-\left[g\left(b\right)-g\left(a\right)\right]{\int }_{a}^{\xi }f\left(t\right)\text{d}t\\ =g\left(a\right){\int }_{a}^{\xi }f\left(x\right)\text{d}x+g\left(b\right){\int }_{\xi }^{b}f\left(x\right)\text{d}x\end{array}$

$\begin{array}{l}|{f}_{n}\left(x\right){g}_{n}\left(x\right)-f\left(x\right)g\left(x\right)|\\ \le |{f}_{n}\left(x\right){g}_{n}\left(x\right)-f\left(x\right){g}_{n}\left(x\right)|+|f\left(x\right){g}_{n}\left(x\right)-f\left(x\right)g\left(x\right)|\\ \le \mathrm{max}\left\{g\left(a\right),g\left(b\right)\right\}|{f}_{n}\left(x\right)-f\left(x\right)|+|f\left(x\right)||{g}_{n}\left(x\right)-g\left(x\right)|\end{array}$

${f}_{n}⇒f,{g}_{n}⇒g,x\in \left[a,b\right]$，且f可积即f有界，易知 ${f}_{n}{g}_{n}⇒fg,x\in \left[a,b\right]$。所有的 ${\xi }_{n}$ 构成有界数列 $\left\{{\xi }_{n}\right\}$，取其收敛子列，不妨仍记为 $\left\{{\xi }_{n}\right\}$，设其收敛于 $\xi \in \left[a,b\right]$，则

$\begin{array}{l}|{\int }_{a}^{{\xi }_{n}}{f}_{n}\left(x\right)\text{d}x-{\int }_{a}^{\xi }f\left(x\right)\text{d}x|\\ =|{\int }_{a}^{{\xi }_{n}}{f}_{n}\left(x\right)\text{d}x-{\int }_{a}^{{\xi }_{n}}f\left(x\right)\text{d}x-{\int }_{{\xi }_{n}}^{\xi }f\left(x\right)\text{d}x|\\ \le {\int }_{a}^{b}|{f}_{n}\left(x\right)-f\left(x\right)|\text{d}x+|{\int }_{{\xi }_{n}}^{\xi }f\left(x\right)\text{d}x|\to 0,n\to \infty \end{array}$

$\underset{n\to \infty }{\mathrm{lim}}{\int }_{a}^{{\xi }_{n}}{f}_{n}\left(x\right)\text{d}x={\int }_{a}^{\xi }f\left(x\right)\text{d}x$

$\underset{n\to \infty }{\mathrm{lim}}{\int }_{{\xi }_{n}}^{b}{f}_{n}\left(x\right)\text{d}x={\int }_{\xi }^{b}f\left(x\right)\text{d}x$

${\int }_{a}^{b}f\left(x\right)g\left(x\right)\text{d}x=g\left(a\right){\int }_{a}^{\xi }f\left(x\right)\text{d}x+g\left(b\right){\int }_{\xi }^{b}f\left(x\right)\text{d}x$

${\int }_{a}^{b}|f\left(x\right)-h\left(x\right)|\text{d}x<\epsilon$。由引理，取 ${\epsilon }_{n}=\frac{1}{n}$，则 $\exists {h}_{n}\in C\left[a,b\right]$，s.t. ${\int }_{a}^{b}|f\left(x\right)-{h}_{n}\left(x\right)|\text{d}x<\frac{1}{n}$

$\underset{n\to \infty }{\mathrm{lim}}{\int }_{a}^{b}|f\left(x\right)-{h}_{n}\left(x\right)|\text{d}x=0$

$0\le |{\int }_{a}^{b}\left[f\left(x\right)-{h}_{n}\left(x\right)\right]\text{d}x|\le {\int }_{a}^{b}|f\left(x\right)-{h}_{n}\left(x\right)|\text{d}x$，故 $\underset{n\to \infty }{\mathrm{lim}}{\int }_{a}^{b}{h}_{n}\left(x\right)\text{d}x={\int }_{a}^{b}f\left(x\right)\text{d}x$，利用上述两个极限式

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