## 2020年1月22日水曜日

### 数学 - 解析学 - 積分の計算 - 定積分の計算 - 連続関数、テイラーの定理、積分の形の剰余項、帰納法と部分積分法

1. $\begin{array}{l}f\left(b\right)=f\left(a\right)+{R}_{1}\\ {R}_{1}=f\left(b\right)-f\left(a\right)\\ \frac{1}{\left(1-1\right)!}{\int }_{a}^{b}f\text{'}\left(x\right)\mathrm{dx}\\ ={\left[f\left(x\right)\right]}_{n}^{b}\\ =f\left(b\right)-f\left(a\right)\end{array}$

よって、

${R}_{1}=\frac{1}{\left(1-1\right)!}{\int }_{a}^{b}{\left(b-x\right)}^{1-1}{f}^{\left(1\right)}\left(x\right)\mathrm{dx}$

また、

$\begin{array}{l}{R}_{n}\\ =f\left(b\right)-\sum _{k=0}^{n-1}\frac{{\left(b-a\right)}^{k}}{k!}{f}^{\left(k\right)}\left(a\right)\\ =\sum _{k=0}^{n-2}\frac{{\left(b-a\right)}^{k}}{k!}{f}^{\left(k\right)}\left(a\right)+{R}_{n-1}-\sum _{k=0}^{n-1}\frac{{\left(b-a\right)}^{k}}{k!}{f}^{\left(k\right)}\left(a\right)\\ ={R}_{n-1}-\frac{{\left(b-a\right)}^{n-1}}{\left(n-1\right)!}{f}^{\left(n-1\right)}\left(a\right)\\ =\frac{1}{\left(n-2\right)!}{\int }_{a}^{b}{\left(b-x\right)}^{n-2}{f}^{\left(n-1\right)}\left(x\right)\mathrm{dx}-\frac{{\left(b-a\right)}^{n-1}}{\left(n-1\right)!}{f}^{\left(n-1\right)}\left(a\right)\\ =\frac{1}{\left(n-1\right)!}\left(-{\left(b-a\right)}^{n-1}{f}^{\left(n-1\right)}\left(a\right)+\left(n-1\right){\int }_{a}^{b}{\left(b-x\right)}^{n-2}{f}^{\left(n-1\right)}\left(x\right)\mathrm{dx}\right)\\ =\frac{1}{\left(n-1\right)!}\left({\left[{\left(b-x\right)}^{n-1}{f}^{\left(n-1\right)}\left(x\right)\right]}_{a}^{b}+\left(n-1\right){\int }_{a}^{b}{\left(b-x\right)}^{n-2}{f}^{\left(n-1\right)}\left(x\right)\mathrm{dx}\right)\\ =\frac{1}{\left(n-1\right)!}{\int }_{a}^{b}{\left(b-x\right)}^{n-1}{f}^{\left(n\right)}\left(x\right)\mathrm{dx}\end{array}$

よって、 帰納法により すべての 正の整数に対して成り立つ。

（証明終）