47203 利用切線段的總長建構弧長的上和

### 第一節 上下和與面積

\begin{align} \hbox{上和}=\,&(x_1-x_0 ) M_1+(x_2-x_1 ) M_2+(x_3-x_2 ) M_3,\label{1}\\ \hbox{下和}=\,&(x_1-x_0 ) m_1+(x_2-x_1 ) m_2+(x_3-x_2 ) m_3;\label{2} \end{align}

\begin{align} \hbox{黎曼和}=(x_1-x_0 )f(z_1)+(x_2-x_1 )f(z_2)+(x_3-x_2 )f(z_3);\label{3} \end{align}

### 第二節 以下和方式定義弧長

\begin{align} [(x_i-x_{i-1} )^2+(y_i-y_{i-1} )^2 ]^{\frac 12}=(x_i-x_{i-1} ) \left[1+\Big(\frac{y_i-y_{i-1}}{x_i-x_{i-1}}\Big)^2 \right]^{\frac 12}.\label{4} \end{align}

$$\frac{y_i-y_{i-1}}{x_i-x_{i-1}}=f' (z_i ),\quad x_{i-1}\le z_i\le x_i.$$

\eqref{4} 式變成

\begin{align} (x_i-x_{i-1} ) \left[1+\big(f' (z_i )\big)^2 \right]^{\frac 12}.\label{5} \end{align}

$$\int_a^b\Big(1+f'(x)^2\Big)^{\frac 12} dx.$$

### 第三節 以上和方式定義弧長

$$\overline{P_0Q_1}+\overline{Q_1Q_2}+\overline{Q_2Q_3}+\overline{Q_3P_3}\ge \overline{P_0P_1}+\overline{P_1P_2}+\overline{P_2P_3},$$

$$\big((u_i-u_{i-1})^2+(v_i-v_{i-1} )^2 \big)^{\frac 12}=(u_i-u_{i-1} ) \left(1+\Big(\frac{v_i-v_{i-1}}{u_i-u_{i-1}}\Big)^2 \right)^{\frac 12}.$$

\begin{align} &\hskip -20pt (u_i-u_{i-1}) \big(1+f'(x_{i-1} )^2 \big)^{\frac 12}.\label{6}\\ {\hbox{而 }} Q_1 P_0=\,&\big((u_1-x_0 )^2+(v_1-y_0 )^2 \big)^{\frac 12}\nonumber\\ =\,&(u_1-x_0 ) \left(1+\Big(\frac{v_1-y_0}{u_1-x_0}\Big)^2 \right)^{\frac 12}\nonumber\\ =\,&(u_1-x_0 ) (1+f'(x_0 )^2 )^{\frac 12}.\label{7} \end{align}

\begin{align} &\hskip -20pt (u_1-x_0 ) \big(1+f' (x_0 )^2 \big)^{\frac 12}+(u_2-u_1 ) \big(1+f' (x_1)^2 \big)^{\frac 12}+\cdots\nonumber\\ &+(u_n-u_{n-1} ) \big(1+f' (x_{n-1} )^2 \big)^{\frac 12}+(x_n-u_n ) \big(1+f' (x_n )^2 \big)^{\frac 12}.\label{8} \end{align}

### 第四節 回顧

\begin{align} L'(x)-f'(x)=f'(c)-f'(x).\label{9} \end{align}

$y=g(x)$ 圖形的反曲點在原點, 並以原點為對稱中心, 即圖形繞原點旋轉 $180^\circ$°, 圖形不變。

$\displaystyle\int_{-1}^0+ \int_0^1= \displaystyle\int_{-1}^1\sqrt{1+f' (x)^2}\,dx.$