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题:\(y=f(x),\;\;f\in\mathscr{C}'([a,b])\,\)的曲线长\(\small\;S=\displaystyle\int_a^b\sqrt{1+(f')^2}.\)
证:对\(\,n\in\mathbb{N}^+,\,\)记\(\,h={\large\frac{b-a}{n}},\,x_k=a+kh,{\small\,M_k=}(x_k,f(x_k)).\)
\(\qquad\)记\(\,\eta(x,h){\small=}\frac{f(x+h)-f(x)}{h}\small-f'(x).\,\)因\(\,f'\,\)在\([a,b]\)一致连续,
\(\qquad\eta_h\small=\max\{|f'(s)-f'(t)|\mid s,t\in[a,b],\;|s-t|\le h\}\overset{(h\to 0)}{\longrightarrow}\ 0\underset{\,}{.}\)
\(\therefore\quad|{\small M_{k-1}M_k}|=\sqrt{h^2+(f(x_k)-f(x_{k-1}))^2}\)
\(\qquad\qquad=h\sqrt{1+(f'(x_{k-1})+\eta(x_{k-1},h))^2}\)
\(\qquad\qquad=h\sqrt{1+(f'(x_{k-1})^2)}+h\Delta_k.\underset{\,}{\;}\)其中
\(\qquad|\Delta_k|=\left|\sqrt{{\small 1+}(f'(x_{k-1})+\eta(x_{k-1},h))^2}-\sqrt{{\small 1+}(f'(x_{k-1})^2)}\right|\)
\(\qquad\qquad=\large\frac{|\eta(x_{k-1},h)(2f'(x_{k-1})+\eta(x_{k-1},h)))|}{\sqrt{1+(f'(x_{k-1})+\eta(x_{k-1},h))^2}+\sqrt{1+(f'(x_{k-1})^2)}}\)
\(\qquad\qquad\le\overset{\,}{\large\frac{1}{2}}\big|\eta(x_{k-1},h)(2f'(x_{k-1})+\eta(x_{k-1},h)))\big|\)
\(\qquad\qquad\le 2\eta_h\max|f'|\)
\(\therefore\quad\displaystyle\big|{\small\sum_{k=1}^n} h\Delta_k\big|\le 2nh\eta_h\max|f'|=2\eta_h(b-a)\max|f'|\overset{(n\to\infty)}{\longrightarrow}\ 0\)
\(\therefore\quad\displaystyle{\small S=\lim_{n\to\infty}\sum_{k=1}^n|M_{k-1}M_k|=\lim_{n\to\infty}\big(\sum_{k=1}^n}h{\small\sqrt{1+(f(x_{k-1}))^2}+\sum_{k=1}^n}h\Delta_k\big)\)
\(\qquad\quad\displaystyle{\small=\int_a^b\sqrt{1+(f'(x))^2}}dx.\quad\small\square\) |
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