91Ó°ÊÓ

The arc length of a sufficiently well behaved function f(x) on an interval [a, b] can be represented by the definite$I={∫}_a^b\sqrt{1+{\left(f\text{'}\left(x\right)\right)}^{2}dx}$.

Recall that an approximation of the arc length of the function f(x) has been defined as the limit of the sum of line segments given by $\underset{k=1}{\overset{n}{∑}}\sqrt{1+{\left(\frac{△y_k}{△x}\right)}^2△x}$

This limit of the sum in equation (2) gives rise to the definite integral (1) in the event when $\frac{â–³y}{â–³x}$

equals f'(x) as n. The Mean Value Theorem is used to prove that$$\begin{gathered} \underset{\mathrm{π}→∞}{\lim }\frac{△\mathrm{y}}{△\mathrm{x}}=\mathrm{f}\text{'}\left(\mathrm{x}\right) \\ \mathrm{which}\mathrm{leads}\mathrm{to}\mathrm{transformation}\mathrm{of}\mathrm{the}\mathrm{limit}\mathrm{of}\mathrm{the}\mathrm{sum}\left(2\right)\mathrm{to}\mathrm{the}\mathrm{definite}\mathrm{integral}\left(1\right). \end{gathered}$$

Use of Mean Value Theorem: Since f(x) is a differentiable function, the Mean Value Theorem guarantees that there exists some point x^k prime on the interval [x _k-1 ,x _k ] at which

$$\begin{gathered} f\text{'}\left(x_k\right)=\frac{f\left(x_k\right)-f\left(x_{k-1}\right)}{x_k-x_{k-1}} \\ =\frac{△y_k}{△x} \\ \mathrm{Therefore},\mathrm{there}\mathrm{is}\mathrm{a}\mathrm{point}{\mathrm{x}}_{\mathrm{k}}\mathrm{in}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{sub}\mathrm{interval}\mathrm{such}\mathrm{that}\mathrm{the}\mathrm{definition}\mathrm{of}\mathrm{the}\mathrm{arc}\mathrm{length}\mathrm{can}\mathrm{be}\mathrm{expressed}\mathrm{as} \\ \underset{\mathrm{n}→∞}{\lim }\underset{\mathrm{k}=1}{\overset{\mathrm{n}}{∑}}\sqrt{1+{\left(\frac{△{\mathrm{y}}_{\mathrm{k}}}{△\mathrm{x}}\right)}^2}=\underset{\mathrm{n}→∞}{\lim }\underset{\mathrm{k}=1}{\overset{\mathrm{n}}{∑}}\sqrt{1+{\left(\mathrm{f}\text{'}\left({\mathrm{x}}_{\mathrm{k}}\right)\right)}^2} \\ \mathrm{The}\mathrm{derivative}\mathrm{f}\text{'}\left(\mathrm{x}\right)\mathrm{has}\mathrm{been}\mathrm{assumed}\mathrm{to}\mathrm{be}\mathrm{continous},\mathrm{accordingly}\mathrm{the}\mathrm{function} \\ {1+\left(\mathrm{f}\text{'}\left({\mathrm{x}}_{\mathrm{k}}\right)\right)}^2\mathrm{is}\mathrm{continuous}.\mathrm{Since}△\mathrm{x}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\mathrm{and}{\mathrm{x}}_{\mathrm{k}}=\mathrm{a}+\mathrm{k}△\mathrm{x},\mathrm{the}\mathrm{limit}\mathrm{of}\mathrm{sums}\mathrm{represents}\mathrm{the}\mathrm{definite}\mathrm{integral}[\mathrm{a},\mathrm{b}].\mathrm{That}\mathrm{is} \\ \underset{\mathrm{n}→∞}{\lim }\underset{\mathrm{k}=1}{\overset{\mathrm{n}}{∑}}\sqrt{1+{\left(\mathrm{f}\text{'}\left({\mathrm{x}}_{\mathrm{k}}\right)\right)}^2}={∫}_{\mathrm{a}}^{\mathrm{b}}\sqrt{1+{\left(\mathrm{f}\text{'}\left(\mathrm{x}\right)\right)}^2\mathrm{dx}} \end{gathered}$$