Q. 76 - Calculus

Question

 Use Definition 6.6, the Mean Value Theorem ,and the definition of the definite integral to prove Theorem 6.7: The arc length of a sufficiently well behaved function f(x) on an interval [a, b] can be represented by the definite $I={\int }_a^b\sqrt{1+{\left(f\text{'}\left(x\right)\right)}^{2}dx}$

Step-by-Step Solution

Answer

$\underset{\mathrm{n}\to \infty }{\lim }\sum _{\mathrm{k}=1}^{\mathrm{n}}\sqrt{1+{\left(\mathrm{f}\text{'}\left({\mathrm{x}}_{\mathrm{k}}\right)\right)}^2}={\int }_{\mathrm{a}}^{\mathrm{b}}\sqrt{1+{\left(\mathrm{f}\text{'}\left(\mathrm{x}\right)\right)}^2\mathrm{dx}}$

Step 1. To proof

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

Step 2. Formula used

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 $\sum _{k=1}^n\sqrt{1+{\left(\frac{△y_k}{△x}\right)}^2△x}$

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

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

Step 3. Use of mean value theorem

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

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