91Ó°ÊÓ

Consider that f(x) is a continuous function on an interval [a,b] and n is a positive integer.

The function y = f(x) has a continuous graph on the interval [a, b] , which has been divided into subintervals with k ^th subdivision point as x_k for k=0,1,2,...,) . Thus the length triangle $â–³x$of each subinterval is defined by$â–³x=\frac{b-a}{n}$ Now, if $x_{k-1}and{x}_k$are the end points of the k^th interval, then role="math" localid="1649458198246" $â–³x=x_{k+1}-x_k$ and the corresponding values of y are f(x_k-1 ) and f(x₂) . So, the length of the line segment joining these points on the graph of the curve is given by the distance formula as

role="math" localid="1649458234113" $\sqrt{{\left(x_k-x_{k-1}\right)}^2+\left(f(x_k\right)-f{\left(x_{k-1}\right)}^2}$

The arc length of the function is then approximated by the sum of all such line segments. That is the arc length is approximated by the expression

$\sqrt{{\left(x_k-x_{k-1}\right)}^2+\left(f(x_k\right)-f{\left(x_{k-1}\right)}^2}$

Replacing $x_{k+1}-x_k$the width of the k ^th subinterval triangle x and $\left(f(x_k\right)-f{\left(x_{k-1}\right)}^2$by the ordinate $â–³y_k$of the k^th subinterval, the above expression equals

$$\begin{gathered} \underset{k=1}{\overset{∞}{∑}}\sqrt{{\left(△x\right)}^2+{\left(△y_k\right)}^2} \\ S\mathrm{ince}\mathrm{triangle}△\mathrm{x}\mathrm{is}\mathrm{a}\mathrm{positive}\mathrm{number},\mathrm{write}\mathrm{the}\mathrm{last}\mathrm{expression}\mathrm{as} \\ \underset{\mathrm{k}=1}{\overset{\mathrm{n}}{∑}}\sqrt{1+{\left(\frac{△{\mathrm{y}}_{\mathrm{k}}}{△\mathrm{x}}\right)}^2△\mathrm{x}} \\ \mathrm{Hence}\mathrm{Proved}\underset{\mathrm{k}=1}{\overset{∞}{∑}}\sqrt{{\left({\mathrm{x}}_{\mathrm{k}}-{\mathrm{x}}_{\mathrm{k}-1}\right)}^2+\left(\mathrm{f}({\mathrm{x}}_{\mathrm{k}}\right)-\mathrm{f}{\left({\mathrm{x}}_{\mathrm{k}-1}\right)}^2}=\underset{\mathrm{k}=1}{\overset{\mathrm{n}}{∑}}\sqrt{1+{\left(\frac{△{\mathrm{y}}_{\mathrm{k}}}{△\mathrm{x}}\right)}^2△\mathrm{x}} \end{gathered}$$