91Ó°ÊÓ

$L={∫}_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx$.

Let us consider a graph.

where$∆x=\frac{b-a}{n},x_k=a+k∆x,∆y_k=f\left(x_k\right)-f\left(x_{k-1}\right)$for all$k=0,1,2,3.....,n$.

$$\begin{gathered} l_k=\sqrt{{\left(∆x\right)}^2+{\left(∆y_k\right)}^2} \\ =\sqrt{{\left(∆x\right)}^2\left(1+\frac{{\left(∆y_k\right)}^2}{{\left(∆x\right)}^2}\right)} \\ =∆x\sqrt{1+{\left(\frac{∆y_k}{∆x}\right)}^2} \end{gathered}$$

And a number of subdivisions $n$, we have defined n subintervals, each with a line segment that approximates a small piece of the curve. Adding up the lengths of these line segments gives us an approximation for the length of the curve:

Length of $f\left(x\right)$on role="math" localid="1649144454037" $\left[a,b\right]≈\underset{k=1}{\overset{n}{∑}}{l}_k$

$=\underset{k=1}{\overset{n}{∑}}\sqrt{1+{\left(\frac{∆y_k}{∆x}\right)}^2}.∆x$.

[As $n→∞,\frac{∆y_k}{∆x}=f^{\text{'}}\left(x\right)$]

role="math" localid="1649144736719" $$\begin{gathered} =\underset{n→∞}{\lim }\underset{k=1}{\overset{n}{∑}}\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}∆x \\ ={∫}_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx \end{gathered}$$

Thus, the definite integral $L={∫}_a^b\sqrt{1+\left(f^{\text{'}}\left(x\right)\right)}dx$might have to done with the distance formula calculation.