Suppose that we are given a function $f$ that is continuous on an interval $[a,b]$ and we want to calculate the length of the curve drawn out by the graph of $f(x)$ from $x=a$ to $x=b$ . If the graph were a straight line this would be easy — the formula for the length of the line is given by Pythagoras' theorem. And if the graph were a piecewise linear function we can calculate the length by adding up the length of each piece.

The problem is that most graphs are not linear. Nevertheless we can estimate the length of the curve by approximating it with straight lines. Suppose the curve $C$ is given by the formula $y=f(x)$ for $a\leq x\leq b$ . We divide the interval $[a,b]$ into $n$ subintervals with equal width $\Delta x$ and endpoints $x_{0},x_{1},\ldots ,x_{n}$ . Now let $y_{i}=f(x_{i})$ so $P_{i}=(x_{i},y_{i})$ is the point on the curve above $x_{i}$ . The length of the straight line between $P_{i}$ and $P_{i+1}$ is

Now this is the definition of the integral of the function $g(x)={\sqrt {1+f'(x)^{2}}}$ between $a$ and $b$ (notice that $g$ is continuous because we are assuming that $f'$ is continuous). Hence

$L=\int \limits _{a}^{b}{\sqrt {1+f'(x)^{2}}}dx$

as claimed.

Example: Length of the curve $y=2x$ from $x=0$ to }$x=1$

As a sanity check of our formula, let's calculate the length of the "curve" $y=2x$ from $x=0$ to $x=1$ . First let's find the answer using the Pythagorean Theorem.

3. Find the circumference of the circle given by the parametric equations $x(t)=R\cos(t)$ , $y(t)=R\sin(t)$ , with $t$ running from $0$ to $2\pi$ .

$2\pi R$

4. Find the length of one arch of the cycloid given by the parametric equations $x(t)=R{\bigl (}t-\sin(t){\bigr )}$ , $y(t)=R{\bigl (}1-\cos(t){\bigr )}$ , with $t$ running from $0$ to $2\pi$ .