
Suppose $f$ is continuously differentiable on the interval $[a,b]$. Let's derive a formula for the length $L$ of the curve on the interval, called the arc length over $[a,b]$.
We'll start by subdividing the interval $[a,b]$ into $n$ subintervals
$[x_0, x_1], [x_1, x_2],~...~, [x_{n1},x_n]$
Introduce the line segments between The resulting polygonal path approximates the curve given by $y=f(x)$, and its length approximates the arc length of $f(x)$ over $[a,b]$. Let's find the length of the polygonal path by adding up the lengths of the individual line segments. The $k$th line segment is the hypotenuse of a triangle with base $\Delta x_k$ and height $f(x_k)f(x_{k1})$, and so has length \[L_k=\sqrt{\left(\Delta x_k \right)^2+\left[f(x_k)f(x_{k1})\right]^2} {\small\textrm{.}} \] By the Mean Value Theorem, there exists $x_k^*\in [x_{k1},x_k]$ such that \[\frac{f(x_k)f(x_{k1})}{x_kx_{k1}}=f'(x_k^*)\] so \[f(x_k)f(x_{k1})=f'(x_k^*)(x_kx_{k1})=f'(x_k^*)\Delta x_k.\] Thus, \[L_k=\sqrt{(\Delta x_k)^2+[f'(x_k^*)\Delta x_k]^2}=\sqrt{1+[f'(x_k^*)]^2}\, \Delta x_k.\] Finally, the length of the entire polygonal path is \[\sum^n_{k=1} L_k=\sum^n_{k=1} \sqrt{1+[f'(x_k^*)]^2}\, \Delta x_k\] which has the form of a Riemann sum. Increasing the number of subintervals such that $\max \Delta x_k \to 0$, $\, \sum^n_{k=1} L_k \to L$. That is, \[L=\lim_{\max \Delta x_k\to 0}\sum^n_{k=1} \sqrt{1+[f'(x_k^*)]^2}\, \Delta x_k=\int^b_a \sqrt{1+[f'(x)]^2}\, dx\] by the definition of the definite integral as a limit of Riemann sums. Thus, we have proved the following:
Arc LengthLet $f(x)$ be continuously differentiable on $[a,b]$. Then the arc length $L$ of $f(x)$ over $[a,b]$ is given by \[L=\int^b_a \sqrt{1+[f'(x)]^2}\, dx.\] Similarly, if $x=g(y)$ with $g$ continuously differentiable on $[c,d]$, then the arc length $L$ of $g(y)$ over $[c,d]$ is given by \[L=\int^d_c \sqrt{1+[g'(y)]^2}\, dy.\] These integrals often can only be computed using numerical methods.
ExampleWe can compute the arc length of the graph of $f(x)=x^{3/2}$ over $[0,1]$ as follows: \begin{eqnarray*} L=\int^1_0 \sqrt{1+[f'(x)]^2}\, dx &=& \int^1_0 \sqrt{1+[3x^{1/2}/2]^2}\, dx\\ &=& \int^1_0 \sqrt{1+9x/4}\, dx\\ &=& \left.\frac{8}{27}(1+9x/4)^{3/2}\right^1_0\\ &=& (1+9/4)^{3/2}(1)^{3/2}\\ &=& (13/4)^{3/2}1\\ &\approx & 1.44. \end{eqnarray*} Key Concepts 