Suppose $f$ is continuously differentiable on the interval $[a,b]$.

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 $[x_0, x_1], [x_1, x_2],~…~, [x_{n-1},x_n]$

where $ a=x_0 < x_1< … < x_{n-1} < x_n=b $.

Introduce the line segments between

$(x_0, f(x_0)) {\small\textrm{ and }} (x_1,
f(x_1)),$

$(x_1, f(x_1)) {\small\textrm{ and }} (x_2, f(x_2)), … ,$

$(x_{n-1}, f(x_{n-1})) {\small\textrm{ and }} (x_n, f(x_n))$.

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 $f(x_k)-f(x_{k-1})$, and so has length

\[L_k=\sqrt{\left(\Delta x_k \right)^2+\left[f(x_k)-f(x_{k-1})\right]^2} {\small\textrm{.}} \] By the Mean Value Theorem, there exists $x_k^*\in [x_{k-1},x_k]$ such that \[\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}}=f'(x_k^*)\] so \[f(x_k)-f(x_{k-1})=f'(x_k^*)(x_k-x_{k-1})=f'(x_k^*)\Delta x_k.\] 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 Length

Let $f(x)$ 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.

#### Example

We 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

Let $f(x)$ 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
\[L = \int^d_c \sqrt{1+[g'(y)]^2}\, dy\]