Arc Length Formula
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 2.4: “Arc Length of a Curve and Surface Area” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/2-4-arc-length-of-a-curve-and-surface-area |
| Textbook used in class | Stewart, Calculus, Section 8.1: “Arc Length” |
Opening Scenario
A straight segment has length $\Delta x$ along the $x$-axis. But if a curve rises by $\Delta y$ while moving $\Delta x$ horizontally, the actual distance along the curve is $\sqrt{(\Delta x)^2 + (\Delta y)^2}$ by the Pythagorean theorem. To find the total length of a curve from $x = a$ to $x = b$, sum these diagonal distances over infinitely many infinitesimal pieces: $\displaystyle\int_a^b \sqrt{1 + [f'(x)]^2}\,dx$.
Quick Reference
Arc length formula: If $f'$ is continuous on $[a, b]$, then the length of the curve $y = f(x)$ from $x = a$ to $x = b$ is: $$L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx.$$
Alternative (curve $x = g(y)$ from $y = c$ to $y = d$): $$L = \int_c^d \sqrt{1 + [g'(y)]^2}\,dy.$$
Key Concepts
1. Derivation
Partition $[a, b]$ into $n$ subintervals. On the $i$-th subinterval, the curve goes from $(x_{i-1}, f(x_{i-1}))$ to $(x_i, f(x_i))$. The straight-line distance between these points is: $$\sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(\Delta x)^2 + (f(x_i) - f(x_{i-1}))^2}.$$
By the Mean Value Theorem, $f(x_i) - f(x_{i-1}) = f'(x_i^*)\,\Delta x$ for some $x_i^* \in (x_{i-1}, x_i)$. So the piece length is: $$\sqrt{(\Delta x)^2 + (f'(x_i^*))^2(\Delta x)^2} = \sqrt{1 + [f'(x_i^*)]^2}\,\Delta x.$$
Summing and taking the limit gives $L = \int_a^b\sqrt{1+[f'(x)]^2}\,dx$.
2. The Integrand Is Often Hard
The arc length integrand $\sqrt{1 + [f'(x)]^2}$ is not usually simple to integrate. Most functions give rise to integrals with no elementary antiderivative. The problems in Stewart that admit exact answers are carefully chosen so that $1 + [f'(x)]^2$ becomes a perfect square under the radical.
Trick: when $1 + (f')^2$ is a perfect square. If $f'(x) = \dfrac{a^{x} - a^{-x}}{2}$ (like the derivative of $\cosh x$), then $(f')^2 = \dfrac{a^{2x} - 2 + a^{-2x}}{4}$, and $1 + (f')^2 = \dfrac{a^{2x} + 2 + a^{-2x}}{4} = \left(\dfrac{a^x + a^{-x}}{2}\right)^2$.
3. Numerical Integration When Exact Fails
When the arc length integral has no elementary antiderivative, apply the Trapezoidal Rule, Simpson’s Rule, or a CAS to obtain a numerical approximation.
“Arc length is just $b - a$ (the horizontal distance).” $b - a$ is the horizontal span of the curve, not the length along the curve. The length is always at least $b - a$ (equality only when $f' = 0$ everywhere, i.e., the curve is horizontal). The $\sqrt{1 + (f')^2}$ factor accounts for the slope at every point.
Worked Example
Find the arc length of $y = \dfrac{x^{3/2}}{3} - x^{1/2}$ from $x = 1$ to $x = 4$.
Step 1 -- Differentiate. $f'(x) = \dfrac{1}{2}x^{1/2} - \dfrac{1}{2}x^{-1/2}$.
Step 2 -- Compute $1 + (f')^2$.
$(f')^2 = \dfrac{1}{4}x - \dfrac{1}{2} + \dfrac{1}{4}x^{-1}$.
$1 + (f')^2 = 1 + \dfrac{x}{4} - \dfrac{1}{2} + \dfrac{1}{4x} = \dfrac{x}{4} + \dfrac{1}{2} + \dfrac{1}{4x} = \left(\dfrac{\sqrt{x}}{2} + \dfrac{1}{2\sqrt{x}}\right)^2$.
Step 3 -- Take the square root.
$\sqrt{1+(f')^2} = \dfrac{\sqrt{x}}{2} + \dfrac{1}{2\sqrt{x}}$ (positive for $x \geq 1$).
Step 4 -- Integrate.
$L = \displaystyle\int_1^4\left(\frac{\sqrt{x}}{2} + \frac{1}{2\sqrt{x}}\right)dx = \left[\frac{x^{3/2}}{3} + \sqrt{x}\right]_1^4 = \left(\frac{8}{3}+2\right) - \left(\frac{1}{3}+1\right) = \frac{7}{3}+1 = \frac{10}{3}$.
Boxed answer: $L = \dfrac{10}{3}$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using $b - a$ as the arc length | $L = 4 - 1 = 3$ for a curve on $[1,4]$ | $L = \int_1^4\sqrt{1+(f')^2}\,dx \neq b-a$ unless $f' = 0$ |
| Forgetting to include the $1 +$ inside the radical | Writing $\int\sqrt{(f')^2}\,dx = \int|f'|\,dx$ | The formula includes 1: $\int\sqrt{1+(f')^2}\,dx$; omitting it gives the integral of $|f'|$, not arc length |
| Not checking whether $1+(f')^2$ simplifies | Attempting to integrate $\sqrt{1+x}$ when $1+(f')^2 = (1+x/4)^2$ is a perfect square | Always simplify $1+(f')^2$ before integrating |
Leveled Practice
Level 1 -- Setup
Problem 1. Write the arc length integral (do not evaluate) for $y = x^2$ from $x = 0$ to $x = 1$.
Show answer
$f'(x) = 2x$, $1+(f')^2 = 1 + 4x^2$.
$L = \displaystyle\int_0^1\sqrt{1+4x^2}\,dx$.
(This integral requires trig substitution or a table; it is typically left in this form or approximated numerically.)
Level 2 -- Perfect Square Trick
Problem 2. Find the arc length of $y = \dfrac{x^2}{4} - \dfrac{\ln x}{2}$ from $x = 1$ to $x = 2$.
Show answer
$f'(x) = \dfrac{x}{2} - \dfrac{1}{2x}$.
$(f')^2 = \dfrac{x^2}{4} - \dfrac{1}{2} + \dfrac{1}{4x^2}$.
$1+(f')^2 = \dfrac{x^2}{4}+\dfrac{1}{2}+\dfrac{1}{4x^2} = \left(\dfrac{x}{2}+\dfrac{1}{2x}\right)^2$.
$\sqrt{1+(f')^2} = \dfrac{x}{2}+\dfrac{1}{2x}$.
$L = \displaystyle\int_1^2\left(\frac{x}{2}+\frac{1}{2x}\right)dx = \left[\frac{x^2}{4}+\frac{\ln x}{2}\right]_1^2 = (1+\frac{\ln 2}{2}) - (\frac{1}{4}+0) = \frac{3}{4}+\frac{\ln 2}{2}$.
Mastery Checklist
Mental Model
Arc length accumulates diagonal steps along the curve. Each infinitesimal step has a horizontal component $dx$ and a vertical component $f'(x)\,dx$. The Pythagorean theorem gives the length of each step as $\sqrt{1+(f')^2}\,dx$. Integrating sums all these steps.
The integrand $\sqrt{1+(f')^2}$ is always at least 1 (since $1 + (f')^2 \geq 1$), so arc length is always at least $b - a$. A steeper curve has larger $|f'|$, so more of the length is “wasted” on vertical distance, and the arc length exceeds $b - a$ by more.
Connections
Looking back
- FTC (Section 4.3): Used to evaluate the arc length integral.
- MVT (Section 3.2): The derivation uses the MVT to write $\Delta y = f'(x_i^*)\,\Delta x$.
Looking ahead
- Arc length function (Section 8.1): The arc length from a fixed point $a$ to a variable point $x$ is a function of $x$.
- Surface area (Section 8.2): The arc length element $ds = \sqrt{1+(f')^2}\,dx$ appears in the surface area formula.
Back to Applications of Integration | Next: Arc Length as a Function