Arc Length

Quick answer

Over a short stretch a smooth curve is nearly straight, so the Pythagorean theorem measures each small piece: its length is √(Δx² + Δy²), which is √(1 + (Δy/Δx)²) Δx. Adding the pieces and taking the limit gives the arc length L = ∫ₐᵇ √(1 + f′(x)²) dx, or ∫ √(1 + g′(y)²) dy for a curve x = g(y). The integral often has no antiderivative in closed form, so a numerical integral finishes many problems; curves where 1 + f′² is a perfect square give exact answers.

What you'll learn

Derive the arc length formula from the Pythagorean theorem
Compute the arc length of y = f(x) or x = g(y)
Recognize integrands that simplify to a perfect square
Approximate arc length numerically and with chords

Measuring a curve with straight pieces

A ruler measures straight lines. To measure a curve, replace it by straight pieces: mark points along it, join neighbors with chords, and add up the chord lengths. More points make a closer fit.

y = x²

Four chords along y = x² from x = 0 to x = 2

The four chords total $4.627$ . The curve’s actual length is $4.647$ : a chord is the shortest path between its ends, so chords always come up a little short.

Why the integrand is √(1 + f′(x)²)

Take one small piece of the curve $y = f(x)$ over a run $\Delta x$ . Its chord rises $\Delta y$ , so by the Pythagorean theorem its length is

\sqrt{\Delta x^2 + \Delta y^2} = \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2}\;\Delta x

By the Mean Value Theorem, $\tfrac{\Delta y}{\Delta x} = f'(x^*)$ at some point $x^*$ of the piece. The chord lengths then form a Riemann sum for $\sqrt{1 + f'(x)^2}$ , and as the pieces shrink the sum becomes the integral. Arc length adds up tiny hypotenuses.

Definition

If $f'$ is continuous on $[a, b]$ , the arc length of $y = f(x)$ from $x = a$ to $x = b$ is

L = \int_a^b \sqrt{1 + f'(x)^2}\,dx

For a curve $x = g(y)$ , $c \le y \le d$ , swap the variables: $L = \int_c^d \sqrt{1 + g'(y)^2}\,dy$ .

When the integral cannot be done by hand

The square root in the formula rarely has an elementary antiderivative. Even the parabola $y = x^2$ leads to $\int \sqrt{1 + 4x^2}\,dx$ , which needs techniques beyond this course. In practice, set up the integral and evaluate it numerically with a calculator. Exact answers come from curves chosen so that $1 + f'(x)^2$ is a perfect square, as in Example 2.

Worked examples

Example 2 — A perfect square under the root

Find the length of $y = \tfrac{x^3}{6} + \tfrac{1}{2x}$ from $x = 1$ to $x = 2$ .

$f'(x) = \tfrac{x^2}{2} - \tfrac{1}{2x^2}$ , and squaring gives $f'(x)^2 = \tfrac{x^4}{4} - \tfrac{1}{2} + \tfrac{1}{4x^4}$ . Adding $1$ flips the middle sign:

1 + f'(x)^2 = \frac{x^4}{4} + \frac{1}{2} + \frac{1}{4x^4} = \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2

L = \int_1^2 \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)dx = \left[\frac{x^3}{6} - \frac{1}{2x}\right]_1^2 = \frac{13}{12} - \left(-\frac{1}{3}\right) = \frac{17}{12}

Common mistakes

Practice problems

Find the length of the line $y = 2x + 1$ from $x = 0$ to $x = 3$ , and check it with the distance formula.

Answer

$3\sqrt{5} \approx 6.71$

Full solution

$\int_0^3 \sqrt{1 + 2^2}\,dx = 3\sqrt{5}$ . The endpoints are $(0, 1)$ and $(3, 7)$ , and $\sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}$ .
Find the length of $y = \tfrac{2}{3}x^{3/2}$ from $x = 0$ to $x = 3$ .

Answer

$\tfrac{14}{3}$

Full solution

$f'(x) = x^{1/2}$ , so $1 + f'^2 = 1 + x$ . $L = \int_0^3 (1 + x)^{1/2}\,dx = \tfrac{2}{3}\big[(1 + x)^{3/2}\big]_0^3 = \tfrac{2}{3}(8 - 1) = \tfrac{14}{3}$ .
Find the length of $y = x^{3/2}$ from $x = 0$ to $x = 1$ .

Answer

$\tfrac{13\sqrt{13} - 8}{27} \approx 1.44$

Full solution

As in Example 1, $L = \tfrac{8}{27}\big[(1 + \tfrac{9}{4}x)^{3/2}\big]_0^1 = \tfrac{8}{27}\left(\left(\tfrac{13}{4}\right)^{3/2} - 1\right) = \tfrac{8}{27}\left(\tfrac{13\sqrt{13}}{8} - 1\right)$ .
Set up an integral for the length of one arch of $y = \sin x$ , $0 \le x \le \pi$ , and evaluate it numerically.

Answer

$\int_0^{\pi} \sqrt{1 + \cos^2 x}\,dx \approx 3.82$

Full solution

$f'(x) = \cos x$ . The integrand has no elementary antiderivative, so a calculator finishes it.
Find the length of $x = \tfrac{2}{3}(y - 1)^{3/2}$ from $y = 1$ to $y = 4$ .

Answer

$\tfrac{14}{3}$

Full solution

$g'(y) = (y - 1)^{1/2}$ , so $1 + g'^2 = y$ . $L = \int_1^4 y^{1/2}\,dy = \tfrac{2}{3}(8 - 1) = \tfrac{14}{3}$ .
Find the length of $y = \tfrac{x^4}{8} + \tfrac{1}{4x^2}$ from $x = 1$ to $x = 2$ .

Answer

$\tfrac{33}{16}$

Full solution

$f'(x) = \tfrac{x^3}{2} - \tfrac{1}{2x^3}$ , and $1 + f'^2 = \left(\tfrac{x^3}{2} + \tfrac{1}{2x^3}\right)^2$ . $L = \big[\tfrac{x^4}{8} - \tfrac{1}{4x^2}\big]_1^2 = \left(2 - \tfrac{1}{16}\right) - \left(\tfrac{1}{8} - \tfrac{1}{4}\right) = \tfrac{33}{16}$ .
Find the length of $y = \ln(\cos x)$ from $x = 0$ to $x = \tfrac{\pi}{4}$ . Use $\int \sec x\,dx = \ln\lvert \sec x + \tan x \rvert + C$ .

Answer

$\ln(\sqrt{2} + 1) \approx 0.881$

Full solution

$f'(x) = -\tan x$ , and $1 + \tan^2 x = \sec^2 x$ , so $L = \int_0^{\pi/4} \sec x\,dx = \ln(\sqrt{2} + 1) - \ln 1$ .
Approximate the length of $y = x^2$ from $x = 0$ to $x = 2$ with two chords, through $x = 0$ , $1$ and $2$ . Is the result an overestimate or an underestimate?

Answer

$\sqrt{2} + \sqrt{10} \approx 4.58$ , an underestimate

Full solution

The chords run from $(0, 0)$ to $(1, 1)$ and from $(1, 1)$ to $(2, 4)$ , with lengths $\sqrt{2}$ and $\sqrt{10}$ . Each chord is shorter than its arc, so the total is below the length $4.647$ .
Find the length of $y = \tfrac{e^x + e^{-x}}{2}$ from $x = 0$ to $x = 1$ .

Answer

$\tfrac{e - e^{-1}}{2} \approx 1.18$

Full solution

$f'(x) = \tfrac{e^x - e^{-x}}{2}$ , and $1 + f'^2 = \left(\tfrac{e^x + e^{-x}}{2}\right)^2$ , since the cross terms flip sign as in Example 2. So $L = \int_0^1 \tfrac{e^x + e^{-x}}{2}\,dx = \big[\tfrac{e^x - e^{-x}}{2}\big]_0^1$ .
A student finds the length of $y = x^2$ from $x = 0$ to $x = 1$ as $\int_0^1 (1 + 2x)\,dx = 2$ . What went wrong?

Hint

What is the length of a tiny piece with run $\Delta x$ and rise $\Delta y$ ?

Answer

The integrand should be $\sqrt{1 + (2x)^2}$ , not $1 + 2x$ . The length is about $1.479$ .

Full solution

$L = \int_0^1 \sqrt{1 + 4x^2}\,dx \approx 1.479$ numerically.

The student’s integrand is too big everywhere: $\sqrt{1 + a^2} < 1 + a$ for every $a > 0$ , because squaring the right side gives $1 + 2a + a^2$ . So $\int_0^1 (1 + 2x)\,dx$ must overshoot the length.

Frequently asked questions

What is the arc length formula?

For y = f(x) from x = a to x = b, the length is the integral from a to b of √(1 + f′(x)²) dx.

Where does the formula come from?

A tiny piece of the curve is nearly a straight segment with run Δx and rise Δy. Its length √(Δx² + Δy²) equals √(1 + (Δy/Δx)²) Δx, and summing the pieces gives the integral.

Why can't I evaluate most arc length integrals by hand?

The square root of 1 + f′² rarely has an elementary antiderivative. Textbook examples are chosen so that it simplifies; in general, use a numerical integral.

How do I find the length of x = g(y)?

Swap the roles of the variables: L = ∫ from c to d of √(1 + g′(y)²) dy, with y-limits.

Do chords overestimate or underestimate the length?

They underestimate it. A straight segment is the shortest path between two points, so each chord is shorter than the arc it cuts off.

What to learn next

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.

Quick answer

What you'll learn

Measuring a curve with straight pieces

Why the integrand is √(1 + f′(x)²)

When the integral cannot be done by hand

Worked examples

Common mistakes

Practice problems

Frequently asked questions

What is the arc length formula?

Where does the formula come from?

Why can't I evaluate most arc length integrals by hand?

How do I find the length of x = g(y)?

Do chords overestimate or underestimate the length?

What to learn next

More Calculus lessons