Arc Length

How long is a curve? Zoom in far enough and every smooth curve looks straight, so each tiny piece is the hypotenuse of a right triangle with legs $dx$ and $dy$. Summing those hypotenuses gives the arc length formula $L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx$.

By the end you'll be able to derive the arc-length integrand from the little $dx$–$dy$–$ds$ triangle, set up $\int \sqrt{1 + [f'(x)]^2}\,dx$ for a given curve, and evaluate it when the algebra cooperates.

Predict: where the curve is steepest (far right here), is $\Delta s$ per $\Delta x$ largest? Drag the point (or use its slider) to check.

The curve is $y = x^2/4$ on $[0, 4]$, chopped into equal-width pieces. At the highlighted piece you see the local right triangle: horizontal leg $\Delta x$, vertical leg $\Delta y$, and hypotenuse $\Delta s = \sqrt{(\Delta x)^2 + (\Delta y)^2}$. The green polyline is all the hypotenuses laid end to end — its total is the approximation $\sum \Delta s$, and cranking up the number of pieces drives it to the true length $L$.

Point position x = 1.00

Pieces n = 8

Highlighted piece: Δx = 0.500 · Δy = 0.313 · Δs ≈ 0.590 (chord slope ≈ 0.625)

Σ Δs ≈ 5.911 over 8 pieces · true length L ≈ 5.916

y = x²/4 — dotted curve, green hypotenuse chain, and the local Δx–Δy–Δs triangle at the draggable point

curve Δs hypotenuses Δx leg Δy leg

The arc length of a smooth curve $y = f(x)$ is $\int \sqrt{1 + [f'(x)]^2}\,dx$, obtained by summing the hypotenuses of infinitesimal right triangles with legs $dx$ and $dy$.

Visual

Break the curve into $n$ short pieces of width $\Delta x$, like the demo above. Over each piece the curve rises by roughly $\Delta y \approx f'(x)\,\Delta x$, so the piece is nearly the hypotenuse of a right triangle with legs $\Delta x$ and $\Delta y$. Pythagoras gives $\Delta s \approx \sqrt{(\Delta x)^2 + (\Delta y)^2}$; factor $(\Delta x)^2$ out of the root and you get $\Delta s \approx \sqrt{1 + (\Delta y/\Delta x)^2}\, \Delta x$. Where the curve is steep, $\Delta y$ is big and each hypotenuse is much longer than its $\Delta x$ — exactly what you saw dragging the point to the right.

Formal

Let $\Delta x \to 0$. The slope ratio $\Delta y/\Delta x$ becomes $f'(x)$ and the Riemann sum $\sum \sqrt{1 + (\Delta y/\Delta x)^2}\,\Delta x$ converges to $$L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx.$$ The 1 under the root is the horizontal leg's own contribution — it comes from the $(\Delta x)^2$ inside Pythagoras, which is why you must never drop it. If the curve is given parametrically as $x = x(t),\ y = y(t)$, the same triangle (legs $x'(t)\,dt$ and $y'(t)\,dt$) gives $L = \int_\alpha^\beta \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt$.

Worked example

Find the arc length of $y = \tfrac{2}{3}x^{3/2}$ from $x = 0$ to $x = 3$. First, $f'(x) = x^{1/2}$, so $[f'(x)]^2 = x$ and $$L = \int_0^3 \sqrt{1 + x}\,dx.$$ Substitute $u = 1 + x$, $du = dx$; when $x = 0$, $u = 1$; when $x = 3$, $u = 4$: $$L = \int_1^4 \sqrt{u}\,du = \left[\tfrac{2}{3}u^{3/2}\right]_1^4 = \tfrac{2}{3}(8 - 1) = \textbf{14/3}.$$

Your turn

Same curve, a longer stretch: $y = \tfrac{2}{3}x^{3/2}$ from $x = 0$ to $x = 8$. As before, $f'(x) = x^{1/2}$, so $L = \int_0^8 \sqrt{1+x}\,dx$. With $u = 1 + x$ the limits become $u = 1$ to $u = 9$, so $L = \int_1^9 \sqrt{u}\,du = \tfrac{2}{3}u^{3/2}\big|_1^9 = \tfrac{2}{3}\,(\,$____$\, - \,$____ $)$ = ____

Reveal the answer

$9^{3/2} = 27$ and $1^{3/2} = 1$, so $L = \tfrac{2}{3}(27 - 1) = \tfrac{2}{3}\cdot 26 = $ $52/3$. The most common slip is forgetting to convert the x-limits to u-limits — integrating $\sqrt{u}$ from 0 to 8 gives the wrong answer.

More info — why exact arc lengths are often hard

$\sqrt{1 + [f'(x)]^2}$ is rarely a "nice" integrand. Even for the humble $y = x^2$, you get $\int \sqrt{1 + 4x^2}\,dx$, which needs a trigonometric substitution and produces answers involving $\ln$ or $\sinh^{-1}$ — that's why the demo above reports the true length of $y = x^2/4$ as a decimal rather than a tidy fraction. Textbook problems are usually engineered (like $y = \tfrac{2}{3}x^{3/2}$ in the worked example) so that $1 + [f'(x)]^2$ collapses to a perfect square or a cleanly integrable power. When it doesn't, a substitution or a numerical estimate is the intended route — see Paul's Online Math Notes in Dive deeper below for both kinds of practice.

Check your understanding

Question 1 of 4

Which integral gives the arc length of $y = x^2$ from $x = 0$ to $x = 1$?

Question 2 of 4

Find the arc length of $y = \tfrac{2}{3}x^{3/2}$ from $x = 0$ to $x = 15$.

Question 3 of 4

A student forgets the 1 and computes $\int_a^b \sqrt{[f'(x)]^2}\,dx = \int_a^b |f'(x)|\,dx$. What does that integral actually measure?

Question 4 of 4

The curve $y = x^2$, $0 \le x \le 2$, bounds a region (down to the x-axis) that you rotate about the y-axis. Which pair correctly gives the shell volume of the solid AND the arc length of the curve?

Recap

Arc length of a smooth curve $y = f(x)$ on $[a, b]$: $L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx$.
It comes from summing hypotenuses of tiny right triangles: $\Delta s = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{1 + (\Delta y/\Delta x)^2}\,\Delta x$.
The 1 under the root is the $(\Delta x)^2$ leg — dropping it turns length into mere vertical rise-and-fall.
Parametric version: $L = \int_\alpha^\beta \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt$.
Most arc-length integrands have no elementary antiderivative — course problems are engineered so $1 + [f'(x)]^2$ simplifies; otherwise estimate numerically.

Dive deeper

Paul's Online Math Notes — Arc Length Review the derivation and practice additional arc-length integrals.

Sources

Arc Length of a Curve