Concept explainerCalculus IIAdvanced10 min read

Where the arc-length formula comes from

Approximate a smooth curve by tiny line segments, then let the partition refine until the polygonal lengths converge.

L=ab1+(f(x))2dxL=\int_a^b\sqrt{1+(f'(x))^2}\,dx
Concept explainer

What the idea means, why its conditions matter, and where it connects.

Reviewed July 11, 2026
FormulaStart here
L=ab1+(f(x))2dx\boxed{L=\int_a^b\sqrt{1+\left(f'(x)\right)^2}\,dx}

Use this form for y = f(x); for x = g(y), swap the roles and integrate with respect to y.

01

From distance formula to integral

A small change along the graph has horizontal component Δx and vertical component approximately f′(x)Δx. The Pythagorean theorem gives a segment length near √(1 + f′(x)²)Δx.

Adding these pieces and taking a limit produces the integral.

02

Why arc-length integrals are often hard

The derivative is squared and placed inside a square root. Only specially structured functions simplify cleanly, so some exact-looking problems still require numerical approximation.

Simplify the radicand before choosing a method; it may become a perfect square.

03

Smoothness and parametrization

Corners can be handled piecewise, but the standard derivation assumes sufficient smoothness. Parametric curves use the same distance idea with both coordinate rates.

Worked exampleA line segment check

Find the length of y = 2x from x = 0 to 3.

1f(x)=2f'(x)=2

The slope is constant.

2L=031+22dxL=\int_0^3\sqrt{1+2^2}\,dx

Apply the graph formula.

3L=5[x]03L=\sqrt5\,[x]_0^3

The integrand is constant.

Result35\boxed{3\sqrt5}
Watch for

Common mistakes

  1. Forgetting to square the derivative.
  2. Using f(x) instead of f′(x) inside the radical.
  3. Assuming every arc-length integral has an elementary antiderivative.
Keep

Three takeaways

  1. The formula is a continuous distance sum.
  2. Simplify the radical before integrating.
  3. Numerical answers are legitimate when no elementary antiderivative exists.