Arc Length Calculator

This arc length calculator measures the length of a curve \( y = f(x) \) between \( x = a \) and \( x = b \). It differentiates your function, writes out the arc length integral, and evaluates it numerically, so you get an answer even when the integral has no closed form (which is most of the time).

It suits calculus students checking homework and anyone who needs the length of a curved path. It is free and needs no sign-up.

How to use the arc length calculator

1. Enter \( f(x) \). Use ^ for powers (fractional powers need parentheses, like x^(3/2)), sqrt(), sin(), ln() and e^(x). Implicit multiplication such as 2x or x ln(x) works.
2. Enter the endpoints \( a \) and \( b \); pi is accepted, for example pi/2.
3. Press Measure length.
4. Read the three rows: the derivative \( f'(x) \), the set-up integral, and the arc length \( L \).

Formula

For a function with a continuous derivative on \( [a, b] \), the arc length is

$$L = \int_a^b \sqrt{1 + \left[f'(x)\right]^2}\,dx$$

It comes from adding up tiny straight segments: over a small step \( \Delta x \), the curve rises about \( f'(x)\,\Delta x \), and by Pythagoras the segment has length \( \sqrt{1 + f'(x)^2}\,\Delta x \). The integral is the limit of that sum. Our arc length formula guide derives it in full and covers parametric curves.

The square root usually prevents a nice antiderivative, so the calculator uses adaptive numerical integration and reports the result to about 8 significant digits.

Worked examples

Example 1 (a curve that simplifies). Find the length of \( y = \frac13 (x^2 + 2)^{3/2} \) from \( x = 0 \) to \( x = 3 \).

By the chain rule, \( f'(x) = x\sqrt{x^2 + 2} \) (the derivative calculator shows this step if you want to check it). Then

$$1 + [f'(x)]^2 = 1 + x^4 + 2x^2 = (x^2 + 1)^2$$

The square root disappears:

$$L = \int_0^3 (x^2 + 1)\,dx = 9 + 3 = 12$$

Type (1/3)(x^2 + 2)^(3/2) with limits 0 and 3 into the calculator above and it returns 12.

Example 2 (another perfect square). For \( y = \frac{x^3}{6} + \frac{1}{2x} \) on \( [1, 2] \), \( f'(x) = \frac{x^2}{2} - \frac{1}{2x^2} \), and \( 1 + f'^2 \) becomes \( \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2 \). So

$$L = \int_1^2 \left(\frac{x^2}{2} + \frac{1}{2x^2}\right) dx = \frac{17}{12} \approx 1.41667$$

Example 3 (no closed form). One arch of \( y = \sin x \), from 0 to \( \pi \), gives \( L = \int_0^{\pi} \sqrt{1 + \cos^2 x}\,dx \), an elliptic integral. The calculator returns \( L \approx 3.8201978 \).

Sanity checks and common pitfalls

• Compare with a straight line. The arc length is never shorter than the straight segment between the endpoints. For \( \sin x \) on \( [0, \pi] \), the chord is \( \pi \approx 3.14159 \), and the arc is a bit longer, as expected.
• Lines are an easy test. For \( y = \frac43 x \) on \( [0, 3] \), the calculator gives 5, the hypotenuse of a 3-4-5 triangle.
• Do not forget the 1. Integrating \( \sqrt{f'(x)^2} \) instead of \( \sqrt{1 + f'(x)^2} \) is a frequent slip.
• Watch for vertical tangents. If \( f'(x) \) blows up at an endpoint (like \( \sqrt{x} \) at 0), the integral is improper. It often still converges, but if a point where \( f \) is not differentiable sits inside the interval, split the interval or rethink the set-up.
• The curve must be a function of x. For a circle or other closed curve, measure one piece that passes the vertical line test and multiply by symmetry.

Related guides

• Arc Length Formula in Calculus: Derivation and Examples
• Chain Rule Explained: Formula, Steps and 8 Examples
• Riemann Sums: Left, Right, Midpoint and Simpson’s Rule

Further reading

• Arc Length (Paul’s Online Math Notes) — derives the formula and works examples in both the \( dx \) and \( dy \) forms.
• Surface Area (Paul’s Online Math Notes) — the next step: rotate the curve about an axis and reuse the same arc length element.

FAQ

How do you calculate the arc length of a curve?

Differentiate \( f \), plug into \( \sqrt{1 + f'(x)^2} \) and integrate from \( a \) to \( b \). The calculator does all three steps and shows the derivative and the integral it used.

Why is my arc length answer a decimal instead of an exact value?

Most arc length integrals have no elementary antiderivative, so the calculator evaluates them numerically. Exact answers appear only for specially designed functions, like the examples above.

Can it find the arc length of parametric or polar curves?

This tool handles curves written as \( y = f(x) \). For a parametric curve, the formula becomes \( \int \sqrt{x'(t)^2 + y'(t)^2}\,dt \), which you can evaluate with the definite integral calculator after renaming \( t \) to \( x \).

What units is the arc length in?

The same units as your x and y axes. If both are in meters, the length is in meters.