Key takeaways
- Arc length: L = ∫ₐᵇ √(1 + [f'(x)]²) dx.
- It comes from Pythagoras applied to tiny pieces of the curve.
- Square the slope f'(x), not the height f(x).
- The arc is always at least as long as the straight line between its ends.
- Many arc length integrals have no neat answer, so they are found numerically.
In this guide
What is arc length?
Imagine laying a piece of string exactly along a curved graph from x = a to x = b. Now pull the string straight and measure it. That length is the arc length. It is always at least as long as the straight line between the two end points, because the curve bends and wanders.
Here is the key idea. Zoom in on a tiny piece of the curve. It looks almost straight, like the long side of a tiny right triangle. If it moves across by dx and up by dy, its length is √(dx² + dy²) by the Pythagorean theorem. Add up all those tiny pieces with an integral, and you get the arc length.
Why does arc length matter?
Real objects follow curves: roads, rails, cables, pipes and rollercoaster tracks. To buy the right amount of material, you need the length along the curve, not straight across. If you measure straight across, you will always come up short.
It also shows how calculus works in general. Almost every formula in integral calculus uses the same trick: cut something into tiny simple pieces, measure each piece, and add.
For y = x^1.5 from x = 0 to x = 4, the curve is about 9.07 units long. The straight line between its ends is about 8.94. Small bends add up.
The formula
| Symbol | What it means | Example |
|---|---|---|
| L | The arc length, in the same units as x and y | 9.0735 m |
| f(x) | The curve. Here f(x) = xⁿ | x^1.5 |
| f'(x) | The slope of the curve at x | 1.5x^0.5 |
| a, b | The start and end x-values | 0 and 4 |
Where does the 1 come from? Divide √(dx² + dy²) by dx and you get √(1 + (dy/dx)²). So the 1 stands for the sideways step and (dy/dx)² for the upward step.
This calculator works with curves of the form f(x) = xⁿ. You choose the power n and the interval [a, b]. Because √(1 + f'²) often has no simple antiderivative, the calculator adds up 100 thin slices (the midpoint rule). For smooth curves this is usually accurate to about four decimal places.
How to find arc length step by step
- 1Find the derivative. For f(x) = xⁿ, the power rule gives
f'(x) = n·xⁿ⁻¹. - 2Square it and add 1. Write
1 + [f'(x)]². - 3Take the square root. This is the length of the curve per unit step in x.
- 4Integrate from a to b. Do it exactly if you can, or numerically if you can't.
Try it yourself below. Pick a power and an interval, and the calculator adds up the tiny pieces for you.
Try it yourself
Pre-filled with the example — change any value.
Common mistakes to avoid
- Forgetting the 1 under the square root. Without it you only measure the vertical change.
- Splitting the square root.
√(1 + f'²)is not1 + f'. The root covers the whole sum. - Using a negative a with a fractional power. x^1.5 is not defined for negative x, so start at a ≥ 0.
- Getting a length shorter than the straight line. That can't happen. Recheck your derivative.
Tips and tricks
- Some powers work out exactly. For y = x^1.5,
1 + f'² = 1 + 2.25x, which you can integrate by hand. From 0 to 4 it gives(8 ÷ 27)(10^1.5 − 1) ≈ 9.0734. - Circle arcs are easier. For part of a circle, use the arc length of a circle calculator: just radius times angle.
- Steeper means longer. Parts of the curve with a big slope add the most length.
Where you'll use it in real life
- Construction: the length of edging or fencing along a curved path.
- Engineering: the length of cable hanging between two poles, or of a bent pipe.
- Transport: the true length of a curved road or railway section.
- Sport and physics: the distance a thrown ball travels along its curved path.
Quick summary
Arc length adds up tiny straight pieces along a curve using L = ∫ √(1 + [f'(x)]²) dx. Differentiate, square, add 1, take the square root and integrate. Use the calculator above to get the length of any curve y = xⁿ with the steps shown.
Worked example
A garden path follows the curve y = x^1.5 (in metres) from x = 0 to x = 4. How many metres of edging do you need along one side? Here n = 1.5, a = 0 and b = 4.
- 01Arc length of f(x) = x^1.5 on [0, 4]
- 02L = ∫[0,4] √(1 + [f'(x)]²) dx
- 03f'(x) = 1.5x^0.5
- 04[f'(x)]² = 2.25x^1
- 05Numerical integration (midpoint, N=100):
- 06L ≈ 9.0735
These numbers are pre-filled in the calculator above.
Frequently asked questions
Why is there a square root in the arc length formula?
Each tiny piece of the curve is the long side of a small right triangle with sides dx and dy. Pythagoras gives its length as √(dx² + dy²), which becomes √(1 + (dy/dx)²) dx.
What is the arc length of y = x² from 0 to 1?
It is about 1.4789 units. That is a little longer than the straight line from (0, 0) to (1, 1), which is √2 ≈ 1.4142.
Can every arc length integral be solved exactly?
No. Many are hard, and some have no closed-form answer at all. That's why numerical methods like the one in this calculator are used.
What units is arc length in?
The same units as x and y. If both are in metres, the arc length is in metres, not square metres.


