Why is there a square root in the arc length formula?

The square root comes from the Pythagorean theorem applied infinitesimally. As you move along the curve, the horizontal change is dx and the vertical change is dy = f'(x) \, dx. These form the legs of a tiny right triangle, and the hypotenuse (the arc element) has length ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{1 + [f'(x)]^2} \, dx. Integrating all these pieces gives the total arc length.

Can I just use the distance formula to find arc length?

No. The distance formula d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} gives the straight-line distance (the chord connecting two points), not the arc length. Arc length follows the actual path of the curve, which is strictly longer unless the curve is already a straight line. For any curved path, you must use the arc length integral.

What if the integral \int \sqrt{1 + [f'(x)]^2} \, dx doesn't have a closed form?

Many arc length integrals have no elementary closed form (for instance, \int \sqrt{1 + x^2} \, dx and \int \sqrt{1 + x^4} \, dx). In such cases, you can leave the answer as the integral, or use numerical integration methods (Simpson's rule, a graphing calculator, or a computer) to approximate the arc length to any desired precision.

When should I use Trigonometric Substitution for arc length?

Use Trigonometric Substitution when 1 + [f'(x)]^2 has the form a^2 + x^2, a^2 - x^2, or x^2 - a^2 (or scalar multiples). For example, if f'(x) = 2x, then 1 + [f'(x)]^2 = 1 + 4x^2, which suits a tangent substitution. The trig substitution converts the radical into a trigonometric function that integrates more easily.

Arc Length of a Curve

Arc length measures the actual distance along a curve between two points, extending the simple distance formula to account for the curve's twists and turns.

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

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What each symbol means

What Arc Length of a Curve takes

Symbol	Meaning
$L$	The arc length, or total distance traveled along the curve from $x = a$ to $x = b$ ; measured in the same units as $a$ and $b$ .
$f$	The function describing the curve; it must be continuous and differentiable on the interval $[a, b]$ for the arc length formula to apply.
$a$	The left endpoint of the interval over which arc length is measured; the starting $x$ -value.
$b$	The right endpoint of the interval; the ending $x$ -value, and arc length is always measured from left to right, so $a < b$ .
$x$	The independent variable; arc length is computed as $x$ ranges from $a$ to $b$ .

When to use it

Use this when you need the actual distance a path travels, such as the length of a hiking trail or the distance a car drives along a curved road.

Level

Usually taught in: Calculus II

Worked examples

1. Arc length of a line segment

Problem

Find the arc length of

f(x) = x

from

x = 0

x = 3

$f'(x) = 1$
Differentiate the function using the power rule.
$1 + \big[f'(x)\big]^2 = 1 + 1^2 = 2$
Compute $1 + [f'(x)]^2$ by substituting the derivative.
$L = \int_0^3 \sqrt{2} \, dx$
Apply the arc length formula by taking the square root and setting up the definite integral.
$L = \left[\sqrt{2} \cdot x\right]_0^3 = \sqrt{2}(3 - 0) = 3\sqrt{2}$
Integrate by recognizing that $\sqrt{2}$ is a constant, then evaluate at the bounds using the Fundamental Theorem of Calculus.

Answer: $L = 3\sqrt{2}$

Since $f(x) = x$ is a straight line, the arc is literally a line segment. The arc length formula correctly gives $3\sqrt{2}$ , which matches the distance formula applied to the segment from $(0, 0)$ to $(3, 3)$ (distance $= \sqrt{3^2 + 3^2} = 3\sqrt{2}$ ).

2. Power function requiring u-substitution

Problem

Find the arc length of

f(x) = \frac{2}{3}x^{3/2}

from

x = 0

x = 4

$f'(x) = \frac{2}{3} \cdot \frac{3}{2} x^{1/2} = x^{1/2}$
Differentiate using the power rule, bringing down the exponent $\frac{3}{2}$ and reducing the power by 1.
$\big[f'(x)\big]^2 = (x^{1/2})^2 = x$
Square the derivative to get $x$ .
$1 + \big[f'(x)\big]^2 = 1 + x$
Add 1 to the squared derivative.
$L = \int_0^4 \sqrt{1 + x} \, dx$
Set up the arc length integral with the square root of $1 + x$ .
$u = 1 + x, \quad du = dx; \quad x = 0 \Rightarrow u = 1, \quad x = 4 \Rightarrow u = 5$
Use $u$ -substitution to transform $\sqrt{1 + x}$ into a simpler form; update the bounds accordingly.
$L = \int_1^5 u^{1/2} \, du = \left[\frac{2}{3}u^{3/2}\right]_1^5$
Rewrite $\sqrt{u}$ as $u^{1/2}$ and integrate using the power rule for integration.
$L = \frac{2}{3}(5^{3/2} - 1^{3/2}) = \frac{2}{3}(5\sqrt{5} - 1)$
Evaluate at the bounds; note that $5^{3/2} = 5 \cdot \sqrt{5}$ and $1^{3/2} = 1$ .

Answer: $L = \frac{2}{3}(5\sqrt{5} - 1)$

This problem showcases $u$ -substitution because the integrand $\sqrt{1 + x}$ does not immediately fit a standard form. The substitution simplifies the square root to a power function, which the power rule handles easily.

3. Word problem: distance on a winding mountain road

Problem

You are on a road trip traveling along a winding mountain road. Your elevation (in meters, relative to the starting point) is given by

f(x) = \frac{1}{3}x^{3/2}

, where

x

is the horizontal distance from the starting point in kilometers. On the first leg, you drive horizontally from

x = 0

x = 3

kilometers. On the second leg, you drive on a flat highway. How far did you actually travel along the first curved road?

$f'(x) = \frac{1}{3} \cdot \frac{3}{2} x^{1/2} = \frac{1}{2}x^{1/2}$
Differentiate to find the instantaneous rate of change of elevation with respect to horizontal distance.
$\big[f'(x)\big]^2 = \left(\frac{1}{2}x^{1/2}\right)^2 = \frac{1}{4}x$
Square the derivative.
$1 + \big[f'(x)\big]^2 = 1 + \frac{1}{4}x = \frac{4 + x}{4}$
Add 1; combining into a single fraction reveals the structure for the next step.
$\sqrt{1 + \big[f'(x)\big]^2} = \sqrt{\frac{4 + x}{4}} = \frac{\sqrt{4 + x}}{2}$
Take the square root of the fraction by taking the square root of numerator and denominator separately.
$L = \int_0^3 \frac{\sqrt{4 + x}}{2} \, dx = \frac{1}{2}\int_0^3 \sqrt{4 + x} \, dx$
Factor out the constant $\frac{1}{2}$ and set up the arc length integral.
$u = 4 + x, \quad du = dx; \quad x = 0 \Rightarrow u = 4, \quad x = 3 \Rightarrow u = 7$
Apply $u$ -substitution to simplify $\sqrt{4 + x}$ .
$L = \frac{1}{2}\int_4^7 u^{1/2} \, du = \frac{1}{2}\left[\frac{2}{3}u^{3/2}\right]_4^7$
Integrate using the power rule after the substitution.
$L = \frac{1}{2} \cdot \frac{2}{3}(7^{3/2} - 4^{3/2}) = \frac{1}{3}(7\sqrt{7} - 8)$
Simplify the product $\frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$ , and compute $7^{3/2} = 7\sqrt{7}$ and $4^{3/2} = 8$ .

Answer: $L = \frac{1}{3}(7\sqrt{7} - 8) \text{ kilometers}$

The actual road distance exceeds the horizontal distance of 3 km because the road curves upward. Arc length captures this by measuring the true path along the winding road. The formula effectively converts a three-dimensional path (horizontal movement plus elevation gain) into a single one-dimensional arc length.

Common mistakes

Where Arc Length of a Curve usually goes wrong

The mistake	Why it is wrong	The fix
Writing $L = \int_0^1 (1 + [f'(x)]^2) \, dx$ when $f(x) = x$ , yielding $L = 2$ instead of $L = \sqrt{2}$ .	The arc length formula requires a square root around $1 + [f'(x)]^2$ ; omitting it computes an entirely different integral that has no geometric meaning as arc length.	Always include the square root: $L = \int_0^1 \sqrt{1 + [f'(x)]^2} \, dx = \int_0^1 \sqrt{2} \, dx = \sqrt{2}$ .
Using the function $f(x)$ instead of the derivative $f'(x)$ , writing $L = \int_0^1 \sqrt{1 + f(x)} \, dx = \int_0^1 \sqrt{1 + x} \, dx$ when $f(x) = x$ .	The arc length formula requires the derivative $f'(x)$ , not the function itself. Substituting the wrong expression leads to a completely incorrect setup.	Find the derivative first: $f'(x) = 1$ , then substitute into the formula: $L = \int_0^1 \sqrt{1 + 1^2} \, dx = \int_0^1 \sqrt{2} \, dx = \sqrt{2}$ .
When $f(x) = \frac{1}{2}x^2$ , squaring the function instead of the derivative and writing $L = \int_0^2 \sqrt{1 + x^4} \, dx$ instead of $L = \int_0^2 \sqrt{1 + x^2} \, dx$ .	You must square the derivative $f'(x) = x$ to get $[f'(x)]^2 = x^2$ , not square the original function to get $[f(x)]^2 = \frac{1}{4}x^4$ .	Differentiate first: $f'(x) = x$ , so $[f'(x)]^2 = x^2$ , and the integrand is $\sqrt{1 + x^2}$ —not $\sqrt{1 + x^4}$ .

Tips and when to use something else

The formula requires the derivative $f'(x)$ : always differentiate the function first, then square it, before substituting into the arc length integral.
When $\int \sqrt{1 + [f'(x)]^2} \, dx$ does not simplify immediately, try $u$ -substitution if $1 + [f'(x)]^2$ is a linear function of $x$ , or Trigonometric Substitution if it has the form $\sqrt{1 + (\text{stuff})^2}$ .
Arc length measures distance along the curve itself. If a problem asks for the straight-line distance between two points, use the distance formula; if it asks for area under the curve, use $\int_a^b f(x) \, dx$ instead.
When the integral does not simplify to an elementary function, leave the answer as the integral itself or use numerical methods to approximate it—do not force a false closed form.

Related formulas

U-SubstitutionU-substitution simplifies integrals by transforming them into easier forms using a change of variables, ideal when you spot a function and its derivative.Trigonometric SubstitutionTrigonometric substitution replaces expressions like \sqrt{a^2-x^2} with trig functions, making integrals with radicals easier to solve.Power Rule for IntegrationPower Rule for Integration instantly finds antiderivatives of polynomial terms; use it to integrate any power of x except x to the negative one.Fundamental Theorem of CalculusCompute the exact value of a definite integral using the antiderivative—connect area under a curve to the accumulation of change.

Frequently asked questions

Why is there a square root in the arc length formula?: The square root comes from the Pythagorean theorem applied infinitesimally. As you move along the curve, the horizontal change is $dx$ and the vertical change is $dy = f'(x) \, dx$ . These form the legs of a tiny right triangle, and the hypotenuse (the arc element) has length $ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{1 + [f'(x)]^2} \, dx$ . Integrating all these pieces gives the total arc length.
Can I just use the distance formula to find arc length?: No. The distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ gives the straight-line distance (the chord connecting two points), not the arc length. Arc length follows the actual path of the curve, which is strictly longer unless the curve is already a straight line. For any curved path, you must use the arc length integral.
What if the integral $\int \sqrt{1 + [f'(x)]^2} \, dx$ doesn't have a closed form?: Many arc length integrals have no elementary closed form (for instance, $\int \sqrt{1 + x^2} \, dx$ and $\int \sqrt{1 + x^4} \, dx$ ). In such cases, you can leave the answer as the integral, or use numerical integration methods (Simpson's rule, a graphing calculator, or a computer) to approximate the arc length to any desired precision.
When should I use Trigonometric Substitution for arc length?: Use Trigonometric Substitution when $1 + [f'(x)]^2$ has the form $a^2 + x^2$ , $a^2 - x^2$ , or $x^2 - a^2$ (or scalar multiples). For example, if $f'(x) = 2x$ , then $1 + [f'(x)]^2 = 1 + 4x^2$ , which suits a tangent substitution. The trig substitution converts the radical into a trigonometric function that integrates more easily.

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Reviewed 2026-09-18