Arc Length of a Curve

The arc length of a smooth curve is the total distance you'd walk along it. We approximate the curve by tiny straight segments, sum up their lengths via Pythagoras, and take the limit — which gives a clean integral formula.

Deriving the Formula

Take a tiny piece of curve. Its horizontal change is $dx$, its vertical change is $dy$. By Pythagoras:

$ds = \sqrt{(dx)^{2} + (dy)^{2}} = \sqrt{1 + \left(\dfrac{dy}{dx}\right)^{2}}\,dx$

Integrating from $a$ to $b$:

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

Step-by-Step Solution

Given: $f(x) = x^{2}$ on $[0, 2]$.

Find: The arc length $L$.

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Step 1 — Compute $f'(x)$.

$f'(x) = 2x$

Step 2 — Plug into the formula.

$L = \int_{0}^{2} \sqrt{1 + (2x)^{2}}\,dx = \int_{0}^{2} \sqrt{1 + 4x^{2}}\,dx$

Step 3 — This integral has a closed form.

The antiderivative of $\sqrt{1 + 4x^{2}}$ involves a hyperbolic substitution or a trig substitution. The closed form is:

$\int\sqrt{1 + 4x^{2}}\,dx = \dfrac{x\sqrt{1 + 4x^{2}}}{2} + \dfrac{1}{4}\ln\!\left(2x + \sqrt{1 + 4x^{2}}\right) + C$

Step 4 — Evaluate at the bounds $x = 0$ and $x = 2$.

At $x = 2$: $\sqrt{1 + 16} = \sqrt{17} \approx 4.123$

$\dfrac{2\sqrt{17}}{2} + \dfrac{1}{4}\ln(4 + \sqrt{17}) = \sqrt{17} + \dfrac{\ln(8.123)}{4}$

$\approx 4.123 + \dfrac{2.0947}{4} \approx 4.123 + 0.5237 \approx 4.647$

At $x = 0$: $\sqrt{1 + 0} = 1$

$\dfrac{0 \cdot 1}{2} + \dfrac{1}{4}\ln(0 + 1) = 0 + 0 = 0$

Step 5 — Subtract.

$L = 4.647 - 0 \approx 4.647\;\text{units}$

Step 6 — Sanity check using a polygonal approximation.

If you split $[0, 2]$ into 100 tiny segments and compute $\sum\sqrt{(dx)^{2} + (\Delta y)^{2}}$, you get approximately $4.6471$ — matching the closed form. The visualization does this on the fly.

Step 7 — Compare to the straight-line distance.

The straight chord from $(0, 0)$ to $(2, 4)$ has length $\sqrt{4 + 16} = \sqrt{20} \approx 4.472$. So the curved path is about $0.175$ longer than the straight one (3.9% longer). The curve doesn't deviate much from the chord because $y = x^{2}$ is fairly mild on this interval.

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Answer: The arc length of $y = x^{2}$ from $x = 0$ to $x = 2$ is

$\boxed{\;L = \sqrt{17} + \dfrac{\ln(4 + \sqrt{17})}{4} \approx 4.647\;\text{units}\;}$

This is about 4% longer than the straight chord from $(0, 0)$ to $(2, 4)$.

Try It

• Adjust the number of segments widget — see how the polygonal approximation gets more accurate as $n$ grows.
• The HUD shows the running sum (the polygonal length) compared to the closed-form value.
• At $n = 50$, the approximation is already accurate to about 4 decimal places.