The length of a curve y=f(x) from x=a to x=b isn't measured directly. It's approximated by a polygonal path connecting points along the curve. That approximation is then pushed to a limit as the segments shrink to zero width. That single idea, polygon-then-limit, reappears completely unchanged once curves gain a third coordinate.
A GPS tracking a hiker's path does not know the trail's true smooth curve. It only knows a string of straight-line hops between logged positions. Logging more often means shorter, straighter hops. Those hops hug the real trail more closely, so the estimated distance creeps toward the true one. Arc length pushes that process to its limit. Imagine infinitely many, infinitesimally short straight hops, each one following the curve's actual slope at that instant.
Arc length is the distance measured along a curve, as opposed to the straight-line distance between its endpoints. For y=f(x) from x=a to x=b, it isn't measured directly. It's approximated by a polygonal path connecting points along the curve, an approximation then pushed to a limit as the segments shrink to zero width. Each polygonal segment between x_i and x_i+\Delta x has length \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{1+(\Delta y/\Delta x)^2}\,\Delta x. The Mean Value Theorem replaces \Delta y/\Delta x with f'(x_i^*) at some point in that subinterval. Summing every segment gives a Riemann sum for \sqrt{1+[f'(x)]^2}, which converges to the stated integral as the number of segments grows without bound. This is the same "sum of thin pieces, then a limit" machinery every integral in this course is built from, arc length included.
Formula
Arc length of a curve
Recalled from Calculus II — Integral Calculus & Infinite Series · Arc length — first taught there.
For a smooth function f with continuous derivative on [a,b], the length of the curve y=f(x) from x=a to x=b is L = \int_a^b \sqrt{1+\big[f'(x)\big]^2}\,dx.
Approximate the curve by a polygonal path connecting points at x_0,x_1,\ldots,x_n. Between consecutive points, the path rises \Delta y_i over a run of \Delta x, so that segment's length is \sqrt{(\Delta x)^2+(\Delta y_i)^2} = \sqrt{1+(\Delta y_i/\Delta x)^2}\,\Delta x. By the Mean Value Theorem, \Delta y_i/\Delta x = f'(x_i^*) for some point x_i^* in that subinterval, so the segment's length is \sqrt{1+[f'(x_i^*)]^2}\,\Delta x. Summing over all n segments gives a Riemann sum for \sqrt{1+[f'(x)]^2}, and letting n\to\infty turns that sum into the stated integral, with the polygonal path's total length converging to the true arc length.
Watch the polygon converge
Drag n up — the polyline's length approaches the true arc length shown alongside it, and the visible gap between polyline and curve shrinks to nothing.
Loading interactive…
A steeper curve needs more segments to converge as fast
Compare this to the widget above at the same n — a steeper curve's polygon has more catching up to do per segment.
Loading interactive…
Worked example
Setting up an arc length integral
-
1
Set up the arc length integral for y=\dfrac{2}{3}x^{3/2} from x=0 to x=3.
-
2
Differentiate: f'(x) = x^{1/2}.
-
3
Substitute into the formula: 1+[f'(x)]^2 = 1+x.
-
4
\boxed{L = \displaystyle\int_0^3 \sqrt{1+x}\;dx}. This particular integral is elementary (a u-substitution with u=1+x finishes it), but the setup step is the one that generalizes. Every arc length problem in this course starts with this substitution into \sqrt{1+[f'(x)]^2}.
Loading interactive…
Where this goes next
Chapter 2: Vector-valued functions and motion
A space curve's arc length swaps \sqrt{1+[f'(x)]^2} for \sqrt{[x'(t)]^2+[y'(t)]^2+[z'(t)]^2}: three derivatives under one square root instead of one. It's still the same polygon-then-limit construction. It's also the very first tool that chapter builds once derivatives of vector-valued functions are defined.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
Write the arc length integrand (the full expression under the definite integral, including the square root) for f(x)=x^2.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
f'(x)=2x, so the integrand is \sqrt{1+[f'(x)]^2}=\sqrt{1+4x^2}.
Correct answer
Also accepted:
Why
Differentiate f(x)=x^2 first, then substitute f'(x) into \sqrt{1+[f'(x)]^2}.
From Calculus II — Integral Calculus & Infinite Series
The arc length integrand \sqrt{1+[f'(x)]^2} can never be less than what value?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Since 1 is always added before the square root, the integrand can never reach 0.
f'(x) can be negative, but the integrand under the square root can't be — this isn't the right bound either.
The integrand does have a firm lower bound, coming from the "+1" under the root.
Why
[f'(x)]^2\ge0 always, so 1+[f'(x)]^2\ge1, hence the square root is at least 1.
From Calculus II — Integral Calculus & Infinite Series
Find the length of f(x)=\frac23 x^{3/2} from x=0 to x=8.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
L=\int_0^8\sqrt{1+x}\,dx = \left[\frac23(1+x)^{3/2}\right]_0^8 = \frac23(27-1)\approx17.33.
Correct answer
Also accepted:
Why
f'(x) = x^(1/2), so 1 + [f'(x)]^2 = 1 + x.