Measure the length of a curve, integrals from easy to advanced lesson

The curve $y = x^{3/2}$ climbs from the origin to the point $(1, 1)$ . The straight segment between those endpoints has length $\sqrt{2}$ , and the curve bends away from it, so the curve is longer. The area under the graph measures something else entirely, so how long is the curve itself? This page answers by adding the lengths of short chords, turning the chord sum into the arc length integral and finding the exact length $(13\sqrt{13} - 8)/27 \approx 1.4397$ .

Key terms

Pythagorean length of a short segment

Definition The segment joining two points $(x_1, y_1)$ and $(x_2, y_2)$ of the plane has length

\Delta s = \sqrt{(\Delta x)^2 + (\Delta y)^2},

where $\Delta x = x_2 - x_1$ and $\Delta y = y_2 - y_1$ . When $\Delta x$ and $\Delta y$ are both nonzero, the segment is the hypotenuse of a right triangle with a horizontal leg of length $\lvert \Delta x \rvert$ and a vertical leg of length $\lvert \Delta y \rvert$ , so this is the Pythagorean theorem. When one of them is zero, the segment is horizontal or vertical, and the formula gives $\lvert \Delta x \rvert$ or $\lvert \Delta y \rvert$ . Suppose both points lie on the graph of a function $f$ that is continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$ , with $x_1 < x_2$ . The mean value theorem gives a point $c$ strictly between $x_1$ and $x_2$ with $\Delta y = f'(c)\,\Delta x$ , and since $\Delta x > 0$ it can be taken out of the square root,

\Delta s = \sqrt{1 + f'(c)^2}\;\Delta x.

In plain words A short chord of a curve is the diagonal of a small rectangle whose sides are the horizontal and vertical changes. Taking out the horizontal width leaves a factor that depends only on how steep the chord is, and that steepness equals the slope of the curve at some point of the piece.

Example On $y = x^{3/2}$ , the chord from $x = \tfrac14$ to $x = 1$ joins $(\tfrac14, \tfrac18)$ to $(1, 1)$ . Its legs are $\Delta x = \tfrac34$ and $\Delta y = \tfrac78$ , so

\begin{aligned} \Delta s &= \sqrt{\tfrac{9}{16} + \tfrac{49}{64}} \\[0.4em] &= \frac{\sqrt{85}}{8} \approx 1.1524. \end{aligned}

The chord has slope $\Delta y / \Delta x = \tfrac76$ . The derivative $\tfrac32 \sqrt{x}$ takes this value at $c = \tfrac{49}{81} \approx 0.605$ , which lies between $\tfrac14$ and $1$ , and indeed $\tfrac34 \sqrt{1 + \tfrac{49}{36}} = \tfrac34 \cdot \tfrac{\sqrt{85}}{6} = \tfrac{\sqrt{85}}{8}$ .

Arc length

Definition Let a curve run from a point $A$ to a point $B$ . Choose points $A = P_0, P_1, \dots, P_n = B$ on the curve, in order along it, and join consecutive points by segments. Write $\lvert P_{i-1} P_i \rvert$ for the length of the segment from $P_{i-1}$ to $P_i$ . The arc length of the curve is the least upper bound of the lengths of all such inscribed polygonal paths,

L = \sup \sum_{i=1}^{n} \lvert P_{i-1} P_i \rvert,

taken over every $n$ and every choice of the points. When this bound is finite the curve is called rectifiable. For the graph of a function $f$ on $[a, b]$ , choosing points in order means choosing $a = x_0 < x_1 < \dots < x_n = b$ and taking $P_i = (x_i, f(x_i))$ . If $f$ is continuous and its graph is rectifiable, the polygon lengths approach $L$ as the widest piece $[x_{i-1}, x_i]$ shrinks to width $0$ .

In plain words A straight segment is the shortest path between its endpoints, so a polygon with its corners on the curve is never longer than the curve. Adding a corner never makes the polygon shorter, and enough corners make it hug the curve. The length of the curve is the number that these polygon lengths climb toward and never exceed.

Example For $y = x^{3/2}$ on $[0, 1]$ , the single chord from $(0, 0)$ to $(1, 1)$ has length $\sqrt{2} \approx 1.4142$ . Adding the corner $(\tfrac14, \tfrac18)$ gives two chords of lengths $\tfrac{\sqrt5}{8}$ and $\tfrac{\sqrt{85}}{8}$ , with total

\frac{\sqrt{5} + \sqrt{85}}{8} \approx 1.4320.

With $4$ , $16$ and $100$ equal pieces the totals are about $1.4362$ , $1.4394$ and $1.43970$ . The curve rises, so every chord has $\Delta y \ge 0$ and is no longer than $\Delta x + \Delta y$ . The widths add to $1$ and the rises add to $1$ , so every polygon has length at most $2$ . Hence $\sqrt{2} \le L \le 2$ , where $2$ is the length of the path that runs one unit right and then one unit up.

Arc length element

Definition For the graph of a differentiable function $y = f(x)$ , traced in the direction of increasing $x$ , the arc length element is

ds = \sqrt{1 + f'(x)^2}\,dx.

It is the limit form of the chord formula. On a piece of width $\Delta x > 0$ the chord has length $\Delta x \sqrt{1 + (\Delta y / \Delta x)^2}$ , and as $\Delta x \to 0$ the ratio $\Delta y / \Delta x$ tends to $f'(x)$ . The shorthand $ds = \sqrt{dx^2 + dy^2}$ records the same Pythagorean picture. For a curve written as $x = g(y)$ and traced with increasing $y$ , the element is $ds = \sqrt{1 + g'(y)^2}\,dy$ . For a curve given by $x(t)$ and $y(t)$ and traced with increasing $t$ , it is $ds = \sqrt{x'(t)^2 + y'(t)^2}\,dt$ .

In plain words The factor $\sqrt{1 + f'(x)^2}$ is the length of curve per unit of horizontal travel. It is never less than $1$ , it equals $1$ exactly where the tangent is horizontal, and it grows as the curve gets steeper. The $1$ accounts for the horizontal motion and the square of the derivative for the vertical motion.

Example For $y = x^{3/2}$ the derivative is $y' = \tfrac32 \sqrt{x}$ , so $(y')^2 = \tfrac94 x$ and

ds = \sqrt{1 + \tfrac94 x}\,dx.

The factor grows from $1$ at $x = 0$ to $\tfrac{\sqrt{13}}{2} \approx 1.803$ at $x = 1$ . Leaving out the $1$ would give $\int_0^1 \tfrac32 \sqrt{x}\,dx = 1$ , which is only the total rise of the curve and ignores its horizontal run. A horizontal line $y = c$ has $f' = 0$ and $ds = dx$ , so its length over $[a, b]$ is $b - a$ , not $0$ .

Continuously differentiable curve

Definition A function $f$ is continuously differentiable on a closed interval $[a, b]$ when it has a derivative at every point of $(a, b)$ , a right derivative at $a$ and a left derivative at $b$ , and the resulting function $f'$ is continuous on $[a, b]$ . Its graph is then a continuously differentiable curve. A curve given by $x(t)$ and $y(t)$ for $\alpha \le t \le \beta$ is continuously differentiable when $x'$ and $y'$ exist and are continuous on $[\alpha, \beta]$ , with one-sided derivatives at the ends.

In plain words The slope exists and is finite at every point of the closed interval, endpoints included, and it changes without jumps. A graph with a corner, a cusp or a vertical tangent fails the test at that point. A graph with finitely many corners can still be measured by splitting it at the corners and adding the lengths of the pieces, as long as each piece is continuously differentiable on its own closed interval.

Example For $f(x) = x^{3/2}$ on $[0, 1]$ , the formula $f'(x) = \tfrac32 \sqrt{x}$ holds for $x > 0$ . At $0$ only the right derivative can be formed, and

\begin{aligned} &\lim_{h \to 0^+} \frac{h^{3/2} - 0}{h} \\[0.4em] &\quad = \lim_{h \to 0^+} h^{1/2} = 0, \end{aligned}

which equals $\tfrac32 \sqrt{0}$ . So $f'$ is continuous on all of $[0, 1]$ and the curve qualifies. Its mirror image in the line $y = x$ , the graph of $y = x^{2/3}$ on $[0, 1]$ , does not, since $h^{2/3}/h = h^{-1/3}$ grows without bound as $h \to 0^+$ and the tangent at the origin is vertical. That curve still has a finite length. Written as $x = y^{3/2}$ for $0 \le y \le 1$ , it is continuously differentiable in $y$ , and its length equals the length of $y = x^{3/2}$ on $[0, 1]$ .

Arc length integral

Definition Let $f$ be continuously differentiable on $[a, b]$ with $a < b$ . Then the graph of $f$ from $(a, f(a))$ to $(b, f(b))$ is rectifiable, and its arc length is

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

The reason is that, by the mean value theorem, an inscribed polygon has length $\sum \sqrt{1 + f'(c_i)^2}\,\Delta x_i$ with each $c_i$ in $(x_{i-1}, x_i)$ . That is a Riemann sum of the continuous function $\sqrt{1 + f'(x)^2}$ , so as the widest piece shrinks the polygon lengths tend to the integral. In the same way, a curve $x = g(y)$ for $c \le y \le d$ with $g'$ continuous on $[c, d]$ has length $\int_c^d \sqrt{1 + g'(y)^2}\,dy$ .

In plain words Differentiate, square the whole derivative, add $1$ , take the square root, and add up the pieces $ds$ from left to right. The order of these steps matters, because $\sqrt{1 + (y')^2}$ is not $1 + y'$ .

Example The curve $y = x^{3/2}$ is continuously differentiable on $[0, 1]$ , including the right derivative $0$ at $x = 0$ , and $1 + (y')^2 = 1 + \tfrac94 x$ . So

L = \int_0^1 \sqrt{1 + \tfrac94 x}\,dx.

A substitution finishes the calculation. Put $u = 1 + \tfrac94 x$ . Then $du = \tfrac94\,dx$ , so $dx = \tfrac49\,du$ , and the bounds $x = 0$ and $x = 1$ become $u = 1$ and $u = \tfrac{13}{4}$ . Hence

\begin{aligned} L &= \frac49 \int_1^{13/4} u^{1/2}\,du \\[0.4em] &= \frac49 \cdot \frac23 \Bigl[u^{3/2}\Bigr]_1^{13/4} \\[0.4em] &= \frac{8}{27}\left[\left(\tfrac{13}{4}\right)^{3/2} - 1\right] \\[0.4em] &= \frac{8}{27}\left(\frac{13\sqrt{13}}{8} - 1\right) \\[0.4em] &= \frac{13\sqrt{13} - 8}{27} \approx 1.4397. \end{aligned}

The answer lies between the chord bound $\sqrt{2} \approx 1.414$ and the bound $2$ , and just above every polygon length found earlier.

Check yourself

Find the length of the curve $y = \tfrac23 x^{3/2}$ from $x = 0$ to $x = 3$ , and compare it with the distance between the endpoints.

Show the answer

The derivative is $y' = \sqrt{x}$ , and it is continuous on $[0, 3]$ , with right derivative $0$ at $x = 0$ . So $1 + (y')^2 = 1 + x$ and

\begin{aligned} L &= \int_0^3 \sqrt{1 + x}\,dx \\[0.4em] &= \tfrac23 \Bigl[(1 + x)^{3/2}\Bigr]_0^3 \\[0.4em] &= \tfrac23 (8 - 1) = \tfrac{14}{3}. \end{aligned}

The endpoints are $(0, 0)$ and $(3, 2\sqrt3)$ , at distance $\sqrt{9 + 12} = \sqrt{21} \approx 4.583$ . The length $\tfrac{14}{3} \approx 4.667$ is a little larger, as it must be.

Find the length of the curve $y = \tfrac{x^2}{4} - \tfrac{\ln x}{2}$ for $1 \le x \le e$ .

Show the answer

The function is continuously differentiable on $[1, e]$ , with $y' = \tfrac{x}{2} - \tfrac{1}{2x}$ . Squaring gives $\tfrac{x^2}{4} - \tfrac12 + \tfrac{1}{4x^2}$ , and adding $1$ changes the sign of the middle term, so

1 + (y')^2 = \left(\frac{x}{2} + \frac{1}{2x}\right)^2.

The square root of a square is its absolute value. Because $\tfrac{x}{2} + \tfrac{1}{2x} > 0$ for $x > 0$ , the square root is $\tfrac{x}{2} + \tfrac{1}{2x}$ itself, and

\begin{aligned} L &= \int_1^e \left(\frac{x}{2} + \frac{1}{2x}\right)\,dx \\[0.4em] &= \left[\frac{x^2}{4} + \frac{\ln x}{2}\right]_1^e \\[0.4em] &= \frac{e^2}{4} + \frac12 - \frac14 \\[0.4em] &= \frac{e^2 + 1}{4} \approx 2.0973. \end{aligned}

The chord between the endpoints $(1, \tfrac14)$ and $(e, \tfrac{e^2}{4} - \tfrac12)$ has length about $2.0387$ , slightly less.

A student computes the length of $y = x^2$ from $x = 0$ to $x = 1$ as $\int_0^1 (1 + 2x)\,dx = 2$ . Find the mistake, and explain why $2$ cannot be the length.

Show the answer

The correct integrand is $\sqrt{1 + (y')^2} = \sqrt{1 + 4x^2}$ , and the square root of a sum is not the sum of the square roots. In fact $(1 + 2x)^2 = 1 + 4x + 4x^2$ , which exceeds $1 + 4x^2$ for every $x > 0$ , so $1 + 2x > \sqrt{1 + 4x^2}$ there and the student’s integral is too large. The value $2$ is the length of the path that runs one unit right and then one unit up, and the curve is strictly shorter than that path. The correct length is

\begin{aligned} L &= \int_0^1 \sqrt{1 + 4x^2}\,dx \\[0.4em] &\approx 1.4789, \end{aligned}

which lies between the chord length $\sqrt2 \approx 1.414$ and $2$ . Its exact value, $\tfrac14\bigl(2\sqrt5 + \ln(2 + \sqrt5)\bigr)$ , comes from the trigonometric substitution $x = \tfrac12 \tan\theta$ .