How can I find the length of a curve, for example $f(x) = x^3$, between two limits on $x$, for example $1$ and $8$?

I was bored in a maths lesson at school and posed myself the question:

What's the perimeter of the region bounded by the $x$-axis, the lines $x=1$ and $x=8$ and the curve $y=x^3$?

Of course, the only "difficult" part of this is finding the length of the part of the curve between $x=1$ and $x=8$.

Maybe there's an established method of doing this, but I as a 16 year-old calculus student don't know it yet.

So my attempt at an approach was to superimpose many triangles onto the curve so that I could sum all of their hypotenuses.

Just use many triangles like the above,

$$ \lim_{\delta x\to 0}\frac{\sqrt{\left(1+\delta x-1\right)^2+\left(\left(1+\delta x\right)^3-1^3\right)^2}+\sqrt{\left(1+2\delta x-\left(1+\delta x\right)\right)^2+\left(\left(1+2\delta x\right)^3-\left(1+\delta x\right)^3\right)^2}+\cdots}{\frac7{\delta x}} $$

I'm not entirely sure if this approach is correct though, or how to go on from the stage I've already got to.

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    $\begingroup$ It would be better to use integration . $\endgroup$
    – Khosrotash
    Feb 3, 2017 at 17:12
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    $\begingroup$ What you search is this: en.wikipedia.org/wiki/Arc_length $\endgroup$ Feb 3, 2017 at 17:14
  • $\begingroup$ @Khosrotash My first first attempt at a solution was to use integration, drawing a parallel to the fact that I know a sphere's volume can be differentiated to get its surface area. I got 1031 as my answer using this method. $\endgroup$ Feb 3, 2017 at 17:17
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    $\begingroup$ @theonlygusti - That "derivative of volume = area" (or for 2D, "derivative of area = perimeter") trick only works for highly regular shapes. For other shapes, the change in thickness due to a change in radius is uneven depending upon the direction, and that uneveness spoils the result. In this situation, it is not even clear what measurement you would call the "radius", which is key even for regular shapes. For example, it only works for squares if you use half the sidelength as the radius. For the full sidelength, the result is off by a factor of 4. $\endgroup$ Feb 3, 2017 at 17:59
  • $\begingroup$ @PaulSinclair yeah, one of my friends pointed that out to me, that it would only works for highly regular shapes. That's why I stopped thinking about it like that and began to try it as I laid out in the question. $\endgroup$ Feb 4, 2017 at 9:21

7 Answers 7


I think it's useful to explore how far you got toward the answer in your own efforts. As it turns out, you were mostly on a good track.

When you get around to computing the lengths of curves in your classes (which you will, if you keep studying calculus and the related mathematics), the formula for the length of the curve will very likely be explained by superimposing many small triangles on the curve and summing the lengths of their hypotenuses, just as you proposed to do.

For a mesh of uniform steps in the $x$ direction starting at the lower left end of your curve, you wrote out the lengths of the hypotenuses of the first two triangles. Let's consider the hypotenuse of the second triangle, since that's where this starts to get complicated (and interesting): $$ \sqrt{\left(1+2\delta x-\left(1+\delta x\right)\right)^2+\left(\left(1+2\delta x\right)^3-\left(1+\delta x\right)^3\right)^2}. $$ We can generalize that to any triangle along the curve by supposing there are $k$ triangles to the left. Then the base of the triangle runs from $x = 1 + k\delta x$ to $x = 1 + (k+1)\delta x,$ and the hypotenuse is $$ \sqrt{\left(1+(k+1)\delta x -\left(1+k\delta x\right)\right)^2 +\left(\left(1+(k+1)\delta x\right)^3-\left(1+k\delta x\right)^3\right)^2}. $$ Your first triangle's hypotenuse was just this with $k=0,$ and the second was this with $k=1.$

We can simplify things a bit by observing that $$1+(k+1)\delta x -\left(1+k\delta x\right) = \delta x.$$ So the hypotenuse of a triangle is $$ \sqrt{\left(\delta x\right)^2 +\left(\left(1+(k+1)\delta x\right)^3-\left(1+k\delta x\right)^3\right)^2}. $$ Next, we recall that $\left(1+k\delta x\right)^3 = f(1+k\delta)$ and $\left(1+(k+1)\delta x\right)^3 = f(1+(k+1)\delta),$ so $$ \left(1+(k+1)\delta x\right)^3-\left(1+k\delta x\right)^3 = f(1+(k+1)\delta) - f(1+k\delta), $$ which (either way you write it) is the height of the small triangle. If you used the small-triangle approximation when studying the derivative of $f,$ you may have written $\delta y$ for the height of the triangle when the base is $\delta x.$ Then $ \left(1+(k+1)\delta x\right)^3-\left(1+k\delta x\right)^3 = \delta y $ and the hypotenuse of the triangle is $$ \sqrt{\left(\delta x\right)^2 + \left(\delta y\right)^2}. $$

Now we come to one of the choices people commonly make in their notation, which is that some people like to denote a small increment in $x$ by the symbol $\delta x,$ whereas others write $\Delta x.$ (The same Greek letter, but upper-case instead of lower-case.) If we recognize that these are just two different styles of writing the same thing, then we see that the length of the hypotenuse is exactly the formula $\sqrt{(\Delta x)^2+(\Delta y)^2}$ that we see at the beginning of the answer by Khosrotash. You could proceed onward from there as that answer does.

The only part where you stumbled a bit was when you divided by $\frac{7}{\delta x}.$ In fact, the numerator in your limit is the total length of the hypotenuses of the small triangles you placed along your curve. The size of your incremental step $\delta x$, and the fact that $x$ increases from $1$ to $8,$ are both represented in the number of terms you would have in your numerator. There will be $\frac{7}{\delta x}$ of those terms (representing that number of triangles), so dividing by $\frac{7}{\delta x}$ will give you the average length of a hypotenuse; but you don't want the average (which is going to go to zero anyway!), you want the total. Therefore, do not divide by the number of terms.

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    $\begingroup$ I want to upvote this multiple times! Super helpful :) $\endgroup$ Feb 4, 2017 at 9:20
  • $\begingroup$ +1 anyway because it's a great introductory answer, but especially for "that's where this starts to get complicated (and interesting)" $\endgroup$
    – Paul Evans
    Feb 4, 2017 at 22:19
  • $\begingroup$ @theonlygusti: Good news! You can upvote it two and a half times, because it was you who asked the question. Just click the tick mark on the left, and David K will get an extra 15 points. $\endgroup$
    – TonyK
    Feb 6, 2017 at 11:21
  • $\begingroup$ @TonyK stop deleting and reposting that. $\endgroup$ Feb 6, 2017 at 11:51
  • $\begingroup$ @theonlygusti: That is the only way to edit a mistake in a comment after the five-minute time limit has expired. I'm sorry if it spoiled your day. (And you still haven't clicked on that tick mark!) $\endgroup$
    – TonyK
    Feb 6, 2017 at 12:35

Element of curve length is $\sqrt{(\Delta x)^2+(\Delta y)^2}$

to find curve length you can add these elements $$\sum\sqrt{(\Delta x)^2+(\Delta y)^2}$$ this is equal to find $$\sum\sqrt{(\Delta x)^2+(\Delta y)^2}=\\\sum\sqrt{(\Delta x)^2\left(1+\left(\dfrac{\Delta y}{\Delta x}\right)^2\right)}\\=\\ \sum\Delta x\sqrt{1+\left(\dfrac{\Delta y}{\Delta x}\right)^2} $$ to better approximation we can calculate limit of $\displaystyle\sum\Delta x\sqrt{1+\left(\dfrac{\Delta y}{\Delta x}\right)^2}$ when $n \to \infty$ so $$\lim_{n \rightarrow \infty}\sum\Delta x\sqrt{1+\left(\dfrac{\Delta y}{\Delta x}\right)^2}=\\\int_{a}^{b}dx\sqrt{1+\left(\dfrac{d y}{d x}\right)^2}=\\\int_{a}^{b}dx\sqrt{1+(f'(x))^2}$$And now , in your case $$y=x^3 \to y'=3x^2\\\int_{1}^{8}dx\sqrt{1+(3x^2)^2}=\\\int_{1}^{8}\sqrt{1+9x^4}dx$$

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    $\begingroup$ I don't know how to find that though. How do I find $\sum\Delta x\sqrt{(1+(\dfrac{\Delta y}{\Delta x})^2}$ $\endgroup$ Feb 3, 2017 at 17:18
  • $\begingroup$ Oh ok, that edit helped a bit, as I know how to find that integral. Still confused about that summation though. $\endgroup$ Feb 3, 2017 at 17:20
  • $\begingroup$ +1 This answer is good but you might want to add (in the beginning of the answer) how the element of curve length is related to the OP's superimposition of many right-angled triangles. I'll think it'll make it really clear and show him that he was on the right track. $\endgroup$ Feb 4, 2017 at 20:52
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    $\begingroup$ What I hate about this (classical) answer is, what if you couldn't just factor out $\Delta x$ like that? Nobody ever mentions how to do it without relying on such a magical factorization... $\endgroup$
    – user541686
    Feb 5, 2017 at 8:54
  • $\begingroup$ You have to remember differentiation from first principles. (dy is the numerator) math2001.github.io/post/length-of-a-curve $\endgroup$
    – math2001
    Mar 9, 2019 at 4:28

How do you measure length when you drive in a car? Right, you just take the speed times the time traveled. In other words: $$ L = \int |\vec{v}(t)| ~ dt $$

In your case you "drive" along the following curve $$ \vec{r} = (y(x), x) = (x^3, x) $$ which has "velocity" (in fact the length is invariant under reparametrisation so we take for brevity x to be our time coordinate) $$ \frac{d \vec{r}}{dx} = (\frac{dy}{dx},\frac{dx}{dx}) = (3x^2,1) $$ Thus $$ L = \int_{x_0}^{x_1} \sqrt{(3x^2)^2+(1)^2} ~ dx = \int_{x_0}^{x_1} \sqrt{9x^4 + 1} ~ dx $$

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    $\begingroup$ Since we're integrating modulus, it's really speed rather than velocity that we're integrating (or multiplying by a short time period). $\endgroup$
    – J.G.
    Feb 3, 2017 at 19:52
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    $\begingroup$ Using a physics argument to explain made it really clear to me! +1 $\endgroup$
    – nluigi
    Feb 3, 2017 at 23:21
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    $\begingroup$ @J.G. Does not speed and magnitude of velocity different ? $\endgroup$
    – user312097
    Feb 5, 2017 at 0:51

If a planar curve in $\mathbb{R}^2$ is defined by the equation $\text{y}=\text{f}\left(x\right)$ where $\text{f}$ is continuously differentiable, then it is simply a special case of a parametric equation where $x=t$ and $\text{y}=\text{f}\left(t\right)$, the arc length is given by:

$$\mathcal{S}=\int_\text{a}^\text{b}\sqrt{1+\left(\frac{\text{d}\space\text{y}}{\text{d}\space x}\right)^2}\space\text{d}x\tag1$$

For the derivation, look at Wikipedia.

So, for example when $\text{y}\left(x\right)=x^3$, $\text{a}=1$ and $\text{b}=8$:

$$\mathcal{S}=\int_1^8\sqrt{1+\left(\frac{\text{d}}{\text{d}\space x}\left(x^3\right)\right)^2}\space\text{d}x=\int_1^8\sqrt{1+9x^4}\space\text{d}x\approx511.1449347\tag2$$

  • $\begingroup$ Do you mind showing a small derivation of this? $\endgroup$ Feb 3, 2017 at 17:19
  • $\begingroup$ @theonlygusti Look at my edit! $\endgroup$ Feb 3, 2017 at 17:22

Your idea to use right triangles is good! It might be easier to look at a general curve than $y=x^3$, though, to build the machinery.

Let's say we wish to measure the arclength between $x=a$ and $x=b$. Choose $N+1$ points $\{P_1,P_2,\dots,P_{N+1}\}$ to partition the curve into $N$ sections. In the graphic I chose $N=6$ such that I had $7$ points.


The distance between two points, $P_{n}$ and $P_{n+1}$ is essentially the Pythagorean theorem.

$$ P_{n}P_{n+1} = \sqrt{(\Delta x)^2 + (\Delta y)^2}$$

We can estimate the arclength by summing the distances between these points.

$$\sum_{n=1}^N P_nP_{n+1} = \sum_{n=1}^{N} \sqrt{(\Delta x_n)^2 + (\Delta y_n)^2}$$

As we take the number of partitioning points to infinity, or $N\to\infty$, we have $\Delta x\to dx$ and $\Delta y\to dy$. We call the resulting differential distance $ds$. It is a measure of infinitesimal length.

\begin{align} ds &= \sqrt{dx^2 + dy^2} \\ &= \sqrt{1 + \left(\frac{dy}{dx}\right)^2}dx \\ &= \sqrt{1 + \left(f'(x)\right)^2}dx \end{align}

In the limit, the sum above becomes the integral you seek (i.e. we are now summing up infinitely many infinitesimal arc length elements along the curve).

$$\sum_{n=1}^{N} P_nP_{n+1} \to \int_{P_1}^{P_{N+1}} ds = \int_a^b \sqrt{1 + \left(f'(x)\right)^2}dx$$

You ought to find a similar dialogue in Stewart's or Thomas's Calculus. Look in the index for arc length integral.


Yes such a method exists, it's called arc length of a curve. As described in other answers, it is an integral. This is for an xy plane. When dealing with vector-valued functions as a function of time, it is the sqrt of the sum of the derivatives of the components squared. I'm sorry for not writing the symbols and the actual formulas. I'm writing from my phone.


Your idea is along the right lines.

To find the arc length from $x=1$ to $x=8$ we first split the interval $[1,8]$ into sub-intervals $[x_0,x_1], \ldots [x_{n-1},x_n]$ where $$1=x_0 < x_1 < x_2 < \ldots < x_n = 8$$

This is called a partition of $[1, 8]$. Like you suggested in your question, we can then superimpose triangles on the curve where the width of each triangle is the width of a sub-interval in our partition.

Summing up the hypotensuse of each of these triangles gives an approximation of the arc length $$\sum_{i=1}^{n}\sqrt{(x_i - x_{i-1})^2 + (f(x_i) - f(x_{i-1}))^2}$$

Now if $f$ is differentiable (which in your case it is) we can apply the mean value theorem to $f$ on each sub-interval $[x_{i-1}, x_i]$ to get that there is $c_i \in [x_{i-1},x_i]$ such that $$ f'(c_i)(x_i - x_{i-1}) = f(x_i) - f(x_{i-1}) $$

Substituting this into our approximation for the arc length and simplifying a bit we get $$ \sum_{i=1}^{n}(x_i - x_{i-1})\sqrt{1 + (f'(c_i))^2} $$

If you've seen the definition of the Reimann integral before you will recognise that this is precisely a Reimann sum for the function $g(x)=\sqrt{1 + f'(x)}$. This means that our approximation of the arc length of $f$ is the same as an approximation for the area under $g$.

Going from this Reimann sum to the integral is exactly the idea of taking '${\delta}x \rightarrow 0$' as you wrote.

Therefore we can say that the arc length is given by $$ \int_{1}^{8}{g(x)dx} = \int_{1}^{8}{\sqrt{1+(f'(x))^2}dx} $$

(provided that $g$ is integrable)

As pointed out in the other answers, in your case the integral is $\int_{1}^{8}{\sqrt{1+9x^4}dx}$.


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