TITLE: How to find length of a part of a curve?
QUESTION [27 upvotes]: 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.

REPLY [6 votes]: 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.