How to find length of a part of a curve? 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.

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 $\int_{1}^{8}{\sqrt{1+9x^4}dx}$.
