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.

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 
$$y=x^3 \to y'=3x^2\\\int_{1}^{8}dx\sqrt{1+(3x^2)^2}=\\\int_{1}^{8}\sqrt{1+9x^4}dx$$
