So I was writing out the derivation of the arc length formula, and I got this far $$\sum_{i=0}^n \sqrt{1+(f'(x))^2} = length $$ With limit of $n$ tending to infinity.
The curve differentiable and continuous in the interval $[a,b]$.
How do I convert this to an integral of the same with limits from $a$ to $b$?
I know the mean value theorem can be used here, but since I haven't learned it at school yet, I wanted to avoid it.
Please give a step-by-step answer as I'm not exactly the strongest guy in math :)
Thanks in advance!
Hint: you dont have to do it this way. Rather think of an infinitesimal piece of your function. If $ds$ is the infinitesimal arclenght then by the theorem of Pythagoras, we know $ds^2=dx^2+dy^2$, now divide by $dx^2$ to get $(ds/dx)^2=1+(dy/dx)^2$, take the squareroot and multiply by $dx$: $ds =\sqrt{1+f'(x)^2}dx$ integrate from $x_1$ to $x_2$ to get the arclength: $s(x_2)-s(x_1)=\int_{x_1}^{x_2}\sqrt{1+f'(x)^2}dx$
If you really want to do it by summation:
A small piece of eh arclength is given by $\Delta s_{i}^2=\Delta x_i^2+\Delta y_i^2 \implies (\Delta s_i/\Delta x_i)^2=1+(\Delta y_i/\Delta x_i)^2 \implies (\Delta s_i/\Delta x_i)=\sqrt{1+(\Delta y_i/\Delta x_i)^2} \implies \Delta s_i=\sqrt{1+(\Delta y_i/\Delta x_i)^2}\Delta x_i$ Now summing over $i$ gives you the Arclength $s^{\{n\}}=\sum_{i=1}^{n}\sqrt{1+(\Delta y_i/\Delta x_i)^2}\Delta x_i$. As $\Delta x_i \to 0$ for $n\to \infty$ this summation will turn into a riemann sum and $\Delta y_i/\Delta x_i \to f'(x)$.