In my textbook the first step in deriving the length of a function curve is to the the sum $\lim_{n\to \infty}\sum_{i=0}^{n}\sqrt{(x_{i}-x_{i-1})^{2}+(f(x_{i})-f(x_{i-1}))^{2}}$ Now my text continues to derive the formula in a clear and understandable way. However i tried to do it myself and this is the logic i used: as $n\to\infty$ then $x_{i}\to x_{i-1}$ and $f(x_{i})\to f(x_{i-1})$ so we can write the previous sum as the following integral $\int_{a}^{b}\sqrt{(x-x)^{2}+(f(x)-f(x))^{2}}$ which is equal zero. So where did i make the mistake? Thank you
Edit sorry to post this question as i was able to see my mistake right after posting it, the first sum does not have a delta x term so it's not reinman sum. I used a similar logic to what's described here here the textbook approximated $x_i$ and $x_{i-1}$ by $x_{i}^*$
If $a$ is approaching $b$, that does not mean $a$ is equal to $b$. In this case, $f(x_i)$ is approaching $f(x_{i-1})$, it does not mean both are equal. The same goes for $x_i$ and $x_{i-1}$.
So their difference is not $0$, instead it is something really small we will call $dy$ and $dx$.
$$f(x_i)-f(x_{i-1}) = dy$$ $$x_i-x_{i-1} = dx$$
Substituting... $$L =\int_{a}^{b} \sqrt{(dx)^2+(dy)^2}$$ Factor out $(dx)^2$... $$L =\int_{a}^{b} \sqrt{(dx)^2(1+(\frac{dy}{dx})^2)}$$ We can take the square root of $(dx)^2$ and change $\frac{dy}{dx}$ to $f'(x)$. $$L =\int_{a}^{b} \sqrt{1+f'(x)^2} \ \ dx$$