can someone teach me how to derive the integration of finding the length of a segment from (x1, y1) to (x2, y2) by spliting the x-axis into steps and using limits and summations?
thanks!
It would be better to ask about a specific example than rather than just the general concept! Also, I take it you are actually asking about the arclength of a curve rather than just the length of a straight line segment since that would not require "limits and summations".
Suppose y= f(x) gives a graph that passes through (x1,y1) and (x2,y2). Divide the x-axis into n smaller segments (not necessarily of the same length but that would simplify calculations if you were actually doing calculations).
If one endpoint of such an interal is "$\displaystyle x_i$" then the other endpoint is "$\displaystyle x_{i+1}$". The corresponding points on the curve are $\displaystyle (x_i, f(x_i))$ and $\displaystyle (x_{i+1}, f(x_{i+1})$ and the straight line distance between them is $\displaystyle \sqrt{(x_{i+1}- x{i})^2+ (f(x_{i+1})- f(x_i))^2}$. The total length of all such segments, and so an approximation to the arclength is $\displaystyle \sum_{i=0}^{n-1} \sqrt{(x_{i+1}- x_i)^2+ (f(x_{i+1}- f(x_i))^2}$.
We can simplify that a little by factoring a "$\displaystyle (x_{i+1}- x_i)^2$" out of the square root- of course, it is no longer squared outside the square root: $\displaystyle \sum_{i=0}^{n-1}\sqrt{1+ \frac{(f(x_{i+1}- f(x_i))^2}{(x_{i+1}- x_i)^2}}(x_{i+1}- x_i)$.
Now let $\displaystyle \Delta x_i= x_{i+1}- x_i$ and $\displaystyle \Delta y_i= f(x_{i+1})- f(x_i)$ and that sum becomes $\displaystyle \sum_{i=0}^{i-1}\sqrt{1+ \left(\frac{\Delta y_i}{\Delta x_y}\right)^2}\Delta x_i$.
Of course, that is a "Riemann sum" and in the limit, n goes to 0 so we are taking more and more small segments, it becomes the integral
$\displaystyle \int_{x_1}^{x_2} \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx$
because the fraction $\displaystyle \frac{f(x_{i+1})- f(x_i)}{x_{i+1}- x_{i}}= \frac{\Delta f(x_i)}{\Delta x_i}$, a "difference quotient", becomes the derivative.