Prove that the length of the curve $y=f(x)$ is given by $\int_a^b \sqrt{1+f'(x)^2}dx$ I'm trying to prove that the length of a curve of equation $y=f(x)$ is given by $$\ell=\int_a^b \sqrt{1+f'(x)^2}dx$$
without using line integral (an using Riemann sum). So I consider only the case where $a=0$ and $b=1$. So I consider a regular partition $$\Pi:0<\frac{1}{n}<\frac{2}{n}<...<1.$$
Then, using a simple draw, I can make the approximation that for $n$ big enough, $$\ell\approx \sum_{k=1}^n \sqrt{\frac{1}{n^2}+\left(f\left(\frac{k}{n}\right)-f\left(\frac{k-1}{n}\right)\right)^2}=\frac{1}{n}\sum_{k=1}^n\sqrt{1+\left(\frac{f\left(\frac{k}{n}\right)-f\left(\frac{k-1}{n}\right)}{\frac{1}{n}}\right)^2}.$$
Now, if $n$ is big enough, then $$\frac{f\left(\frac{k}{n}\right)-f\left(\frac{k-1}{n}\right)}{\frac{1}{n}}\approx -f'\left(\frac{k}{n}\right),$$
and thus,
$$\ell=\lim_{n\to \infty }\frac{1}{n}\sum_{k=1}^n\sqrt{1+\left(\frac{f\left(\frac{k}{n}\right)-f\left(\frac{k-1}{n}\right)}{\frac{1}{n}}\right)^2}\approx\lim_{n\to \infty }\frac{1}{n}\sum_{k=1}^n\sqrt{1+f'\left(\frac{k}{n}\right)^2}=\int_0^1\sqrt{1+f'(x)^2}\mathrm d x.$$
My question : How can I do it rigorously ?
In other words, how can I get $$\lim_{n\to \infty }\frac{1}{n}\sum_{k=1}^n\sqrt{1+\left(\frac{f\left(\frac{k}{n}\right)-f\left(\frac{k-1}{n}\right)}{\frac{1}{n}}\right)^2}=\int_0^1\sqrt{1+f'(x)^2}\mathrm d x.$$
 A: Let's assume that $f'^{2}$ is Riemann integrable on $[0, 1]$ which implies that $g(x) = \sqrt{1 + \{f'(x)\}^{2}}$ is Riemann integrable on $[0, 1]$ and hence by definition of Riemann integral as a limit of a sum we have $$\int_{0}^{1}\sqrt{1 + \{f'(x)\}^{2}}\,dx = \lim_{n \to \infty}\frac{1}{n}\sum_{k = 1}^{n}\sqrt{1 + \left\{f'(t_{k})\right\}^{2}}\tag{1}$$ where $t_{k}$ is any arbitrary point in $[(k - 1)/n, k/n]$. Now the approximation of arc-length (as mentioned in question) is given by $$F(n) = \frac{1}{n}\sum_{k = 1}^{n}\sqrt{1 + \left(\dfrac{f\left(\dfrac{k}{n}\right) - f\left(\dfrac{k - 1}{n}\right)}{\dfrac{1}{n}}\right)^{2}}\tag{2}$$ By mean value theorem we have $$\dfrac{f\left(\dfrac{k}{n}\right) - f\left(\dfrac{k - 1}{n}\right)}{\dfrac{1}{n}} = f'(t_{k})\tag{3}$$ where $t_{k}$ is some point in $((k - 1)/n, k/n)$. The arc-length $l$ is given by $\lim_{n \to \infty}F(n)$ and using equations $(1)-(3)$ we see that $$l = \lim_{n \to \infty}F(n) = \lim_{n \to \infty}\frac{1}{n}\sum_{k = 1}^{n}\sqrt{1 + \left\{f'(t_{k})\right\}^{2}} = \int_{0}^{1}\sqrt{1 + \{f'(x)\}^{2}}\,dx$$
