Arc Length Formula: Understanding the Proof

[Dethrone]

I'm reading about Line Integrals, so I thought I'd review the proof for the arc length formula. However, there's something I don't quite understand about the proof that I either overlooked or understood before.View attachment 4468View attachment 4469

From what I see, the arc length formula holds because the MVT guarantees that there exists an $x_i^* \in [x_{i-1},x_i]$ for all $i$. This means that the integral depends on the sample points $x_i^*$ - the sample points must be the same $x_i^*$ that the MVT guarantees. If the value of the integral depends on the sample points, how does the integral exist?

 

[Euge]

Hi Rido12,

How does your book define the definite integral? There are various (but equivalent) formulations that used by different textbooks. It'll be helpful if you could show your book's definition.

From what I see though, my guess is that the formulation they use is the following. Let $f$ be a continuous function on $[a,b]$. Partition the interval $[a,b]$ into $n$ subintervals of equal length $\Delta x$ and choose a point $x_i^*$ from each subinterval. Then $\int_a^b f(x)\, dx$ is defined as $\lim\limits_{n\to \infty} \sum\limits_{i = 1}^n f(x_i^*)\, \Delta x$. If this is the formulation being used, then your question has to do why this formulation is well-defined, not about the arclength. The definition of the definite integral is independent of the choice of sample points, which can be proven by using the uniform continuity of $f$ on $[a,b]$.

 

[Dethrone]

Hi Euge,

In class, we were taught the partition definition of the definite integral, but the definition that the book uses is the following:

View attachment 4473

Yes, my question is the last sentence of yours. From the definition of the integral provided above, doesn't the arc length formula depend on the sample points chosen and thus, not integrable?

 

[Euge]

The definition used matches the one I wrote. As long as $f'$ is continuous on $[a,b]$, so is $\sqrt{1 + (f')^2}$. So the Riemann sums $\sum\limits_{i = 1}^n \sqrt{1 + [f'(x_i^*)]^2}\,\Delta x$ converge to $\int_a^b \sqrt{1 + [f'(x)]^2}\, dx$.