Concern regarding a proof from brilliant

Question

Here in the proof part they deduced that the arc length is $$\lim_{n \to \infty}\sum_{n=1}^{\infty}\sqrt{1+(f'(x_{i}*))^2}\Delta x$$ then said that it was equal to the definite integral. But I didn't understand why. Shouldn't it be $f(x_{i})$ for the limit to be an integral? Since the $x_{i}*$ are not continuous and evenly spaced as $x_i$. Please explain.

Answer 1

Different authors define Riemann integration in different ways. In the definition that I think I remember from high school, to integrate $f$ over the interval $[a,b]$ we define an approximate integral of $n$ steps by setting $x_k = a + k\frac{b-a}{n}$ for $k = 0, 1, 2, \ldots, n.$ This divides the interval $[a,b]$ into $n$ equal-sized intervals between consecutive subscripted $x$ values, $[x_{k-1}, x_k].$ Then the $n$th approximation of the integral is $$ \frac{b-a}{n} \sum_{k=1}^n f(x_k). \tag1$$ Or perhaps it was $$ \frac{b-a}{n} \sum_{k=1}^n f(x_{k-1}). \tag2$$ There actually are distinct names for these sums: $(1)$ is the "right endpoint" sum because you can get it by evaluating $f$ at the right-hand end of each interval $[x_{k-1}, x_k],$ while $(2)$ is the "left endpoint" sum because you can get it by evaluating $f$ at the left-hand end of each interval $[x_{k-1}, x_k].$

But there are other ways to define the integral. You could evaluate $f$ at an $x$-value midway between the two ends of each interval. That's the "midpoint" sum. You could average the values of $f$ at the left and right ends of each interval. That's a "trapezoid" rule.

You could take the "upper" sum, which means you look for the greatest value of $f$ in each interval $[x_{k-1}, x_k].$ This largest value might occur at $x_{k-1}$ (left end), or at $x_k$ (right end), or (in cases where the function has a local maximum in the interval) anywhere between $x_{k-1}$ and $x_k.$

You could take the "lower" sum, which means you look for the least value of $f$ in each interval $[x_{k-1}, x_k].$ Again this might use an $x$-value at the left end, the right end, or somewhere in between.

All of these kinds of sums (except one) are illustrated in an answer to What does it mean for a function to be Riemann integrable? And as it turns out, for functions that satisfy some reasonable criteria, it really does not matter which of these definitions we use--in every case we are going to to define the value of the integral as the limit of the sequence of sums produced by letting $n$ increase without bound, and the limit always comes out to the same number.

Now we can do one better: do not even make a rule that says which $x$ to pick within each interval. The only constraint on the $k$th term of the sum is it has to use an $x$ from somewhere in the interval $[x_{k-1}, x_k].$ If we say that $\xi_k$ represents a value somewhere in the interval $[x_{k-1}, x_k],$ the $n$th approximate integral looks like this: $$ \frac{b-a}{n} \sum_{k=1}^n f(\xi_k).$$ Some authors use a different symbol than $\xi_k$. For example, the symbol might be $x_k^*.$

We can go even further than this, though. Why does each $x_k$ have to follow the rigid formula $x_k = a + k\frac{b-a}{n}$? Instead, we merely insist that $a = x_0 < x_1 < x_2 < \cdots < x_{n-1} < x_n = b.$ Now the sum looks like this: $$ \sum_{k=1}^n (x_k - x_{k-1}) f(\xi_k).$$ So now even the $x_k$s are not necessarily uniformly spaced. This means we cannot rely on the number $\frac{b-a}{n}$ to be the "grid size": instead, we choose the largest of all interval lengths $x_k - x_{k-1}$ as the "grid size". This also means it is not good enough to say $n$ increases without bound; what we really need is to know that the grid size decreases to zero. That's the definition shown in On definition of Riemann integral.

It appears that the definition of integration that was used in that answer on brilliant.com was different from the one you learned.

Answer 2

I would point you to another link. But the confusion arises from the fact that your definition of integral admits only right-Riemann sum. Whereas, you may pick any point in $\Delta x$ and in the $ limit \rightarrow \infty $, it will work out fine. So you may assume it's $x_i$ and the derivation still works.