In Simmons' Calculus with Analytic Geometry, on pg. 221, the following formula is provided to evalute the area under some curve:

$$ \text{Area} = \lim_{n\rightarrow\infty}\sum_{k=1}^nf(\bar{x}_k)\Delta{x_k}$$

where $\Delta{x_k}=x_k-x_{k-1}$ and $f(\bar{x}_k)$ is the minimum value of $f(x)$ on the $k\text{th}$ subinterval $[x_{k-1},x_k]$, as in this figure:

enter image description here

But then the book goes on to state that that formula is equivalent to:


My question is: why are these formulas equivalent? I understand why the first one yields the desired result, but I can't see how we can evaluate the area under the curve by the second one. Any clarification will be appreciated. Thanks.


The first one is not entirely clear because it doesn't say what $x_k$ actually are. In the figure it looks like you have $x_k=a+\frac{k(b-a)}{n}$ (the uniform partition), but that isn't made explicit.

The second one is saying that it doesn't really matter what $x_k$ actually are: as long as you partition $[a,b]$ into subintervals and the lengths of all those subintervals go to zero, you will recover the same limit (for a nice enough function, in particular for a continuous function).

What's going on here is the following. I won't go into too much detail because exactly these things are discussed in any undergraduate-level real analysis text. (I learned from Strichartz The Way of Analysis, which has a few serious flaws but I generally like.)

We define Riemann integrability in the absolute most restrictive way we can think of. We define an "evaluation partition" of $[a,b]$ to be a pair of finite sequences $(\{ x_k \}_{k=0}^n,\{ y_k \}_{k=1}^n)$, subject to the following conditions. (Note that this is not a standard term in the literature.)

  • $x_k$ is strictly increasing
  • $x_0=a$
  • $x_n=b$
  • $x_{k-1} \leq y_k \leq x_k$.

We define the Riemann sum for an evaluation partition $P$ and a bounded function $f$ to be:

$$R(f,P)=\sum_{k=1}^n f(y_k) (x_k-x_{k-1}).$$

We define the mesh $m(P)$ of an evaluation partition $P$ to be $\max_{k=1,\dots,n} x_k-x_{k-1}$. (This is a standard term.) We say a function is Riemann integrable if for any sequence of evaluation partitions $P_n$ whose mesh goes to zero, $R(f,P_n)$ converges to a single universal limit.

Now this definition is horrible, because it tells us to check for evaluation at totally arbitrary points. We can make it easier to use by focusing only on the partitions with evaluation points $y_k=\arg \min_{x \in [x_{k-1},x_k]} f(x)$ and $z_k=\arg \max_{x \in [x_{k-1},x_k]} f(x)$. (The former is what is suggested in your definition). The former sum is called the lower sum; the latter sum is called the upper sum. Restricting attention to just these evaluation partitions (still leaving the $x_k$ part arbitrary) gives us a criterion for so-called "Darboux integrability".

The obvious question is now: is Darboux integrability the same as Riemann integrability? The answer is yes. The gist of the reason is that if we fix any partition $x_k$ and we look at evaluation partitions $(x_k,y_k)$ and $(x_k,z_k)$, the difference between their Riemann sums is less than the difference between the lower and upper sums for the partition $x_k$.

Finally if we want to make use of this, we should prove that continuous functions are Riemann/Darboux integrable. This requires an intermediate lemma, called the Heine-Cantor theorem, which says that continuous functions on a compact set like $[a,b]$ are uniformly continuous. With this in hand, we have a uniform modulus of continuity $\delta(\epsilon)$, and now we can ensure that the upper and lower sums are within $\epsilon$ of each other provided that our partition has mesh less than $\delta(\epsilon/(b-a))$. Note that in the special but common case of continuously differentiable $f$, $\delta(\epsilon)$ can be taken to be $\epsilon/M$ where $M=\max_{x \in [a,b]} |f'(x)|$.

  • $\begingroup$ Oh, my mistake. So, the idea is as follows: we let $n$ be a positive integer and divide the interval $[a,b]$ (in which we want to evaluate the area under the curve) into $n$ equal subintervals by inserting $n-1$ equally spaced points of division $x_1, x_2,...,x_{n-1}$ between the endpoints $a$ and $b$. We then denote $a$ and $b$ by $x_0$ and $x_n$ and the lenght of the $k$th interval by $\Delta{x_k}$. I understand the geometric intuition behind the first formula, but I really don't get it in the second one. I'd really appreciate it if you could clarify it. Thanks for you patience. $\endgroup$ – daniels Mar 14 '17 at 9:20
  • $\begingroup$ By the way, I think just some intuition would be useful already, but if you can actually prove that the two limits are indeed equal, that would be even more appreciated. $\endgroup$ – daniels Mar 14 '17 at 9:42
  • $\begingroup$ @daniels Let me know if my answer helps now. (It may be more technical/advanced than you wanted...) $\endgroup$ – Ian Mar 14 '17 at 12:59
  • $\begingroup$ First, thx again for your answer. So, as far as I understoond your answer it seems to have showed that lower, upper sums and sums with completely arbitrary evaluation points are all equivalent. I already had some intuition on that, but it was nice to see it demonstrated more rigorously anyway. Now, let's stick with the lower sum and take the limit of it as $n\rightarrow\infty$. Such a limit would yield the area under $f$ between $a$ and $b$, right? What I fail to understand is why we can take the same limit, but as $\max{\Delta{x_k}}\rightarrow0$, and still get the same result. $\endgroup$ – daniels Mar 15 '17 at 19:39
  • $\begingroup$ I'm not sure it can be demonstrated rigorously. However, any intuition on why this is the case would be greatly appreciated already. $\endgroup$ – daniels Mar 15 '17 at 19:45

Ian's answer is very good. I will provide a slightly more particular point of view, assuming $f$ is continuously differentiable and $\Delta x_i$ is a constant $\Delta x$.

Note (from your geometric intuition) that there exists numbers $c_i$ (depending on $n$) such that $x_{i-1} < c_i < x_{i}$ and $$ \text{Area} = \sum_{i=1}^n f(c_i) \Delta x. $$ Because this equality is true for all $n$, you also have $$ \text{Area} = \lim_{n\rightarrow \infty}\sum_{i=1}^n f(c_i) \Delta x. $$

Now, my choice of points $c_i$ at which to evaluate $f$ is rather particular and may differ from your choice $\bar{x}_i$. I'll show that, as long as $\Delta x \rightarrow 0$, the difference in calculated area coming from different choices of points is negligible.

Let $M = \max_{x \in [a,b]} |f'(x)|$, so that $$ |f(c_i) - f(\bar{x_i})| \le M |c_i - \bar{x_i}|. $$ We can compare the two different ways of approximating the area: $$ |\sum_{i=1}^n f(c_i) \Delta x - \sum_{i=1}^n f(\bar{x}_i) \Delta x| \le \sum_{i=1}^n M|c_i-\bar{x}_i| \Delta x \le M (b-a) \Delta x, $$ using the fact that $\sum_{i=1}^n |c_i - \bar{x}_{i}| \le (b-a)$.

As $n \rightarrow \infty$, $\Delta_x \rightarrow 0$ and we find $$ \text{Area} = \lim_{n \rightarrow \infty} \sum_{i=1}^n f(c_i) \Delta x = \lim_{n \rightarrow \infty} \sum_{i=1}^n f(\bar{x}_i) \Delta x . $$

  • $\begingroup$ Good idea to explicitly mention the $C^1$ case, seeing as this is the familiar one. I added that to my answer. $\endgroup$ – Ian Mar 14 '17 at 14:46

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