# The sequence in the definition of the integral

In my high school Calculus class, we learned this definition of the definite integral: $$\int_a^b f(x)dx = \lim_{n\to \infty} \sum_{i=1}^n f(x_i) \frac{b-a}{n}$$

Now that I know more about sequences and series than I did back then, I realize that $x_i$ must be talking about a sequence that generates the real numbers between a and b. I ALSO know that the real numbers are "dense," meaning there's "no next real number." These facts seem to contradict one another. Can anyone resolve this for me? Can anyone give me a formal definition of the $x_i$ that this series refers to?

EDIT: I can see by the definition Spencer gave that the sequence starts at a and goes to b as i gets large. BUT if I write out the first couple of terms, they're all a. In fact, if I try to pick any term besides the nth term, it's a. What's going on? Can anyone prove the sequence is increasing?

• It doesn't make sense to talk about a sequence that "generates" the real numbers between $a$ and $b$. The point is that the $x_i$ should be allowed to be arbitrary "sample" points, which RRL denotes by $\xi_k$ in his/her answer. Jun 14, 2014 at 13:48

A more general definition of the Riemann integral of a bounded function $f:[a,b] \rightarrow \mathbb{R}$ makes reference to a partition of the interval and a Riemann sum. A partition $P$ is a set of points

$$a = x_0 < x_1 < \ldots< x_{n-1} <x_n = b,$$

and a Riemann sum is defined as

$$S(P;f) = \sum_{k=0}^{n}f(\xi_k)(x_k-x_{k-1}),$$

where $\xi_k$ can be any point in the interval $[x_{k-1},x_k].$

The function is said to be integrable if there exists a real number $I$ such that for any $\epsilon >0$ there exists $\delta >0$ such that if $P$ is any partition with $\max\{|x_k-x_{k-1}|:k=1,2,\ldots,n\}<\delta$ then

$$|S(P;f)-I|<\epsilon$$

for any choice of intermediate points in the Riemann sum.

The integral exists when $f$ is continuous. It also exists for discontinuous functions that are not too "badly" behaved. Suppose, in particular, we choose the partition with

$$x_k = a + \frac{b-a}{n}k \\(k=0,1,\ldots,n),$$

meaning $x_0=a$, $x_1=a+(b-a)/n$, $x_2 = a+2(b-a)/n$, etc.,

Then $\max\{|x_k-x_{k-1}|:k=1,2,\ldots,n\}=(b-a)/n$ and we can make this quantity less than $\delta$ by choosing $n$ sufficiently large.

It will be the case that

$$I=\lim_{n \rightarrow \infty}\sum_{k=0}^{n}f(\xi_k)(x_k-x_{k-1})=\lim_{n \rightarrow \infty} \frac{b-a}{n}\sum_{k=0}^{n}f(\xi_k).$$

The limit is unique regardless of how the points $\xi_k \in [x_{k-1},x_k]$ are chosen. We could for example choose the points $\xi_k = x_k$.

• Thanks for answering! I will probably be chewing on this for the next few days... Jun 14, 2014 at 1:18
• It might be a bit advanced but it can't hurt to see it for the first time. To get a better idea of what is going on, try forming that sum and taking the limit when $f(x) =x$.
– RRL
Jun 14, 2014 at 1:24
• Will do! thanks for the tip Jun 14, 2014 at 1:51

The usual definition in a freshmen calculus course is $x_i=a+i \Delta x$ where $\Delta x = (b-a)/n$.

The $x_i$ are called abcissa and the purpose of the formula above is to produce $x$ coordinates which are evenly distributed throughout the interval. This isn't the only way to specify the $x_i$.

• Thanks for answering. Any other definitions you would like to explain or turn me on to would be appreciated :) Jun 14, 2014 at 0:30
• I can see by this that the sequence starts at a and goes to b as i gets large. BUT if I write out the first couple of terms, they're all a. In fact, if I try to pick any term besides the nth term, it's a. What's going on? Can you prove the sequence is increasing? Jun 14, 2014 at 0:42
• For a fixed $i$ every term eventually gets closer and closer to $a$, but also notice that the total number of terms increases which each step. For each $n$ the $x_i$ are evenly distributed throughout the interval. What I recommend you do is write out the explicit forms of the $x_i$ for $n=3,4,5,$ and $6$. While doing this pay attention to what $i=2$ is doing. Also note that the terms closest to $b$ always correspond to an $i$ that wasn't in the previous partition. Jun 14, 2014 at 0:52
• Oh! aha moment achieved Thank you very much Jun 14, 2014 at 1:43
• This definition (which, as you mention, is unfortunately given in many calculus courses) is not appropriate. For instance, according to this the characteristic function of $\mathbb Q \cap [0,1]$ would be integrable. You really have to allow arbitrary sample points $x_i$ to get the correct definition. Jun 14, 2014 at 13:41