The sequence in the definition of the integral

Question

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?

Answer 1

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

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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 $x_i$.

Answer 2

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 $\xi_k = x_k$.