Given function $f$ continuous on $[a,b]$, the area of the 'region' between the horizontal axis and the line described by $f$ can be approximated by the sum of multiple rectangles of equal width. This width, $w$, is equal to $\frac{b-a}{n}$, where $b-a$ describes the length of the segment between $b$ and $a$ and $n$ is the number of equal sub-segments which that segment is divided into.

In approximating the area of the region between the line described by $f$ on $[a,b]$ and the horizontal axis, the area of some $n$ amount of rectangles of width $\frac{b-a}{n}$ are considered. The height of these rectangles is determined by $f$.

The formula for the area of this region using $n$ rectangles (call this an $n$-approximation) is as follows: $$(\frac{b-a}{n})\sum_{i=0}^{n-1}{f(a+i(\frac{b-a}{n})})$$ Where the multiple $\frac{b-a}{n}$ is the width of each rectangle, and each iteration of the series describes height (or value of $f$) as the input is increased from $a$ (which is the height of the initial triangle with its top-left corner on $(a, f(a))$) by increments of $\frac{b-a}{n}$ until $a+i(\frac{b-a}{n})=b$; this can also be said as 'up until the sum of the previous interation and $\frac{b-a}{n}$ is equal to $b$.

As the numbers of rectangles considered is increased/as $n$ is increased/as the width of each rectangle is decreased, the amount of area under the graph not accounted for decreases. In other words, approximations considering more rectangles of less width (whose total width is still $b-a$) are more accurate than approximations using fewer rectangles of greater width.

So, the area between the line of $f$ and the horizontal axis on $[a,b]$ can be calculated as: $$\lim_{n \to \infty}(\frac{b-a}{n})\sum_{i=0}^{n-1}{f(a+i(\frac{b-a}{n})})$$ Which is the Riemann Sum.

The proper notation of a Riemann sum is as follows: $$\sum_{k=1}^n f(c_k)\Delta x_k$$

As I understand it: the interval $[a,b]$ is decomposed into $n$ sub-intervals; these subintervals can be described by the collection (sequence? set?) of intervals $\{ [x_0,x_1],[x_1,x_2]...[x_{n-1},x_n] \}$. $x_0=a, x_n=b$. $c_k$ is any point on $[x_{k-1},x_k]$, and $\Delta x_k$ is the width of the interval $[x_{k-1},x_k]$. So, $f(c_k)$$\Delta x_k$ is the area of the $k^{th}$ rectangle, and the sum is the total area of the rectangles under the function $f$ on the interval $[a,b]$.

For some function $f$ continuous on $[a,b]$ for which $c_k$ is any point on $[x_{k-1},x_k]$. $[a,b]$ is, of course, partitioned into $n$ closed sub-intervals, where $x_k$ is part of the collection of points (set? sequence?) $ \{ {x_0,x_1,x_2...x_n} \}$

My question is whether or not my understanding of Riemann Sums is correct; especially my understanding of its proper notation.


2 Answers 2


Your understanding is quite solid. I do have a couple of comments:

  1. The set of intervals $\lbrace [x_0, x_1] ,\ldots, [x_{n-1}, x_n]\rbrace$ is often called a (topological) "partition". A partition of a subset $S$ of $\mathbb{R}$ (or a topological space, more generally) is a set of subsets that that union to $S$ and have disjoint interiors. That is, they cover all of $S$, and they can overlap on the edges.

  2. Your first spiel about Riemann sums are specifically related to uniform Riemann sums, where the intervals in your partition are all of the same length: $\frac{b - a}{n}$. Riemann sums need not be uniform (some rectangles can be thinner than others), though for an integrable functions, the uniform Riemann sums will approach the integral.

  3. The notation $\Delta x_k$, with its dependence on $k$, I think is supposed to reference non-uniform Riemann sums. The $\Delta x_k$ is the width of the rectangle, and the notation reserves the right to have this change with $k$. (That said, I don't like the notation, and would favour $\Delta_k$ instead, since the length primarily depends on $k$, not the right endpoint of the interval $[x_{k - 1}, x_k]$!)


I think your reasoning is sound. The $c_k$ can be viewed as just shorthand for $a+\frac{k}{n}(b-a)$, and the $\Delta x_k$ can be viewed as $\frac{b-a}{n}$. As for your other questions as to what $\{x_0,x_1,...,x_n\}$ is, it's a set. Here's a more set-theory-heavy definition of the definite integral.

Let the function $f:\Bbb R\mapsto \Bbb R$ be continuous on $I=[a,b]$. Let $A=\{0,1,...,n\}$ for any $n\in \Bbb N$. Let $I$ be partitioned into $n$ sub-intervals $S_k=[a_k,b_k]$ such that the following properties hold: $$\bigcup_{k\in A}S_k=I$$ $$\bigcap_{k\in A}S_k=\emptyset$$ $$\forall k\in A\setminus\{n\}, S_{k}\cap S_{k+1}=\{b_{k}\}=\{a_{k+1}\}$$ $$\forall i,j\in A, S_i\cap S_j=\emptyset\iff i\notin \{j-1,j,j+1\}$$ $$a_0=a, b_n=b$$ $$a<b$$ $$\forall k\in A, a_k<b_k$$ Suppose that $\forall k\in A, x_k\in (a_k,b_k)$.

Therefore:$$\int_{a}^{b}f(x)dx=\lim_{n\to \infty}\sum_{k\in A}f(x_k)(b_k-a_k)$$.

This definition generalizes the integral to a wider variety of sub-intervals of $[a,b]$.


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