MAT236H5F, University of Toronto, Mississauga. Instructor: Ke Zhang.
Let $f = f(x)$ be a function on $[a, b]$, we set:
The Riemann sums (for this partition and choice of inner points $x_i^*$ is: $$ \sum_{i = 1}^n f(x_i^*) \Delta x. $$ Remark: The proper definition of the Riemann sums allows artibrary partiion of $[a, b]$ into $[x_{i-1}, x_i]$. Here we only look at the particular case of equal-size partitions.
The following Theorem is fundamental in the theory of Riemann integral:
Theorem 1 Suppose $f: [a, b] \to \mathbb{R}$ is continuous, then $$ \lim_{n \to \infty} \sum_{i = 1}^n f(x_i^*) \Delta x $$ exists. It is then defined as the definite integral $\int_a^b f(x)dx $.
We visualize it by appoximating the area under a graph using rectangles of width $\Delta x$ and height $f(x_i^*)$.
Two choices of the "sampling points" $x_i^*$ are particularly relavant. We define:
In the following example $a = 0$, $b = 2\pi$ and $f(x) = \sin(x) + 1$.
The way Theorem 1 is proven is a bit delicate and is usually covered in an analysis course. A key idea behind the proof is that the difference between the Darboux upper sums and the Darboux lower sums converges to $0$ as $n \to \infty$. This shows the limit of the Riemann sums is independent of how the sample points are chosen. The idea of the proof is the following:
Visually, this means the area of the blue bars shown in the picture goes to $0$ as the partition gets finer and finer. If the heights of the rectangles are at most $\epsilon$, then the total area is at most $(b-a)\epsilon$. A great way to visualize it is to slide all the rectangles to the bottom, then the total area is smaller than the area of a rectangle with width $b-a$ and height $\epsilon$.