## Riemann Summation

1. Estimating Area of Shapes. We can consider limits of finite sums when estimating the areas of shapes in the plane. So for example, the triangle:

Note that
(1)$\displaystyle f(x) = h\left(1-\frac{x}{b}\right)$
describes the triangle’s hypoteneuse.

What can we do? Well, we can pick a number of points ${0=x_{0}. The points are equidistant from each other, so
(2)$\displaystyle x_{j}=\frac{j(b-0)}{n}.$
Then we form a number of rectangles:

We have shaded the area of the rectangles. Note there is some defecit area. What is the area of one of these rectangles? It is
(3)$\displaystyle A_{k} = f(x_{k})\Delta x_{k}$
where
(4)$\displaystyle \Delta x_{k}=x_{k}-x_{k-1}.$
Pop Quiz: is $\Delta x_{k}\in\mathcal{O}(n^{-1})$? $\mathcal{O}(n^{-2})$? $\mathcal{O}(\mathrm{e}^{-n})$? $\mathcal{O}(\ln n)$?

The reader may verify these ${A_{k}}$ describe the rectangle’s area. Thus
(5)$\displaystyle \begin{array}{rl} A_{k} &= \displaystyle\left[h\left(1-\frac{x_{k}}{b}\right)\right]\cdot(x_{k}-x_{k-1})\\ &=\displaystyle h\left(1-\frac{k}{n}\right)\cdot\frac{b}{n}. \end{array}$
Thus we find the area ${A_{tot}}$ of all the rectangles by
(6)$\displaystyle A_{tot}=\sum^{n}_{k=1}A_{k}=\sum^{n}_{k=1}\frac{hb}{n}\left(1-\frac{k}{n}\right)$
Factoring out ${hb/n}$, which is constant with respect to the dummy index ${k}$, we get
(7)$\displaystyle \sum^{n}_{k=1}\frac{hb}{n}\left(1-\frac{k}{n}\right)=\frac{hb}{n}\sum^{n}_{k=1}\left(1-\frac{k}{n}\right)$
Using the sum rule yields
(8)$\displaystyle \frac{hb}{n}\sum^{n}_{k=1}\left(1-\frac{k}{n}\right)=\frac{hb}{n}\left(\sum^{n}_{k=1}1-\sum^{n}_{k=1}\frac{k}{n}\right)$
The sum ${\sum^{n}_{k}1=n}$ transforms our problem
(9)$\displaystyle \frac{hb}{n}\left(\sum^{n}_{k=1}1-\sum^{n}_{k=1}\frac{k}{n}\right)=\frac{hb}{n}\left(n - \frac{1}{n}\sum^{n}_{k=1}k\right)$
Using Gauss’ sum we find
(10)$\displaystyle \frac{hb}{n}\left(n - \frac{1}{n}\sum^{n}_{k=1}k\right)=\frac{hb}{n}\left(n - \frac{1}{n}\frac{n(n+1)}{2}\right)$
Basic arithmetic yields
(11)$\displaystyle \frac{hb}{n}\left(n - \frac{1}{n}\frac{n(n+1)}{2}\right) = \frac{hb}{n}\left(n-\frac{n+1}{2}\right) = \frac{hb}{2n}(n-1).$
So we get
(12)$\displaystyle A_{tot} = \frac{hb}{2n}(n-1).$
What’s the error of this approximation? It’s the unshaded regions of the triangle, which is about ${hb/2n}$.

We can refine the approximation, taking twice as many points. So this changes ${n\mapsto 2n}$. The approximation gets better, but still only approximates the area. We can take the limit
(13)$\displaystyle \lim_{n\rightarrow\infty}A_{tot}=\lim_{n\rightarrow\infty}\frac{hb}{2n}(n-1)=\frac{hb}{2}.$
This gives us the area of the triangle.

2. Riemann Summation. We generalize this scheme approximating the area between the curve ${f(x)}$ and the ${x}$ axis. We consider a closed interval ${[a,b]}$. We choose an arbitrary partition of the interval, i.e., a set of points ${a. They don’t have to be equally spaced. Notice this consists of closed subintervals! There are ${n}$ intervals and the width of the ${k}$-th interval is
(14)$\displaystyle \Delta x_{k}=x_{k}-x_{k-1}$
The Riemann sum approximating the area is
(15)$\displaystyle \sum^{n}_{k=1}f(c_{k})\Delta x_{k} = \mbox{Riemann Sum}$
where $x_{k-1}\leq c_{k}\leq x_{k}$.

How do we get the area? We take the limit ${n\rightarrow\infty}$. This gives us the “Riemann Integral”. This is the definite integral
(16)$\displaystyle \int^{b}_{a}f(x)\,\mathrm{d}{x} = \lim_{n\rightarrow\infty}\sum^{n}_{k=1}f(x_{k})\Delta x_{k}.$
Note if ${F(x)}$ is the antiderivative of ${f(x)}$, then
(17)$\displaystyle \int^{b}_{a}f(x)\,\mathrm{d}{x} = F(b)-F(a)$
is the definite integral from ${a}$ to ${b}$.

In a sense, what happens is our limit $n\to\infty$ makes the following transformation
(18)$\displaystyle\begin{array}{rcl} \displaystyle\sum & \to & \displaystyle\int\\ \displaystyle f(c_{k}) & \to & \displaystyle f(x)\\ \displaystyle\Delta x & \to & \displaystyle\mathrm{d}x\end{array}$
The sum $\sum$ is transformed into a “continuous sum” (i.e., integral) $\int$.
So we can keep the intuition of adding up a bunch of rectangles, but now the rectangles have width $\mathrm{d}x$ (an “infinitesimally” thin rectangle!) and height $f(x)$.

Of course, infinitesimals do not exist. But it is a useful intuitive picture to have. The idea that $\Delta x\to\mathrm{d}x$ invokes fond memories of the derivative. We will show in future posts (viz. the Fundamental Theorem of Calculus) this is no coincidence!

Exercise 1. Consider the Riemann sum for $f(x)=x^{2}$ on the domain $[0,1]$. What is its limit as the number of intervals $n\to\infty$? [Hint: see exercise 1 of the finite series post.]

Exercise 2. Consider the Riemann sum for $f(x)=x^{3}$ on the domain $[0,1]$. What is its limit as the number of intervals $n\to\infty$? [Hint: see exercise 2 of the finite series post.]

Exercise 3. Consider the Riemann sum for $f(x)=x^{-1}$ on the domain $[1,3]$. What is its limit as the number of intervals $n\to\infty$?