## Sigma Notation and Riemann Sums

Read this section to learn about area. Work through practice problems 1-9.

### Area Under A Curve – Riemann Sums

Suppose we want to calculate the area between the graph of a positive function $f$ and the interval $[a, b]$ on the x–axis (Fig. 7). The Riemann Sum method is to build several rectangles with bases on the interval $[a, b]$ and sides that reach up to the graph of f (Fig. 8). Then the areas of the rectangles can be calculated and added together to get a number called a Riemann Sum of f on [a, b]. The area of the region formed by the rectangles is an approximation of the area we want.

Example 5: Approximate the area in Fig. 9a between the graph of $f$ and the interval $[2, 5]$ on the x–axis by summing the areas of the rectangles in Fig. 9b.

Solution: The total area of rectangles is $(2)(3) + (1)(5) = 11$ square units.

In order to effectively describe this process, some new vocabulary is helpful: a partition of an interval and the mesh of the partition.

A partition $P$ of a closed interval $[a,b]$ into $n$ subintervals is a set of $n+1$ points $\left\{x_{0}=\mathrm{a}, x_{1}, x_{2}, x_{3}, \ldots, x_{\mathrm{n}-1}, x_{\mathrm{n}}=\mathrm{b}\right\}$ in increasing order, $\mathrm{a}=x_{0}$.

(A partition is a collection of points on the axis and it does not depend on the function in any way.)

The points of the partition $P$ divide the interval into $n$ subintervals (Fig. 10): $\left[x_{0}, x_{1}\right],\left[x_{1}, x_{2}\right],\left[x_{2}, x_{3}\right], \ldots, \text { and }\left[x_{\mathrm{n}-1}, x_{\mathrm{n}}\right]$ with lengths $\Delta x_{1}=x_{1}-x_{0}, \Delta x_{2}=x_{2}-x_{1}, \Delta x_{3}=x_{3}-x_{2}, \ldots$, and $\Delta x_{\mathrm{n}}=x_{\mathrm{n}}-x_{\mathrm{n}-1}$. The points $x_{\mathrm{k}}$ of the partition P are the locations of the vertical lines for the sides of the rectangles, and the bases of the rectangles have lengths $\Delta x_{\mathrm{k}} \text { for } \mathrm{k}=1,2,3, \ldots, \mathrm{n}$.

The mesh or norm of partition $P$ is the length of the longest of the subintervals $\left[x_{\mathrm{k}-1}, x_{\mathrm{k}}\right]$, or, equivalently, the maximum of $\Delta x_{\mathrm{k}} \text { for } \mathrm{k}=1,2,3, \ldots, \mathrm{n}$.

For example, the set $\mathrm{P}=\{2,3,4.6,5.1,6\}$ is a partition of the interval $[2,6]$ (Fig. 11) and divides the interval $[2,6]$ into 4 subintervals with lengths $\Delta x_{1}=1, \Delta x_{2}=1.6, \Delta x_{3}=.5 \text { and } \Delta x_{4}=.9$. The mesh of this partition is 1.6, the maximum of the lengths of the subintervals. (If the mesh of a partition is "small," then the length of each one of the subintervals is the same or smaller.)

Practice 6: $\mathrm{P}=\{3,3.8,4.8,5.3,6.5,7,8\}$ is a partition of what interval? How many subintervals does it create? What is the mesh of the partition? What are the values of $x_{2}$ and $\Delta x_{2}$?

A function, a partition, and a point in each subinterval determine a Riemann sum.

Suppose $f$ is a positive function on the interval $[a,b]$

$\mathrm{P}=\left\{x_{0}=\mathrm{a}, x_{1}, x_{2}, x_{3}, \ldots, x_{\mathrm{n}-1}, x_{\mathrm{n}}=\mathrm{b}\right\}$ is a partition of $[a,b]$.

$\mathbf{c}_{\mathbf{k}}$ is an x–value in the kth subinterval $\left[x_{\mathrm{k}-1}, x_{\mathrm{k}}\right]: \quad x_{\mathrm{k}-1} \leq \mathrm{c}_{\mathrm{k}} \leq x_{\mathrm{k}}$.

Then the area of the kth rectangle is $\mathrm{f}\left(\mathrm{c}_{\mathrm{k}}\right) \cdot\left(x_{\mathrm{k}}-x_{\mathrm{k}-1}\right)=\mathrm{f}\left(\mathrm{c}_{\mathrm{k}}\right) \cdot \Delta x_{\mathrm{k}}$. (Figure 12)

Definition: A summation of the form $\sum_{\mathbf{k}=1}^{\mathbf{n}} \mathbf{f}\left(\mathbf{c}_{\mathbf{k}}\right) \cdot \Delta x_{\mathbf{k}}$ is called a Riemann Sum of $f$ for the partition $P$.

This Riemann sum is the total of the areas of the rectangular regions and is an approximation of the area between the graph of f and the x–axis.

Example 6:

Find the Riemann sum for $f(x)=1 / x$ and the partition ${1, 4, 5}$ using the values $\mathrm{c}_{1}=2$ and $c_{2}=5$. (Fig. 13)

Solution: The 2 subintervals are $[1,4]$ and $[4,5]$ so $\Delta x_{1}=3$ and $\Delta x_{2}=1$.

Then the Riemann sum for this partition is

\begin{aligned} \sum_{k=1}^{n} f\left(c_{k}\right) \cdot \Delta x_{k} &=\sum_{k=1}^{2} \mathrm{f}\left(c_{k}\right) \cdot \Delta x_{k}=f\left(c_{1}\right) \cdot \Delta x_{1}+f\left(c_{2}\right) \cdot \Delta x_{2}=f(2) \cdot(3)+f(5) \cdot(1) \\ &=\frac{1}{2}(3)+\frac{1}{5}(1)=1.7 \end{aligned}.

Practice 7:

Calculate the Riemann sum for $\mathrm{f}(x)=1 / x$ on the partition ${1, 4, 5}$ using the values $c_{1}=3, c_{2}=4$.

Practice 8: What is the smallest value a Riemann sum for $\mathrm{f}(x)=1 / x$ and the partition ${1, 4, 5}$ can have? (You will need to select values for $\mathrm{c}_{1}$ and $\mathrm{c}_{2}$.) What is the largest value a Riemann sum can have for this function and partition?

Table 2 shows the results of a computer program that calculated Riemann sums for the function $f(x)=1 / x$ with different numbers of subintervals and different ways of selecting the points $\mathrm{c}_{\mathrm{i}}$ in each subinterval.

When the mesh of the partition is small (and the number of subintervals large), all of the ways of selecting the $\mathrm{c}_{\mathrm{i}}$ lead to approximately the same number for the Riemann sums. For this decreasing function, using the left endpoint of the subinterval always resulted in a sun that was larger than the area. Choosing the right end point gave a value smaller that the area. Why?

Table 2: Riemann sums for $f(x)=1 / x$ on the interval $[1,5]$

Values of the Riemann sum for different choices of $\mathrm{c}_{\mathrm{k}}$

number of subintervals mesh $\mathrm{c}_{\mathrm{k}}$ = left edge
= $x_{\mathrm{k}-1}$
$\mathrm{c}_{\mathrm{k}}$ = "random" point in
$\left[x_{\mathrm{k}-1}, x_{\mathrm{k}}\right]$
$\mathrm{c}_{\mathrm{k}}$ = right edge
= $x_{\mathrm{k}}$
4 1 2.083333 1.473523 1.283333
8 .5 1.828968 1.633204 1.428968
16 .25 1.714406 1.577806 1.514406
40 .1 1.650237 1.606364 1.570237
400 .01 1.613446 1.609221 1.605446
4000 .001 1.609838 1.609436 1.609038

As the mesh gets smaller, all of the Riemann Sums seem to be approaching the same value, approximately $\text { 1.609. }(\ln 5=1.609437912)$.

Example 7: Find the Riemann sum for the function $f(x)=\sin (x)$ on the interval $[0, \pi]$ using the partition $\{0, \pi / 4, \pi / 2, \pi\}$ with $\mathrm{c}_{1}=\pi / 4, \mathrm{c}_{2}=\pi / 2, \mathrm{c}_{3}=3 \pi / 4$.

Solution: The 3 subintervals (Fig. 14) are $[0, \pi / 4],[\pi / 4, \pi / 2], \text { and }[\pi / 2, \pi]$, and $[\pi / 2, \pi]$ so $\Delta \mathrm{x}_{1}=\pi / 4, \Delta \mathrm{x}_{2}=\pi / 4$ and $\Delta \mathrm{x}_{3}=\pi / 2$. The Riemann sum for this partition is

\begin{aligned} \sum_{\mathrm{k}=1}^{3} \mathrm{f}\left(\mathrm{c}_{\mathrm{k}}\right) \cdot \Delta \mathrm{x}_{\mathrm{k}} &=\sin (\pi / 4) \cdot(\pi / 4)+\sin (\pi / 2) \cdot(\pi / 4)+\sin (3 \pi / 4) \cdot(\pi / 2) \\ &=\frac{\sqrt{2}}{2} \cdot \frac{\pi}{4}+1 \cdot \frac{\pi}{4}+\frac{\sqrt{2}}{2} \cdot \frac{\pi}{2} \approx 2.45148 . \end{aligned}.

Practice 9: Find the Riemann sum for the function and partition in the previous example, but use $\mathrm{c}_{1}=0, \mathrm{c}_{2}=\pi / 2, \mathrm{c}_{3}=\pi / 2$.