MA005: Calculus I

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 k^th 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 k^th 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:

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 $\mathrm{c}_{\mathrm{k}}$

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 $\mathrm{c}_{1}=0, \mathrm{c}_{2}=\pi / 2, \mathrm{c}_{3}=\pi / 2$.