# Math Insight

### Riemann sums

Math 1241, Fall 2020
Name:
ID #:
Due date: Nov. 13, 2020, 11:59 p.m.
Table/group #:
Group members:
Total points: 3
1. Suppose we want to know how much water flows through a channel in an hour. We could, theoretically, find a container to hold that water so that we can measure it, but chances are we don't want to interrupt the flow downstream for the hour. Instead, we can measure the rate of flow every few minutes during the hour and use that to estimate the total flow. Let's look at this in a little more detail.
. These two assumptions result in the left-hand and right-hand estimates, respectively.

Compute the left-hand estimate of the total water flow from the above data. We have $3$ time intervals, and on each we assume the flow is constant based on the starting value for that time interval. To get the total flow, we need to multiply each of our starting values by the size of the time interval,
minutes, and then
them together. Remember not to include the last measurement. So the left-hand estimate is
${\rm m}^3$.

To compute the right-hand estimate of the total water flow from the above data, we'll do the same thing but take the values at the end of each time interval. This time, we won't include the first measurement. The right-hand estimate is
${\rm m}^3$.

2. We came up with two fairly different estimates in part a, and there's no way to tell which one is better. How could we improve our estimates?

Suppose we took measurements every $10$ minutes instead of every $20$ minutes.

We can come up with new estimates using these measurements. Now, we have
, and instead of multiplying by
, we need to multiply by
.

Compute the left-hand and the right-hand estimates using the new data.
Left-hand estimate:
${\rm m}^3$
Right-hand estimate:
${\rm m}^3$

3. The estimates in part b were still not great, but they were closer to each other than those in part a. If we took progressively more measurements, the left- and right-hand estimates would
. What if, instead of a set of measurements, we had a function that described the flow rate at any given time? Let's say the water flows through the channel at a rate of $f(t)=\frac{t^{2}}{2500} - \frac{t}{20} + 2$ ${\rm m}^3 / {\rm min}$. We can use this to get measurements at any time we want.

In order to represent a general estimate, we need to use summation notation. Here are a few examples of summation notation, where $x_1=4$, $x_2=2$, and $x_3=5$:
$\displaystyle \sum_{i=1}^{4} i = 1 + 2 + 3 + 4=10$
$\displaystyle \sum_{i=1}^{3} x_{i} = x_{1} + x_{2} + x_{3} = 4 + 2 + 5 = 11$
$\displaystyle \sum_{i=1}^{3} x_{i}^{2} = x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = 4^{2} + 2^{2} + 5^{2} = 16 + 4 + 25 = 45$
$\displaystyle \sum_{i=1}^{3} \left(x_{i}^{3} - x_{i}\right) =$

The idea behind summation notation is that the index $i$ tells us what we are adding together. The expression $\displaystyle \sum_{i=1}^{3} x_{i}$, which we read as "the sum from $i$ equals $1$ to $3$ of $x$ sub $i$", indicates that we will let $i$ be each of the integers from $1$ to $3$ and add together the corresponding values of $x_i$.

Using this notation, we can describe a general estimate. First, let $\Delta t$ denote the width of the interval, and $t_i$ be the time at the end of the $i$th time interval (with $t_0$ being the starting time). Then we can write our estimates from part a as $\displaystyle \sum_{i=1}^{3} f{\left (t_{i - 1} \right )} \Delta t$ and $\displaystyle \sum_{i=1}^{3} f{\left (t_{i} \right )} \Delta t$ (where $\Delta t=20$). The estimates from part b are $\displaystyle \sum_{i=1}^{6} f{\left (t_{i - 1} \right )} \Delta t$ and $\displaystyle \sum_{i=1}^{6} f{\left (t_{i} \right )} \Delta t$ (where $\Delta t=10$). These are left- and right-hand Riemann sums.

The general left- and right-hand Riemann sums for this are $\displaystyle \sum_{i=1}^{n} f{\left (t_{i - 1} \right )} \Delta t$ and $\displaystyle \sum_{i=1}^{n} f{\left (t_{i} \right )} \Delta t$, where $n$ is the number of time intervals and $\Delta t=60/n$.

4. As the number of intervals goes to infinity, the values of the left-hand and right-hand Riemann sums approach each other. The value they approach is the total change. In this case, that is the actual volume of water that flows through the channel. We define the definite integral $$\int_0^{60} \frac{t^{2}}{2500} - \frac{t}{20} + 2 \, dt$$ to be this limit.

More generally, the total change from time $a$ to time $b$ is given by the definite integral $$\int_a^b f(t) \, dt$$ which is the limit as $n$ goes to infinity of both $\sum_{i=1}^{n} f{\left (t_{i - 1} \right )} \Delta t$ and $\sum_{i=1}^{n} f{\left (t_{i} \right )} \Delta t$ where $\Delta t= \frac{b-a}{n}$ and $t_i$ are the right end-points of the intervals of size $\Delta t$.

We read $\displaystyle \int_a^b f(t) \, dt$ as "the integral from a to b of f(t) dt". The values $a$ and $b$ are called the limits of integration. Don't worry about how to evaluate a definite integral yet. We'll come back to that later.

Note that the definite integral is a
, while the indefinite integral is a
.

5. Compute the left-hand and the right-hand Riemann sums with intervals of size $5$ minutes to estimate the total flow of water through the channel. Recall that the function describing the flow rate is $f(t)=\frac{t^{2}}{2500} - \frac{t}{20} + 2$ ${\rm m}^3 / {\rm min}$. Use the following table to record the values at the endpoints, accurate to $5$ significant figures.

Time $\quad$ Rate
$\phantom{1} 0 \quad$

$\phantom{1} 5 \quad$

$10 \quad$

$15 \quad$

$20 \quad$

Time $\quad$ Rate
$25 \quad$

$30 \quad$

$35 \quad$

$40 \quad$

Time $\quad$ Rate
$45 \quad$

$50 \quad$

$55 \quad$

$60 \quad$

The left-hand Riemann sum is
.
The right-hand Riemann sum is
.

2. Suppose the speed of a car is given by $- 40 t^{2} + 80 t + 30$ miles per hour ($t$ is measured in hours).
1. Estimate how far the car has travelled in an hour using a left-hand Riemann sum with $5$ intervals. Include at least 4 significant figures in your answer.

2. Estimate how far the car has travelled in an hour using a right-hand Riemann sum with $5$ intervals. Include at least 4 significant figures in your answer.

3. Estimate how far the car has travelled in an hour using a left-hand Riemann sum with $10$ intervals. Include at least 4 significant figures in your answer.

3. Suppose a population is increasing at a rate of $50 t e^{- t}$ per year ($t$ is measured in years).
1. Estimate the total change in the population size over the first ten years using a left-hand Riemann sum with $5$ intervals. Include at least 4 significant figures in your answer.

2. Estimate the total change in the population size over the first ten years using a right-hand Riemann sum with $5$ intervals. Include at least 4 significant figures in your answer.

3. Estimate the total change in the population size over the first ten years using a left-hand Riemann sum with $10$ intervals. Include at least 4 significant figures in your answer.

4. Use the following applets to visualize the Riemann sums. In each, as you increase the nRectangles slider, you will only be able to modify the most recently added rectangle, so you will have to increase it one number at a time to build the rectangles until you have reach the correct number. Set each rectangle by moving the corners so that one corner that has a point on it is on the graph at the left (or right) side of the interval, while the opposite corner is on the axis at the other end of the interval.

Use this applet to represent the results of part a, with a left-hand Riemann sum with $5$ intervals.

Feedback from applet
number of rectangles:
rectangles:

Use this applet to represent the results of part b, with a right-hand Riemann sum with $5$ intervals.

Feedback from applet
number of rectangles:
rectangles:

Use this applet to represent the results of part c, with a left-hand Riemann sum with $10$ intervals.

Feedback from applet
number of rectangles:
rectangles:

Why are the estimates with $5$ intervals so similar?
Even though the two estimates with $5$ intervals give similar answers, we can see from the applets that the estimate with $10$ intervals captures the full height of the graph, which is missed by the sums with $5$ intervals. It is a much better estimate, although smaller intervals would still improve the estimate.