The Fundamental Theorem of Calculus is, well, fundamental. It relates the derivative and the integral.
Writing a Riemann sum with all that fancy notation is tedious. To speed things up a special notation is used to replace it. The limit of the Riemann sum for a function on an interval [a, b] is written as its definite integral:
The (called the integrand) is the function with no fancy notation and the dx, called differential x replaces the . The a and b, called the lower and upper limit of integration respectively, show you the interval the Riemann sum was formed on (which the Riemann sum does not).
Keep in mind that behind every definite integral is a Riemann sum. Therefore, all the properties of limits apply to definite integrals. They can be added and subtracted, a constant may be factored out, and so on.
The Fundamental Theorem of Calculus, the FTC, tells you how to evaluate a definite integral (and therefore its Riemann sum): Simply evaluate the function of which is the derivative at the endpoints of the interval and subtract.
To keep this in mind you can write the FTC like this considering the integrand as the derivative (of something):
.
For example, since ,
That’s all there is to it!
But wait! There’s more! This reveals another important idea: Since derivatives are rates of change, the FTC says that the integral of a rate of change is the net amount of change over the interval. Also called the accumulated change.
Well, okay, there is the problem of finding the function whose derivative is the integrand which is not always easy. This function is called theantiderivative of the integrand; another name is the indefinite integral. (The notation for an antiderivative or indefinite integral is the same as for a definite integral without the limits of integration). The truth is that finding the antiderivative is not as straightforward as finding the derivative. We will tackle that soon.
You are now ready to move into the study of integration, the other “half” of calculus. To integrate is defined as “to bring together or incorporate parts into a whole” (Dictionary.com).
The initial problem in integral calculus is to find the area of a region between the graph of a function and the x-axis with vertical sides. This is done by lining up very thin rectangles, finding their individual areas and incorporating them into a whole by adding their areas.
The way the rectangles areas are found and added is to use a Riemann sum. The width of each rectangle is a small distance along the x-axis and the length is the distance from the x-axis to the curve. As you use more rectangles over the same interval, their width decreases, and the approximation of the area becomes better.
Yes, that’s limits again. As the number of rectangles increases (), their width decreases () and the (Riemann) sum approaches the area.
You will start by setting up some of these Riemann sums with a small number of rectangles to help you get the idea of what’s happening. (Lots of arithmetic here.)
Written in mathematical notation, a Riemann sum looks like this . The interval on the x-axis is divided into subintervals of width ; these do not have to be the same, but almost always are. The is the function’s value at some point, , in each interval. So, is the area of the rectangle for that subinterval. The sigma sign sums them up.
And the gives the area.
Most of the time the limit will not be easy to find, so you’ll avoid it! Soon you will learn a quick and efficient way to find the limits.
Riemann sums can be used in many other applications as you will soon learn.
The question below appears in the new Course and Exam Description (CED) for AP Calculus, and has caused some questions since it is not something included in most textbooks and has not appeared on recent exams.
Example 1
Which of the following integral expressions is equal to
There were 4 answer choices that we will consider in a minute.
To the best of my recollection the last time a question of this type appeared on the AP Calculus exams was in 1997, when only about 7% of the students taking the exam got it correct. Considering that by random guessing about 20% should have gotten it correct, this was a difficult question. This question, the “radical 50” question is at the end of this post.
The first key to answering the question is to recognize the limit as a Riemann sum. In general, a right-side Riemann sum for the function f on the interval [a, b] with n equal subdivisions has the form:
To evaluate the limit and express it as an integral, we must identify, a, b, and f. I usually begin by looking for (b – a)/n. In this problem (b – a)/n = 1/n and from this conclude that b – a = 1, so b = a + 1.
Then rewriting the radicand as
It appears that the function is
and the limit is
.This is the first answer choice. The choices are:
In this example, choices B, C, and D can be eliminated as soon as we determine that b = a + 1, but that is not always the case.
Let’s consider another example:
Example 2:
As before consider (b – a)/n = 3/n implies that b = a + 3. And the function appears to be
on the interval [0, 3], so the limit is
BUT
What is we take a = 2. If so, the limit is
And now one of the “problems” with this kind of question appears: the answer written as a definite integral is not unique!
Not only are there two answers, but there are many more possible answers. These two answers are horizontal translations of each other, and many other translations are possible, such as
Returning to example 1, using something like a u-substitution, we can rewrite the original limit as .
Now b = a + 3 and the limit could be either
You will probably have your students write Riemann sums with a small value of n when you are teaching Riemann sums leading up to the Fundamental Theorem of Calculus. You can make up problems like this these by stopping after you get to the limit, giving your students just the limit, and have them work backwards to identify the function(s) and interval(s). You could also give them an integral and ask for the associated Riemann sum. Question writes call a question like this a reversal question, since the work is done in reverse of the usual way.
Another example appears in the 2016 “Practice Exam” available at your audit website. It is question AB 30. That question gives the definite integral and asks for the associate Riemann sum; a slightly different kind of reversal. Since this type of question appears in both the CED examples and the practice exam, chances of it appearing on future exams look good.
Critique of the problem
I’m not sure if this type of problem has any practical or real-world use. Certainly, setting up a Riemann sum is important and necessary to solve a variety of problems. After all, behind every definite integral there is a Riemann sum. But starting with a Riemann sum and finding the function and interval does not seem to me to be of practical use.
The CED references this question to MPAC 1: Reasoning with definitions and theorems, and to MPAC 5: Building notational fluency. They are appropriate, but still is the question ever done outside a test or classroom setting?
Another, bigger, problem is that the answer choices to Example 1 force the student to do the problem in a way that gets one of the answers. It is perfectly reasonable for the student to approach the problem a different way, and get another correct answer that is not among the choices. This is not good. The question could be fixed by giving the answer choices as numbers. These are the numerical values of the 4 choices:
As you can see that presents another problem.
Finally, here is the question from 1997, for you to try:
Answer B. Hint n = 50
_______________________________
Note: The original of this post was lost somehow. I’ve recreated it here. Sorry if anyone was inconvenienced. LMc May 5, 2024
One of the main uses of the definite integral is summed up (pun intended) in the idea of accumulation. When you integrate a rate of change you get the (net) amount of change. This important idea is often treated very lightly, if at all, in textbooks.
Here are a series of past posts that use, explain, and illustrate that concept.
Accumulation – Need an Amount? The Fundamental Theorem of Calculus says that the integral of a rate of change (a derivative) is the net amount of change. This post shows how that works in practice.
AP Accumulation Questions and Good Question 7 – 2009 AB 3 the “Mighty Cable Company” show how accumulation is tested on the AP Calculus exams. The “Mighty Cable Company” question is a particularly good and difficult example.
The next two posts show how to use the concept of accumulation to analyze a function and its graph without reference to the derivative. The graphical idea of a Riemann sum rectangle moving across the interval of integration makes the features of function much more intuitive than the common approach. You will not find these ideas in textbooks. Nevertheless, a lesson on this idea may help your students.
Graphing with Accumulation 1 explains how to analyze the derivative to determine when a function is increasing or decreasing and finding the locations of extreme values. By thinking of the individual Rieman sum rectangles moving across the interval the features of the function are easy to see and easier to remember. Once understood, this method will help students with their graph analysis work.
Graphing with Accumulation 2 continues the idea of using accumulation to determine information about the concavity of a function.
I haven’t missed Unit 7! This unit seems to fit more logically after the opening unit on integration (Unit 6). The Course and Exam Description (CED) places Unit 7 Differential Equations before Unit 8 probably because the previous unit ended with techniques of antidifferentiation. My guess is that many teachers will teach Unit 8: Applications of Integration immediately after Unit 6 and before Unit 7: Differential Equations. The order is up to you. Unit 7 will post next Tuesday.
Unit 8 includes some standard problems solvable by integration (CED – 2019 p. 143 – 161). These topics account for about 10 – 15% of questions on the AB exam and 6 – 9% of the BC questions.
Topics 8.1 – 8.3 Average Value and Accumulation
Topic 8.1 Finding the Average Value of a Function on an Interval Be sure to distinguish between average value of a function on an interval, average rate of change on an interval and the mean value
Topic 8.2 Connecting Position, Velocity, and Acceleration of Functions using Integrals Distinguish between displacement (= integral of velocity) and total distance traveled (= integral of speed)
Topic 8. 3 Using Accumulation Functions and Definite Integrals in Applied Contexts The integral of a rate of change equals the net amount of change. A really big idea and one that is tested on all the exams. So, if you are asked for an amount, look around for a rate to integrate.
Topics 8.4 – 8.6 Area
Topic 8.4 Finding the Area Between Curves Expressed as Functions of x
Topic 8.5 Finding the Area Between Curves Expressed as Functions of y
Topic 8.6 Finding the Area Between Curves That Intersect at More Than Two Points Use two or more integrals or integrate the absolute value of the difference of the two functions. The latter is especially useful when do the computation of a graphing calculator.
Topics 8.7 – 8.12 Volume
Topic 8.7 Volumes with Cross Sections: Squares and Rectangles
Topic 8.8 Volumes with Cross Sections: Triangles and Semicircles
Topic 8.9 Volume with Disk Method: Revolving around the x– or y-Axis Volumes of revolution are volumes with circular cross sections, so this continues the previous two topics.
Topic 8.10 Volume with Disk Method: Revolving Around Other Axes
Topic 8.11 Volume with Washer Method: Revolving Around the x– or y-Axis See Subtract the Hole from the Whole for an easier way to remember how to do these problems.
Topic 8.12 Volume with Washer Method: Revolving Around Other Axes. See Subtract the Hole from the Whole for an easier way to remember how to do these problems.
Topic 8.13 Arc LengthBC Only
Topic 8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled BC ONLY
Timing
The suggested time for Unit 8 is 19 – 20 classes for AB and 13 – 14 for BC of 40 – 50-minute class periods, this includes time for testing etc.
Previous posts on these topics for both AB and BC include:
I think that the path leading up to and including the Fundamental Theorem of Calculus (FTC) is one of the most beautiful walks in mathematics. I have written several posts about it. You will soon be ready to travel that path with your students. (I always try to post on topics shortly before most teachers will get to them, so that you have some time to consider them and work the ideas you like into your lessons.)
Here is an annotated list of some of the posts to guide you on your journey.
Working Towards Riemann Sums gives the preliminary definitions you will need to define and discuss Riemann sums.
Riemann Sums defines the several Riemann sums often used in the calculus left-side sums, right-side sums, midpoint sums and the trapezoidal sums. “The Area Under a Curve” in the iPad app A Little Calculusis a great visual display of these and shows what happens as you use more subintervals.
The Definition of the Definite Integral gives the definition of the definite integral as the limit of any Riemann sum. As with any definition, there is nothing to prove or argue about here. The thing to remember is that the limit of the Riemann sum and the definite integral are the same thing. Behind any definite integral is a Riemann sum. The advantage of the definition’s integral notation is that it shows the interval involved which the Riemann sum does not. (Any Riemann sum may be represented by many definite integrals. See Good Question 11 – Riemann Reversed.)
Foreshadowing the FTC is an example of how a definite integral may be evaluated. It is long and has a lot of notation, so you may not want to use this.
The Fundamental Theorem of Calculus is where the path leads. This post develops the FTC based on the other “big” idea of the calculus: the Mean Value Theorem. (I think the form here is preferable to the usual book notation that uses F(x) and its derivate f (x).)
Y the FTC? Tries to answer the question of what’s so important about the FTC. Example 1: The verbal interpretation of the FTC (the integral of a rate of change is the net amount of change over the interval.) will soon be used in many practical applications. While example 2 shows how the FTC allows one to evaluate a definite integral and, therefore the Riemann sum it represents, by evaluating a function whose derivative is the integrand (its antiderivative).
More About the FTC presents examples leading up to the other form of the FTC: the derivative of the integral is the integrand).
At this point you may go in the direction of learning how to find antiderivatives or working on applications. (See Integration itinerary.)
A tank is being filled with water using a pump that is old and slows down as it runs. The table below gives the rate at which the pump pumps at ten-minute intervals. If the tank initially has 570 gallons of water in it, approximately how much water is in the tank after 90 minutes?
Elapsed time (minutes)
0
10
20
30
40
50
60
70
80
90
Rate (gallons / minute)
42
40
38
35
35
32
28
20
19
10
And so, integration begins.
Ask your students to do this problem alone. When they are ready (after a few minutes) collect their opinions. They will not all be the same (we hope, because there is more than one reasonable way to approximate the amount). Ask exactly how they got their answers and what assumptions they made. Be sure they always include units (gallons). Here are some points to make in your discussion – points that we hope the kids will make and you can just “underline.”
Answers between 3140 and 3460 gallons are reasonable. Other answers in that range are acceptable. They will not use terms like “left-sum”, “right sum” and “trapezoidal rule” because they do not know them yet, but their explanations should amount to the same thing. An answer of 3300 gallons may be popular; it is the average of the other two, but students may not have gotten it by averaging 3140 and 3460.
Ask if they think their estimate is too large or too small and why they think that.
Ask what they need to know to give a better approximation – more and shorter time intervals.
Assumptions: If they added 570 + 42(10) + 40(10) + … +19(10) they are assuming that the pump ran at each rate for the full ten minutes and then suddenly dropped to the next. Others will assume the rate dropped immediately and ran at the slower rate for the 10 minutes. Some students will assume the rate dropped evenly over each 10-minute interval and use the average of the rates at the ends of each interval (570 + 41(10) + 39(10) + … 14.5(10) = 3300).
What is the 570 gallons in the problem for? Well, of course to foreshadow the idea of an initial condition. Hopefully, someone will forget to include it and you can point it out.
With luck someone will begin by graphing the data. If no one does, you should suggest it; (as always) to help them see what they are doing graphically. They are figuring the “areas” of rectangles whose height is the rate in gallons/minute and whose width is the time in minutes. Thus the “area” is not really an area but a volume (gal/min)(min) = gallons). In addition to unit analysis, graphing is important since you will soon be finding the area between the graph of a function and the x-axis in just this same manner.
), their width decreases (
) and the (Riemann) sum approaches the area.
. The interval on the x-axis is divided into subintervals of width 
is the function’s value at some point, 
, in each interval. So, 
is the area of the rectangle for that subinterval. The sigma sign sums them up.
gives the area. 
Teaching Calculus 