# The Definite Integral and the FTC

The Definition of the Definite Integral.

The definition of the definite integrals is: If f is a function continuous on the closed interval [a, b], and $a={{x}_{0}}<{{x}_{1}}<{{x}_{2}}<\cdots <{{x}_{{n-1}}}<{{x}_{n}}=b$  is a partition of that interval, and $x_{i}^{*}\in [{{x}_{{i-1}}},{{x}_{i}}]$, then

$\displaystyle \underset{{\left| {\left| {\Delta x} \right|} \right|\to 0}}{\mathop{{\lim }}}\,\sum\limits_{{i=0}}^{n}{{f\left( {x_{i}^{*}} \right)}}\left( {{{x}_{i}}-{{x}_{{i-i}}}} \right)=\int\limits_{a}^{b}{{f\left( x \right)dx}}$

The left side of the definition is, of course, any Riemann sum for the function f on the interval [a, b]. In addition to being shorter, the right side also tells you about the interval on which the definite integral is computed. The expression $\left\| {\Delta x} \right\|$  is called the “norm of the partition” and is the longest subinterval in the partition. Usually, all the subintervals are the same length, $\frac{{b-a}}{n}$, and this is the last your will hear of the norm. With all the subdivisions of the same length this can be written as

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=0}}^{n}{{f\left( {x_{i}^{*}} \right)}}\frac{{b-a}}{n}=\int\limits_{a}^{b}{{f\left( x \right)dx}}$

Other than that, there is not much more to the definition. It is simply a quicker and more efficient notation for the sum.

The Fundamental Theorem of Calculus (FTC).

First recall the Mean Value Theorem (MVT) which says: If a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) then there exist a number, c, in the open interval (a, b) such that ${f}'\left( c \right)\left( {b-a} \right)=f\left( b \right)-f\left( a \right)$.

Next, let’s rewrite the definition above with a few changes. The reason for this will become clear.

$\int\limits_{a}^{b}{{{f}'\left( x \right)dx}}=\underset{{\left| {\left| {\Delta x} \right|} \right|\to 0}}{\mathop{{\lim }}}\,\sum\limits_{{i=0}}^{n}{{{f}'\left( {{{c}_{i}}} \right)\left( {{{x}_{i}}-{{x}_{{i-i}}}} \right)}}$

Since every function is the derivative of another function (even though we may not know that function or be able to write a closed-form expression for it), I’ve expressed the function as a derivative, I’ve also chosen the point in each subinterval, ${{c}_{i}}$, to be the number in each subinterval guaranteed by the MVT for that subinterval.

Then, $\displaystyle {f}'\left( {{{c}_{i}}} \right)\left( {{{x}_{i}}-{{x}_{{i-i}}}} \right)=f\left( {{{x}_{i}}} \right)-f\left( {{{x}_{{i-1}}}} \right)$. Making this substitution, we have

$\int\limits_{a}^{b}{{{f}'\left( x \right)dx}}=\underset{{\left| {\left| {\Delta x} \right|} \right|\to 0}}{\mathop{{\lim }}}\,\sum\limits_{{i=0}}^{n}{{\left( {f\left( {{{x}_{i}}} \right)-f\left( {{{x}_{{i-1}}}} \right)} \right)}}$

$\displaystyle =f\left( {{{x}_{1}}} \right)-f\left( {{{x}_{0}}} \right)+f\left( {{{x}_{2}}} \right)-f\left( {{{x}_{1}}} \right)+f\left( {{{x}_{3}}} \right)-f\left( {{{x}_{2}}} \right)+\cdots +f\left( {{{x}_{n}}} \right)-f\left( {{{x}_{{n-1}}}} \right)$

$\displaystyle =f\left( {{{n}_{n}}} \right)-f\left( {{{x}_{0}}} \right)$

And since ${{x}_{0}}=a$ and  ${{x}_{n}}=b$,

$\displaystyle \int_{a}^{b}{{{f}'\left( x \right)dx}}=f\left( b \right)-f\left( a \right)$

This equation is called the Fundamental Theorem of Calculus. In words, it says that the integral of a function can be found by evaluating the function of which the integrand is the derivative at the endpoints of the interval and subtracting the values. This is a number that may be positive, negative, or zero depending on the function and the interval. The function of which the integrand is the derivative, is called the antiderivative of the integrand.

The real meaning and use of the FTC is twofold:

1. It says that the integral of a rate of change (i.e. a derivative) is the net amount of change. Thus, when you want to find the amount of change – and you will want to do this with every application of the derivative – integrate the rate of change.
2. It also gives us an easy way to evaluate a Riemann sum without going to all the trouble that is necessary with a Riemann sum; simply evaluate the antiderivative at the endpoints and subtract.

At this point I suggest two quick questions to emphasize the second point:

1. Find $\int_{3}^{7}{{2xdx}}$.

Ask if anyone knows a function whose derivative is 2x? Your students will know this one. The answer is x2, so

$\displaystyle \int_{3}^{7}{{2xdx}}={{7}^{2}}-{{3}^{2}}=40$.

Much easier than setting up and evaluating a Riemann sum!

2. Then ask your students to find the area enclosed by the coordinate axes and the graph of cos(x) from zero to $\frac{\pi }{2}$. With a little help they should arrive at

$\displaystyle \int_{0}^{{\pi /2}}{{\cos \left( x \right)dx}}$.

Then ask if anyone knows a function whose derivative is cos(x). it’s sin(x), so

$\displaystyle \int_{0}^{{\pi /2}}{{\cos \left( x \right)dx}}=\sin \left( {\frac{\pi }{2}} \right)-\sin \left( 0 \right)=1-0=1$.

At this point they should be convinced that the FTC is a good thing to know.

There is another form of the FTC that is discussed in More About the FTC.

# Good Question 7 – 2009 AB 3

Another in my occasional series on Good Questions to teach from. This is the Mighty Cable Company question from the 2009 AB Calculus exam, number 3

This question presented students with a different situation than had been seen before. It is a pretty standard “in-out” question, except that what was going in and out was money. Students were told that the Mighty Cable Company sold its cable for $120 per meter. They were also told that the cost of the cable varied with its distance from the starting end of the cable. Specifically, the cost of producing a portion of the cable x meters from the end is $6\sqrt{x}$ dollars per meter. Profit was defined as the difference between the money the company received for selling the cable minus the cost of producing the cable. Students had a great deal of trouble answering this question. (The mean was 1.92 out of a possible 9 points. Fully, 36.9% of students earned no point; only 0.02% earned all 9 points.) This was probably because they had difficulty in interpreting the question and translating it into the proper mathematical terms and symbols. Since economic problems are not often seen on AP Calculus exams, students needed to be able to use the clues in the stem: • The$120 per meter is a rate. This should be deduced from the units: dollars per meter.
• The cost of producing the portion of cable x meters from one end cable is also a rate for the same reason. In economics this is called the marginal cost; the students did not need to know this term.
• The profit is an amount that is a function of x, the length of the cable.

Part (a): Students were required to find the profit from the sale of a 25-meter cable. This is an amount. As always, when asked for an amount, integrate a rate. In this case integrate the difference between the rate at which the cable sells and the cost of producing it.

$\displaystyle P(25)=\int_{0}^{25}{\left( 120-6\sqrt{x} \right)dx}=\2500$

or

$\displaystyle P(25)=120(25)-\int_{0}^{25}{6\sqrt{x}\ dx}=\2500$

Part (b): Students were asked to explain the meaning of $\displaystyle \int_{25}^{30}{6\sqrt{x}dx}$ in the context of the problem. Since the answer is probably not immediately obvious, here is the reasoning involved.

This is the integral of a rate and therefore, gives the amount (of money) needed to manufacture the cable. This can be found by a unit analysis of the integrand: $\displaystyle \frac{\text{dollars}}{\text{meter}}\cdot \text{meters}=\text{dollars}$ .

Let C be the cost of production, so $\displaystyle \frac{dC}{dx}=6\sqrt{x}$, and therefore, $\displaystyle \int_{25}^{30}{6\sqrt{x}dx}=C\left( 30 \right)-C\left( 25 \right)$ by the Fundamental Theorem of Calculus (FTC).

Therefore, $\displaystyle \int_{25}^{30}{6\sqrt{x}dx}$ is the difference in dollars between the cost of producing a cable 30 meters long, C(30), and the cost of producing a cable 25 meters long, C(25). (Another acceptable is that the integral is the cost in dollars of producing the last 5 meters of a 30 meter cable.)

Part (c): Students were asked to write an expression involving an integral that represents the profit on the sale of a cable k meters long.

Part (a) serves as a hint for this part of the question. Here the students should write the same expression as they wrote in (a) with the 25 replaced by k.

$\displaystyle P(k)=\int_{0}^{k}{\left( 120-6\sqrt{x} \right)dx}$

or

$\displaystyle P(k)=120k-\int_{0}^{k}{6\sqrt{x}\ dx}$

Part d: Students were required to find the maximum profit that can be earned by the sale of one cable and to justify their answer. Here they need to find when the rate of change of the profit (the marginal profit) changes from positive to negative.

Using the FTC to differentiate either of the answers in part (c) or by starting fresh from the given information:

$\displaystyle \frac{dP}{dx}=120-6\sqrt{x}$

$\displaystyle \frac{dP}{dx}=0$ when x = 400 and P(400)= $16,000. Justification: The maximum profit on the sale of one cable is$16,000 for a cable 400 meters long. For $00$ and for $x>400,{P} '(x)<0$ therefore, the maximum profit occurs at x = 400. (The First Derivative Test).

Once students are familiar with in-out questions, this is a good question to challenge them with. The actual calculus is not that difficult or unusual but concentrating on the translation of the unfamiliar context into symbols and calculus ideas is different. Show them how to read the hints in the problem such as the units.

Steel cable or steel wire rope as it is called also has some interesting geometry in its construction. You can find many good illustrations of this, such as the ones below, by Googling “steel wire rope.”

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# Proof

When math books present a theorem, they almost always immediately present its proof. I tend to skip the proofs. I assume they are correct. I want to get on with the ideas in the text. Later I may come back and read through them. Is this a good thing to advise students to do? I don’t know.

There are reasons to read proofs. One reason is to help understand why a theorem is true, by seeing the reasoning that leads to the result. Another is to check the reasoning yourself. A third is to learn how to do proofs.

Learning to write original proofs is not usually one of the goals of a beginning calculus course. That comes later in a course with “analysis” in its title. There are many theorems that involve some one-off that rarely will be used again. I’m thinking of a proof like that of the sum of the limits is equal to the limit of the sums, where you add and subtract the same expression and this more complicated form allows you to group and factor the terms of the numerator and arrive at the result. Another example is in the Mean Value Theorem where you consider a new function that gives the vertical distance between a function and its secant line. These always bring the question, “How did you know to do that?”

If a student can accept things like that, then the proof is usually easy enough to follow. But I would never spend a lot of time making every student fight his or her way through each and every proof.

On this other hand, I would never just present a theorem and not give some explanation as to why it is true (and why it is important enough to mention). Unfortunately, I have seen teachers write the Fundamental Theorem of Calculus on the board and proceed to show how to use it to evaluate definite integrals, with no hint of why this important theorem is true. Sure kids can memorize it and use it, but it seems to me they should also have a hint as to why it is true.

Some theorems are easy to understand if explained in ways other than giving a proof. For an example of this, see my post of October 1, 2012 on the Mean Value Theorem. Almost every book will bail out on the Intermediate Value Theorem by claiming (quite rightly) that, “the proof is beyond the scope of this book,” or they give the proof in an appendix. But a simple drawing will convince you that it is true.

So my feeling is that you do not need to labor over a proof for every theorem, BUT, big BUT, you should provide a good explanation of why it is true.

This is important for all students and especially for young women. Jo Boaler writes

“As I interviewed more and more boys and girls, I noticed that the desire to know why was something that separated the girls from the boys. The girls were able to accept the method that were shown them and practice them, but they wanted to know why they worked, where they came from, and how they connected with other methods…. When they could not get access to the depth of understanding they wanted, the girls started to turn away from the subject…. Classes in which students discuss concepts, giving them access to a deep and connected understanding of math, are good for boys and girls. Boys may be willing to work in isolation on abstract rules, but such approaches do not give many students, girls or boys, access to the understanding they need. In addition, high-level work in mathematics, science and engineering is not about isolated, abstract rule following, but about collaboration and connection making.”

[Jo Boaler, What’s Math Got to Do with It? Helping Children Learn to Love Their Most Hated Subject – And Why It’s Important for America, © 2008 Penguin Group, New York. From Chapter 6]

# More About the FTC

In keeping with my idea from the last post of sneaking up on ideas, here is a way to sneak up on the other part of the FTC.

Consider these three functions F1(t) = 3, F2(t) = 2t and F3(t) = 2t + 3

For each of these three functions do the following:

1. Graph each function on the interval [0, t] for $t\ge 0$.
2. Let x be a value of t and let x move to the right along the t-axis starting at zero. Using area formulas, write functions, A1(x), A2(x), and A3(x), for the areas of the region between the graph and the t-axis. Make a table of values for x = 1, 2, 3, …, 10. Do not use any “calculus”; the area expressions are easy to find from geometry.  (A1(t) = 3t; A2(t) = t2; A3(t) = t2 + 3)
3. Consider what happens as the as x moves from zero to the right.  The expressions are a function that gives the area for each x. Now write expressions for the three areas with an integral that gives the same area. The lower limit will be 0 and the upper limit will be x.
4.  These are examples of functions defined by integrals. Each different value of x as the upper limit of integration, will give a unique value (area).
5. Differentiate the area functions and compare them to the functions from which they came. What do you notice?

Consider the FTC where the integral has an upper limit of x. That is, a function defined by an integral. (The lower limit does not have to be zero.)

$\displaystyle A\left( x \right)=\int_{a}^{x}{{f}'\left( t \right)dt}=f\left( x \right)-f\left( a \right)$

What is ${A}'\left( x \right)$?

$\displaystyle {A}'\left( x \right)=\frac{d}{dx}\int_{a}^{x}{{f}'\left( t \right)dt}=\frac{d}{dx}\left( f\left( x \right)-f\left( a \right) \right)={f}'\left( x \right)-0$

or

$\displaystyle \frac{d}{dx}\int_{a}^{x}{{f}'\left( t \right)dt}={f}'\left( x \right)$

The is the other part of the FTC which says that the derivative of a function defined by an integral is the integrand, f (x) evaluated at x.

A good way to demonstrate this to your students is to do a simple integral like

$\displaystyle \int_{\pi /4}^{x}{\cos \left( t \right)dt}=\sin \left( x \right)-\tfrac{\sqrt{2}}{2}$,

and then differentiate to show that

$\displaystyle \frac{d}{dx}\int_{\pi /4}^{x}{\cos \left( t \right)dt}=\frac{d}{dx}\left( \sin \left( x \right)-\tfrac{\sqrt{2}}{2} \right)=\cos \left( x \right)$

This part of the FTC says, roughly, very roughly, that the derivative of an integral is this integrand. The other part, discussed in the last post, just as roughly, says the integral of the derivative is the antiderivative.

Keep your notes on this activity. In my next post I will use these functions to demonstrate some of the important properties of integrals.

# The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus or FTC, as its name suggests, is a very important idea. It is not sufficient to present the formula and show students how to use it. Show them where it comes from.

Here is an approach to demonstrate the FTC. I try to sneak up on the result by proposing a problem and then solving it. Here is the outline.

Suppose we have a differentiable function f that goes from $\left( a,f\left( a \right) \right)$ to $\left( b,f\left( b \right) \right)$. What is the net change in f over this interval? Easy it’s $f\left( b \right)-f\left( a \right)$.  No problem, but way too easy for a calculus class. So let’s try a harder way!

Partition the interval [a, b] as you would for a Riemann sum, and calculate the change in f on each subinterval. The subintervals may be the same width or not. The change in y on the general subinterval [xi-1, xi] is $f\left( {{x}_{i}} \right)-f\left( {{x}_{i-1}} \right)$.

Approximate the net change over the whole interval by adding these $\displaystyle \sum\limits_{i=1}^{n}{\left( f\left( {{x}_{i}} \right)-f\left( {{x}_{i-1}} \right) \right)}$ .
Is this a Riemann sum? No it is not! There is no $\Delta x$ part. What to do?

The expression $f\left( {{x}_{i}} \right)-f\left( {{x}_{i-1}} \right)$ looks familiar.
It is part of the equation for the Mean Value Theorem: $\displaystyle {f}'\left( c \right)=\frac{f\left( b \right)-f\left( a \right)}{b-a}$ or ${f}'\left( c \right)\left( b-a \right)=f\left( b \right)-f\left( a \right)$.
If we adapt this to the subinterval letting ci be the number guaranteed by the MVT on each subinterval  [xi-1, xi], then $f\left( {{x}_{i}} \right)-f\left( {{x}_{i-1}} \right)={f}'\left( {{c}_{i}} \right)\left( {{x}_{i}}-{{x}_{i-1}} \right)$

We can rewrite the sum in step 3 as $\displaystyle \sum\limits_{i=1}^{n}{\left( f\left( {{x}_{i}} \right)-f\left( {{x}_{i-1}} \right) \right)}=\sum\limits_{n=1}^{n}{\left( {f}'\left( {{c}_{i}} \right)\left( {{x}_{i}}-{{x}_{i-1}} \right) \right)}$ .

1. This is a Riemann sum and therefore, $\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{n=1}^{n}{\left( {f}'\left( {{c}_{i}} \right)\left( {{x}_{i}}-{{x}_{i-1}} \right) \right)}=\int_{a}^{b}{{f}'\left( c \right)dx}$ .
2. So what is this equal to? We have already found what this limit is in step 1, so we now have:

$\displaystyle \int_{a}^{b}{{f}'\left( x \right)dx}=f\left( b \right)-f\left( a \right)$.

This is called the FTC. And it is important.

The first thing it tells us is that the integral of a rate of change is the (net) amount of change. This will help us do a variety of problems.

The second thing it tells us is that, if we can find a function of which the integrand is the derivative (i.e. its antiderivative), then we can find the value of a definite integral by evaluating an antiderivative at the endpoints and subtracting. No more struggling with trying to find the limit of Riemann sums or graphing the function and hoping you can break it into regions with easy to find areas. All we need is an antiderivative and then one quick computation will do the trick from now on.

There is more to the FTC. This will be the subject of the next post.

# The Definition of the Definite Integral

From the last post, it seems pretty obvious that as the number of rectangles in a Riemann sum increases or, what amounts to the same thing, the width of the sub-intervals decreases, the Riemann sum approaches the area of the region between a graph and the x-axis. The figures below show left-Riemann sums for the function $f\left( x \right)=1+{{x}^{2}}$ on the interval [1, 4]. Hover and click on the figure below.

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As the number of rectangles increases, they fill up more and more of the region. The sums increase, yet, their sum is always less than the area. They form an increasing sequence which is bounded above and therefore approaches its least upper bound (the area) as a limit.

Right-side rectangles work almost the same way. The sums form a decreasing sequence that is bounded below by the area and thus they approach the same limit.

All of the other Riemann sums must be between these two (at least for this example) and thus all approach the same limit.

This limit is called the definite integral for f on the interval [a, b], denoted by the new symbol below.

$\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{f\left( x_{i}^{*} \right)\left( {{x}_{i}}-{{x}_{i-1}} \right)}=\underset{\Delta x\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{f\left( x_{i}^{*} \right)\Delta x=\int_{a}^{b}{f\left( x \right)dx}}$

The notation has the advantage of being simpler to write than all the limit stuff and it shows us the interval which the limits do not. (For now consider the dx as a stand-in for $\Delta x$.)

The disadvantage of the notation is that, as we use it for real applications, the concept of Riemann sums often gets lost. The integrals become formulas and they get memorized but not understood.

Remember: behind every (any, all) definite integral is a Riemann sum.  As we look at applications, we should always look for the Riemann sum and how it is set up. This will tell us what the definite integral should be. We will not need to be too fussy about the subscripts and such; in fact, we will almost ignore them, but we will look carefully at the Riemann sum rectangles.