Good Question 13

Let’s end the year with this problem that I came across a while ago in a review book:

Integrate \int{x\sqrt{x+1}dx}

It was a multiple-choice question and had four choices for the answer. The author intended it to be done with a u-substitution but being a bit rusty I tried integration by parts. I got the correct answer, but it was not among the choices. So I thought it would make a good challenge to work on over the holidays.

    1. Find the antiderivative using a u-substitution.
    2. Find the antiderivative using integration by parts.
    3. Find the antiderivative using a different u-substitution.
    4. Find the antiderivative by adding zero in a convenient form.

Your answers for 1, 3, and 4 should be the same, but look different from your answer to 2. The difference is NOT due to the constant of integration which is the same for all four answers. Show that the two forms are the same by

  1. “Simplifying” your answer to 2 and get a third form of the answer.
  2. “Simplifying” your answer to 1, 3, 4 and get that third form again.

Give it a try before reading on. The solutions are below the picture.

Method 1: u-substitution

Integrate \int{x\sqrt{x+1}dx}


\int{x\sqrt{x+1}dx=}\int{\left( u-1 \right)\sqrt{u}}du=\int{{{u}^{3/2}}-{{u}^{1/2}}}du=\tfrac{5}{2}{{u}^{5/2}}-\tfrac{3}{2}{{u}^{3/2}}

\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C

Method 2: By Parts

Integrate \int{x\sqrt{x+1}dx}


dv=\sqrt{x+1}dx,v=\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}

\int{x\sqrt{x+1}dx}=\tfrac{2}{3}x{{\left( x+1 \right)}^{3/2}}-\int{\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}dx}=

\tfrac{2}{3}x{{\left( x+1 \right)}^{3/2}}-\tfrac{4}{15}{{\left( x+1 \right)}^{5/2}}+C

Method 3: A different u-substitution

Integrate \int{x\sqrt{x+1}dx}


du=\tfrac{1}{2}{{\left( x+1 \right)}^{-1/2}}dx,dx=2udu

2{{\int{\left( {{u}^{2}}-1 \right){{u}^{2}}}}^{{}}}du=2\int{{{u}^{4}}-{{u}^{2}}}du=\tfrac{2}{5}{{u}^{5}}-\tfrac{2}{3}{{u}^{3}}=

\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C

This gives the same answer as Method 1.

Method 4: Add zero in a convenient form.

Integrate \int{x\sqrt{x+1}dx}

\int{x\sqrt{x+1}}dx=\int{x\sqrt{x+1}+\sqrt{x+1}-\sqrt{x+1} dx=}

\int{\left( x+1 \right)\sqrt{x+1}-\sqrt{x+1}}dx=

\int{{{\left( x+1 \right)}^{3/2}}-{{\left( x+1 \right)}^{1/2}}dx}=

\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C

This, also, gives the same answer as Methods 1 and 3.

So, by a vote of three to one Method 2 must be wrong. Yes, no, maybe?

No, all four answers are the same. Often when you get two forms for the same antiderivative, the problem is with the constant of integration. That is not the case here. We can show that the answers are the same by factoring out a common factor of {{\left( x+1 \right)}^{3/2}}. (Factoring the term with the lowest fractional exponent often is the key to simplifying expressions of this kind.)

Simplify the answer for Method 2:

\tfrac{2}{3}x{{\left( x+1 \right)}^{3/2}}-\tfrac{4}{15}{{\left( x+1 \right)}^{5/2}}+C=

\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}\left( x-\tfrac{2}{5}\left( x+1 \right) \right)+C=

\displaystyle \tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}\left( \frac{5x-2x-2}{5} \right)+C=

\tfrac{2}{15}{{\left( x+1 \right)}^{3/2}}\left( 3x-2 \right)+C

Simplify the answer for Methods 1, 3, and 4:

\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C=

\tfrac{2}{15}{{\left( x+1 \right)}^{3/2}}\left( 3\left( x+1 \right)-5 \right)+C=

\tfrac{2}{15}{{\left( x+1 \right)}^{3/2}}\left( 3x-2 \right)+C

So, same answer and same constant.

Is this a good question? No and yes.

As a multiple-choice question, no, this is not a good question. It is reasonable that a student may use the method of integration by parts. His or her answer is not among the choices, but they have done nothing wrong. Obviously, you cannot include both answers, since then there will be two correct choices. Moral: writing a multiple-choice question is not as simple as it seems.

From another point of view, yes, this is a good question, but not for multiple-choice. You can use it in your class to widen your students’ perspective. Give the class a hint on where to start. Even better, ask the class to suggest methods; if necessary, suggest methods until you have all four (… maybe there is even a fifth). Assign one-quarter of your class to do the problem by each method. Then have them compare their results. Finally, have them do the simplification to show that the answers are the same.

My next post will be after the holidays.

Happy Holidays to Everyone!



Starting Integration

Behind every definite integral is a Riemann sums. Students need to know about Riemann sums so that they can understand definite integrals (a shorthand notation for the limit if a Riemann sun) and the Fundamental theorem of Calculus. Theses posts help prepare students for Riemann sums.

  1. The Old Pump Where I start Integration
  2. Flying into Integrationland Continues the investigation in the Old Pump – the airplane problem
  3. Working Towards Riemann Sums
  4. Definition of the Definite Integral and the FTC – a more exact demonstration from last Friday’s post and The Fundamental Theorem of Calculus –  an older demonstration
  5. More about the FTC The derivative of a function defined by an integral – the other half of the FTC.
  6. Good Question 11 Riemann Reversed – How to find the integral, given the Riemann sum
  7. Properties of Integrals
  8. Variation on a Theme – 2 Comparing Riemann sums
  9. Trapezoids – Ancient and Modern – some history






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 12 – Parts with a Constant?

partsSomeone asked me about this a while ago and I thought I would share it with you. It may be a good question to get your students thinking about; see if they can give a definitive answer that will, of course, include a justification.

Integration by Parts is summarized in the equation

\displaystyle \int_{{}}^{{}}{udv}=uv-\int_{{}}^{{}}{vdu}

To use the equation, you choose part of a given integral (left side) to be u and part to be dv, both functions of x. Then you differentiate u and integrate dv and use them on the right side to obtain a simpler integral that you can integrate.

The question is this: When you integrate dv, should you, can you, have a constant of integration, the “+ ” that you insist upon in other integration problems? Why don’t you use it here? Or can you?

Scroll down for my answer.

Answer: Let’s see what happens if we use a constant. Assume that \displaystyle \int_{{}}^{{}}{dv}=v+C. Then

\displaystyle \int_{{}}^{{}}{udv}=u\cdot \left( v+C \right)-\int_{{}}^{{}}{\left( v+C \right)du}

\displaystyle =uv+Cu-\left( \int_{{}}^{{}}{vdu}+\int_{{}}^{{}}{Cdu} \right)

\displaystyle =uv+Cu-\int_{{}}^{{}}{vdu}-Cu

\displaystyle =uv-\int_{{}}^{{}}{vdu}

So, you may use a constant if you want, but it will always add out of the expression.

For more on integration by parts see here for the basic idea, here for the tabular method, here for a quicker way than the tabular method, and here for more on the tabular method and reduction formulas.

Good Question 11 – Riemann Reversed

Good Question 11 – or not. double-riemann


The question below appears in the 2016 Course and Exam Description (CED) for AP Calculus (CED, p. 54), and has caused some questions since it is not something included in most textbooks and has not appeared on recent exams. The question gives a Riemann sum and asks for the definite integral that is its limit. Another example appears in the 2016 “Practice Exam” available at your audit website; see question AB 30. This type of question asks the student to relate a definite integral to the limit of its Riemann sum. These are called reversal questions since you must work in reverse of the usual order. Since this type of question appears in both the CED examples and the practice exam, the chances of it appearing on future exams look good.

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.

Example 1

Which of the following integral expressions is equal to \displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\left( \sqrt{1+\frac{3k}{n}}\cdot \frac{1}{n} \right)} ?

There were 4 answer choices that we will consider in a minute.

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:

\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\left( f\left( a+\frac{b-a}{n}\cdot k \right)\cdot \frac{b-a}{n} \right)}=\int_{a}^{b}{f\left( x \right)dx}

To evaluate the limit and express it as an integral, we must identify, a, b, and f. I usually begin by looking for \displaystyle \frac{b-a}{n}. Here \displaystyle \frac{b-a}{n}=\frac{1}{n} and from this conclude that ba = 1, so b = a + 1.

Usually, you can start by considering a = 0 , which means that the \displaystyle \frac{b-a}{n}\cdot k becomes the “x.”. Then rewriting the radicand as \displaystyle 1+3\frac{1}{n}k=1+3\left( a+\frac{1}{n}\cdot k \right), it appears the function is \sqrt{1+3x} and the limit is \displaystyle \int_{0}^{1}{\sqrt{1+3x}}dx=\frac{14}{9}.

The answer choices are

(A)  \displaystyle \int_{0}^{1}{\sqrt{1+3x}}dx        (B)    \displaystyle \int_{0}^{3}{\sqrt{1+x}}dx      (C)    \displaystyle \int_{1}^{4}{\sqrt{x}}dx     (D)   \displaystyle \tfrac{1}{3}\int_{0}^{3}{\sqrt{x}}dx

The correct choice is (A), but notice that choices B, C, and D can be eliminated as soon as we determine that b = a + 1. That is not always the case.

Let’s consider another example:

Example 2: \displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\left( {{\left( 2+\frac{3}{n}k \right)}^{2}}\left( \frac{3}{n} \right) \right)}=

As before consider \displaystyle \frac{b-a}{n}=\frac{3}{n}, which implies that b = a + 3. With a = 0,  the function appears to be {{\left( 2+x \right)}^{2}} on the interval [0, 3], so the limit is \displaystyle \int_{0}^{3}{{{\left( 2+x \right)}^{2}}}dx=39


What if we take a = 2? If so, the limit is \displaystyle \int_{2}^{5}{{{x}^{2}}dx}=39.

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 \displaystyle \int_{-25.65}^{-22.65}{{{\left( 27.65+x \right)}^{2}}dx}=39.

The same thing can occur in other ways. Returning to example 1,and using something like a u-substitution, we can rewrite the original limit as \displaystyle \frac{1}{3}\cdot \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\left( \sqrt{1+\frac{3k}{n}}\cdot \frac{3}{n} \right)}.

Now b = a + 3 and the limit could be either \displaystyle \frac{1}{3}\int_{0}^{3}{\sqrt{1+x}}dx=\frac{14}{9} or \displaystyle \frac{1}{3}\int_{1}^{4}{\sqrt{x}}dx=\frac{14}{9}, among others.

My opinions about this kind of question.

The real problem with the answer choices to Example 1 is that they force the student to do the question 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 a different correct answer that is not among the choices. This is not good.

The problem could be fixed by giving the answer choices as numbers. These are the numerical values of the 4 choices:(A) 14/9   (B) 14/3   (C)  14/3   (D)    2\sqrt{3}/3. As you can see that presents another problem. Distractors (wrong answers) are made by making predictable calculus mistakes. Apparently, two predictable mistakes give the same numerical answer; therefore, one of them must go.

A related problem is this: The limit of a Riemann sum is a number; a definite integral is a number. Therefore, any definite integral, even one totally unrelated to the Riemann sum, which has the correct numerical value, is a correct answer.

I’m not sure if this type of question 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,and the questions do make students unpack the notation.

My opinions notwithstanding, it appears that future exams will include questions like these.

These questions are easy enough to make up. 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 these by stopping after you get to the limit, giving your students just the limit, and having 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 writers call questions like these reversal questions since the work is done in reverse of the usual way.

Here is the question from 1997, for you to try. The answer is below.






Answer B. Hint n = 50






Revised 5-5-2022


Parts and More Parts

At an APSI this summer the participants and I got to discussing the “tabular method” for integration by parts. Since we were getting far from what is tested on the BC Calculus exams, I ended the discussion and said for those that were interested I would post more on the tabular method this blog going farther than just the basic set up. So here goes.

Here are some previous posts on integration by parts and the tabular method

 Integration by Parts 1 discusses the basics of the method. This is as far as a BC course needs to go.

Integration by Parts 2 introduces the tabular method

Modified Tabular Integration presents a very quick and slick way of doing the tabular method without making a table. This is worth knowing.

There is also a video on integration by parts here. Scroll down to “Antiderivatives 5: A BC topic – Integration by parts.” The tabular method is discussed starting about time 15:16. There are several ways of setting up the table; one is shown here and a slightly different way is in the Integration by Parts 2 post above. There are others.

Going further with the tabular method.

The tabular method works well if one of the factors in the original integrand is a polynomial; eventually its derivative will be zero and you are done. These are shown in the examples in the posts above and Example 1 below. To complete the topic, this post will show two other things that can happen when using integration by parts and the tabular method.

First we look at an example with a polynomial factor and learn how to stop midway through. Why stop? Because often there will be no end if you don’t stop. There are ways to complete the integration as shown in the examples.

Example 1:  Find \displaystyle \int_{{}}^{{}}{\left( 4{{x}^{3}} \right)\cos \left( x \right)dx} by the tabular method (See Integration by Parts 2 for more detail on how to set the table up)


Adding the last column gives the antiderivative:

\displaystyle \int_{{}}^{{}}{\left( 4{{x}^{3}} \right)\cos \left( x \right)dx}=4{{x}^{3}}\sin \left( x \right)+12{{x}^{2}}\cos \left( x \right)-24x\sin \left( x \right)-24\sin \left( x \right)+C

Now say you wanted to stop after 12{{x}^{2}}\cos \left( x \right). Example 2 shows why you want (need) to stop. In Example 1 you will have

\displaystyle \int_{{}}^{{}}{\left( 4{{x}^{3}} \right)\cos \left( x \right)dx}=4{{x}^{3}}\sin \left( x \right)+12{{x}^{2}}\sin \left( x \right)+\int_{{}}^{{}}{-24x\cos \left( x \right)}dx

The integrand on the right is the product of the last column in the row at which you stopped and the first two columns in the next row, as shown in yellow above.

Example 2 Find \displaystyle \int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx}

Tabular 4

As you can see things are just repeating the lines above sometimes with minus signs. However, if we stop on the third line we can write:

\displaystyle \int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx={{e}^{x}}\sin \left( x \right)}+{{e}^{x}}\cos \left( x \right)-\int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx}

The integral at the end is identical to the original integral.  We can continue by adding the integral to both sides:

\displaystyle 2\int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx={{e}^{x}}\sin \left( x \right)}+{{e}^{x}}\cos \left( x \right)

Finally, we divide by 2 and have the antiderivative we were trying to find:

\displaystyle \int_{{}}^{{}}{{{e}^{x}}\cos \left( x \right)dx=\tfrac{1}{2}{{e}^{x}}\sin \left( x \right)}+\tfrac{1}{2}{{e}^{x}}\cos \left( x \right)+C

In working this type of problem you must be aware of that the original integrand showing up again can happen and what to do if it does. As long as the coefficient is not +1, we can proceed as above. The same thing happens if we do not use the tabular method. (If the coefficient is +1 then the other terms on the right will add to zero and you need to make different choices for u and dv.)

Reduction Formulas.

Another use of integration by parts is to produce formulas for integrals involving powers. An integral whose integrand is of less degree than the original, but of the same form results. The formula is then iterated to continually reduce the degree until the final integral can be integrated easily.

Example 3: Find \displaystyle \int_{{}}^{{}}{{{x}^{n}}{{e}^{x}}dx}

Let u={{x}^{n}},\ du=n{{x}^{n-1}}dx,\ dv={{e}^{x}}dx,\ v={{e}^{x}}

\displaystyle \int_{{}}^{{}}{{{x}^{n}}{{e}^{x}}dx}={{x}^{n}}{{e}^{x}}-n\int_{{}}^{{}}{{{x}^{n-1}}{{e}^{x}}dx}

This is a reduction formula; the second integral is the same as the first, but of lower degree. Here is how it is used.  At each step the integrand is the same as the original, but one degree lower. So the formula can be applied again, three more times in this example.

Tabular 7

Most textbooks have a short selection of reduction formulas.

Final Thoughts.

Back in the “old days”, BC (before calculators), beginning calculus courses spent a lot of time on the topic of “Techniques of Integration.” This included integration by parts, algebraic techniques, techniques known as trig-substitutions, and others. Mathematicians and engineers had tables of integrals listing over a thousand forms and students were taught how to use the tables and distinguish between similar forms in the tables. (See the photo below from the fourteenth edition of the CRC tables (c) 1965.) Current textbooks often contain such sections still.

Today, none of this is necessary. CAS calculators can find the antiderivatives of almost any integral. Websites such as WolframAlpha are also available to do this work.

I’m not sure why the College Board recently expanded slightly the list of types of antiderivatives tested on the exams. Certainly a few of the basic types should be included in a course, but what students really need to know is how to write the integral appropriate to a problem, and what definite and indefinite integrals mean. This, in my opinion, is far more important than being able to crank out antiderivatives of increasingly complicated expressions: let technology do that – or buy yourself an integral table. Just saying … .


Trapezoids – Ancient and Modern

The other day, in the course of about 10 minutes, I came across two interesting things about Trapezoidal approximations that I thought I would share with you.

cuneifirm tablet

Cuneiform writing

The first was a link to a story about how the ancient Babylonian astronomers sometime between 350 and 50 BCE used trapezoids to, in effect, find the area under a velocity-time graph tracking Jupiter’s motion. This was an NPR story based on a January 2016 Science magazine article in which the author, Mathieu Ossendrijver discusses his work deciphering cuneiform tablets written over 1,400 years before the technique showed up in Europe.

The second was a question asked on the AP Calculus bulletin board. A teacher asked, “Can someone please help me answer this question a student posed the other day. We were comparing left, right and midpoint and trapezoidal approximations. He asked since the trapezoidal calculation is the best estimate what is the use of LRAM and RRAM?” Here is an expanded form of my answer.

There are several things to consider here.

  1. First, if all you need is an estimate of the area or integral of a continuous function then a Trapezoid sum is certainly better than the left Riemann sum (left-RΣ) or the right-RΣ. Better, yes, the “best” maybe not: midpoint sums are about as good and parabola sums (Simpson’s Rule) are better.

2. Another reason to do left RΣ and right RΣ with small values of n is simply to give students practice in setting up Riemann sums so that they will be familiar with them when they move on to finding their limits and getting ready to define definite integrals.

3. A RΣ for a function f on a closed interval [a, b] is formed by partitioning the interval into subintervals and taking exactly one function value from each closed subinterval, multiplying the value by the width of that subinterval and adding these results. You may pick the function value any way you want – left end, middle, right end, any place at random in the subintervals and someplace else in the next subinterval. One way is to pick the smallest function value in each subinterval; this gives a RΣ called the lower RΣ. Likewise, you could pick the largest value in each subinterval; this gives the upper RΣ. Now it is true that

lower RΣ ≤ (any/all other RΣs) ≤ upper RΣ

Then as you add more partition points (n approaches infinity, or Δx approaches 0, etc.) the lower sum increases and the upper sum decreases. The series of lower sums is increasing and bounded above (by the upper sum) and therefore converges to its least upper bound. The upper sum decreases forming a decreasing series that is bounded below and therefore converges to its greatest lower bound.

If the lower RΣ and the upper RΣ approach the same value, then ALL the other RΣs approach that same value by the Squeeze theorem. This value is then defined as the definite integral of f from a to b.

In most AP calculus course, the textbooks do not deal with upper and lower sums. Instead, they deal with left RΣ and right RΣ on intervals on which f is only increasing (or only decreasing). In this case the lower RΣ = left RΣ and the upper RΣ = the right RΣ (or the other way around for decreasing functions).

So, this is why you need the left RΣ and right RΣ; not so much to approximate, but to complete the theory leading to the definite integral.