2019 CED Unit 6: Integration and Accumulation of Change

Unit 6 develops the ideas behind integration, the Fundamental Theorem of Calculus, and Accumulation. (CED – 2019 p. 109 – 128 ). These topics account for about 17 – 20% of questions on the AB exam and 17 – 20% of the BC questions.

Topics 6.1 – 6.4 Working up to the FTC

Topic 6.1 Exploring Accumulations of Change Accumulation is introduced through finding the area between the graph of a function and the x-axis. Positive and negative rates of change, unit analysis.

Topic 6.2 Approximating Areas with Riemann Sums Left-, right-, midpoint Riemann sums, and Trapezoidal sums, with uniform partitions are developed. Approximating with numerical methods, including use of technology are considered. Determining if the approximation is an over- or under-approximation.

Topic 6.3 Riemann Sums, Summation Notation and the Definite Integral. The definition integral is defined as the limit of a Riemann sum.

Topic 6.4 The Fundamental Theorem of Calculus (FTC) and Accumulation Functions Functions defined by definite integrals and the FTC.

Topic 6.5 Interpreting the Behavior of Accumulation Functions Involving Area Graphical, numerical, analytical, and verbal representations.

Topic 6.6 Applying Properties of Definite Integrals Using the properties to ease evaluation, evaluating by geometry and dealing with discontinuities.

Topic 6.7 The Fundamental Theorem of Calculus and Definite Integrals Antiderivatives. (Note: I suggest writing the FTC in this form \displaystyle \int_{a}^{b}{{{f}'\left( x \right)}}dx=f\left( b \right)-f\left( a \right) because it seems more efficient than using upper case and lower-case f.)

Topics 6.5 – 6.14 Techniques of Integration

Topic 6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation. Using basic differentiation formulas to find antiderivatives. Some functions do not have closed-form antiderivatives. (Note: While textbooks often consider antidifferentiation before any work with integration, this seems like the place to introduce them. After learning the FTC students have a reason to find antiderivatives. See Integration Itinerary

Topic 6.9 Integration Using Substitution The u-substitution method. Changing the limits of integration when substituting.

Topic 6.10 Integrating Functions Using Long Division and Completing the Square 

Topic 6.11 Integrating Using Integration by Parts (BC ONLY)

Topic 6.12 Integrating Using Linear Partial Fractions (BC ONLY)

Topic 6.13 Evaluating Improper Integrals (BC ONLY) Showing the work requires students to show correct limit notation.

Topic 6.14 Selecting Techniques for Antidifferentiation This means practice, practice, practice.


Timing

The suggested time for Unit 6 is  18 – 20 classes for AB and 15 – 16 for BC of 40 – 50-minute class periods, this includes time for testing etc.


Previous posts on these topics include:

Introducing the Derivative

Integration Itinerary

The Old Pump and Flying to Integrationland   Two introductory explorations

Working Towards Riemann Sums

Riemann Sums

The Definition of the Definite Integral

Foreshadowing the FTC 

The Fundamental Theorem of Calculus

More About the FTC

Y the FTC?

Area Between Curves

Under is a Long Way Down 

Properties of Integrals 

Trapezoids – Ancient and Modern  On Trapezoid sums

Good Question 9 – Riemann Reversed   Given a Riemann sum can you find the Integral it converges to?  A common and difficult AP Exam problem

Accumulation

Accumulation: Need an Amount?

Good Question 7 – 2009 AB 3

Good Question 8 – or Not?  Unit analysis

AP Exams Accumulation Question    A summary of accumulation ideas.

Graphing with Accumulation 1

Graphing with Accumulation 2

Accumulation and Differential Equations 

Painting a Point

Techniques of Integrations (AB and BC)

Antidifferentiation

Why Muss with the “+C”?

Good Question 13  More than one way to skin a cat.

Integration by Parts – a BC Topic

Integration by Parts 1

Integration by Part 2

Parts and More Parts

Good Question 12 – Parts with a Constant?

Modified Tabular Integration 

Improper Integrals and Proper Areas

Math vs the Real World Why \displaystyle \int_{{-\infty }}^{\infty }{{\frac{1}{x}}}dx does not converge.


Here are links to the full list of posts discussing the ten units in the 2019 Course and Exam Description.

2019 CED – Unit 1: Limits and Continuity

2019 CED – Unit 2: Differentiation: Definition and Fundamental Properties.

2019 CED – Unit 3: Differentiation: Composite , Implicit, and Inverse Functions

2019 CED – Unit 4 Contextual Applications of the Derivative  Consider teaching Unit 5 before Unit 4

2019 – CED Unit 5 Analytical Applications of Differentiation  Consider teaching Unit 5 before Unit 4

2019 – CED Unit 6 Integration and Accumulation of Change

2019 – CED Unit 7 Differential Equations  Consider teaching after Unit 8

2019 – CED Unit 8 Applications of Integration   Consider teaching after Unit 6, before Unit 7

2019 – CED Unit 9 Parametric Equations, Polar Coordinates, and Vector-Values Functions 

2019 CED Unit 10 Infinite Sequences and Series


Good Question 14

Good Question 14 – The Integral Test

I have no criteria for what constitutes a “Good Question” for this series of occasional posts. They are just questions that I found interesting, or that seem more than usually instructive, or that I learn something from. I cannot quote this question (2016 BC 92) since it is on a secure exam. What made it interesting is that to answer it students pretty much needed to know the proof of the Integral Test and the figures that go with it.

I recall only one AP question from many years ago that asked students to “prove” something – usually students are asked to show that a result was true by citing the theorem that applied and showing the hypotheses were met. The directions are often “justify your answer.”

Doing an original proof is not, in my opinion, a fair question and proving some known theorem is just a matter of memorization. For these reasons, students are not asked to prove things on the exams. So, should you prove things in class? Probably, yes.

Here is the usual proof of the integral test. Afterwards I’ll discuss the question from the exam.

The Integral Test

Hypotheses: Let f\left( x \right) be a function that is positive, decreasing, and continuous for x\ge 1 ; and let {{a}_{n}}=f\left( n \right) for x\ge 1

In the first drawing the rectangles have a height of an and a width of 1. The area is of each is an, and the sum of their areas the series is \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}.

Part 1: Notice that \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}>\int_{1}^{\infty }{{f\left( x \right)dx}} Assume that the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}} diverges.

  • Conclusion 1: If the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}}  diverges, then the series \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}} diverges.
  • Conclusion 2: (The contrapositive of conclusion 1) If the series \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}} converges, then the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}} converges.

Part 2: In the second drawing below, assume that the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}} converges. The sum of the areas of the rectangles is \sum\limits_{{n=2}}^{\infty }{{{{a}_{n}}}}. (NB: this series starts at n = 2.) Since \sum\limits_{{n=2}}^{\infty }{{{{a}_{n}}}} is less than the convergent improper integral it will also converge. Adding {{a}_{1}} to this gives the original series, \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}; this series also converges.

  • Conclusion 3: If the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}} converges, then the series \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}} converges.
  • Conclusion 4: (The contrapositive of conclusion 3) If the series   \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}} diverges, then the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}} diverges.

Putting the four conclusions together is the Integral Test: If the hypotheses above are met, then the series and the improper integral will both converge, or both diverge.


To answer the multiple-choice question (2106 BC 92) on the exams students were told that the improper integral converges. Therefore, the associated series converges. They then had to determine whether the series or the improper integral has the greater value. Stop here and see if you can figure that out.


Return to the first figure above, only this time assume that the improper integral and the series converge. It is pretty obvious that \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}>\int_{1}^{\infty }{{f\left( x \right)dx}}.

So, even though students were not asked to prove anything, a familiarity with the proof and its figures is necessary to answer the question. That’s why I liked it,

On the other hand, it is kind of an obscure point and I’m not sure it has any practical value.


Math vs. the “Real World”

There is a difference between mathematics and the “real world”: In the real world you are allowed to do whatever you want, as long as there is no law against doing it; in mathematics, you cannot do something unless there is a law that says you may.

A question that comes up often on the AP Calculus Community bulletin board concerns the divergence of the improper integral \displaystyle \int_{{-1}}^{1}{{\frac{1}{x}dx}}.

There are several mistakes students make when computing this integral.

First, they may not realize this is an improper integral and compute incorrectly:

\displaystyle \int_{{-1}}^{1}{{\frac{1}{x}dx}}=\ln \left| 1 \right|-\ln \left| {-1} \right|=0-0=0

Since the laws concerning improper integrals do not allow this, you may not do it.

Or, they might think that it does converge to zero by the symmetry of the graph. There is no law (theorem) that permits calculating limits based on the appearance of a graph.

Finally, and most often, they may start out following the rules but go astray. The law says you must find deal with the discontinuity at x = 0 by using one-sided limits:

\displaystyle \int_{{-1}}^{1}{{\frac{1}{x}dx}}=\underset{{a\to 0-}}{\mathop{{\lim }}}\,\int_{{-1}}^{a}{{\frac{1}{x}dx}}+\underset{{b\to 0+}}{\mathop{{\lim }}}\,\int_{b}^{1}{{\frac{1}{x}dx}}

=\underset{{a\to 0-}}{\mathop{{\lim }}}\,\left( {\ln \left| a \right|-\ln \left| 1 \right|} \right)+\underset{{b\to 0+}}{\mathop{{\lim }}}\,\left( {\ln \left| 1 \right|-\ln \left| b \right|} \right)

=\left( {-\infty -0} \right)+\left( {0-(-\infty )} \right)=\infty -(\infty )=0

The mistake is subtler here. It is correct to say that  =\underset{{a\to 0-}}{\mathop{{\lim }}}\,\ln \left| a \right|=-\infty , but what that really means is that the limit does not exist (DNE). Then in the last line above you cannot say

(does not exist) – (does not exist) = 0.

You cannot subtract something that does not exist from something else that does not exist. As soon as you see that one of the limits does not exist, the entire limit does not exist. (That’s the law.) There is no algebra/calculus theorem that permits the addition of two divergent integrals, therefore, it is not correct to add them.

Students need help in understanding this. Here are three ways to think about it.

  1. Infinity is not a number. When you say the integrals equal infinity and negative infinity, you must stop. Just because it looks like something minus the same thing is zero, you cannot do this, because you’re not working with numbers. In fact, the integrals do not exist, so you cannot add them – there’s nothing to add.
  2. “Infinity” is a short, and correct, way of expressing the limits as you approach zero for this function from the right or left. But, you must remember that infinity is a shorthand for DNE, not for some really large number and its opposite.
  3. Infinity minus infinity (\infty -\infty ) is an indeterminate form. Some indeterminate forms of this type converge, if you can find some additional algebra/calculus to do on them (such as L’Hospital’s Rule in some cases). For this example, such algebra/calculus does not exist (no pun intended)

So, in conclusion \displaystyle \int_{{-1}}^{1}{{\frac{1}{x}dx}} does not converge!


This question was discussed recently on the AP Calculus bulletin board. The two items below were included and may help your students understand what’s going on with infinity. The are by Stu Schwartz.

Thank You Stu!



Improper Integrals and Proper Areas

A few years ago, on the old AP Calculus discussion group a teacher asked a question about this improper integral:

\displaystyle \int_{0}^{\infty }{\frac{1}{1+{{x}^{2}}}dx}=\underset{b\to \infty }{\mathop{\lim }}\,\int_{0}^{b}{\frac{1}{1+{{x}^{2}}}dx}

=\underset{b\to \infty }{\mathop{\lim }}\,\left. \left( {{\tan }^{-1}}\left( x \right) \right) \right|_{0}^{b}

=\underset{b\to \infty }{\mathop{\lim }}\,\left( {{\tan }^{-1}}\left( b \right)-{{\tan }^{-1}}\left( 0 \right) \right)=\frac{\pi }{2}  

His (quite perceptive) student pointed out that the range of the inverse tangent function is arbitrarily restricted to the open interval \left( -\tfrac{\pi }{2},\tfrac{\pi }{2} \right). The student asked if some other range would affect the answer to this problem. The short answer is no, the result is the same. For example, if range were restricted to say \left( \tfrac{5\pi }{2},\tfrac{7\pi }{2} \right), then in the computation above:

\underset{b\to \infty }{\mathop{\lim }}\,\left( {{\tan }^{-1}}\left( b \right)-{{\tan }^{-1}}\left( 0 \right) \right)=\tfrac{7\pi }{2}-3\pi =\tfrac{\pi }{2}

The value is the same. While that is pretty straightforward, there are other things going on here which may be enlightening. The original indefinite integral represents the area in the first quadrant between the graph of y=\frac{1}{1+{{x}^{2}}} and the x-axis. Let’s consider the function that gives the area between the y-axis and the vertical line at various values of x.

A \displaystyle\left( x \right)=\int_{0}^{x}{\frac{1}{1+{{t}^{2}}}}\ dt

Pretending for the moment that we don’t know the antiderivative, we can use a calculator to graph the area function. Improper integralOf course, we recognize this as the inverse tangent function, but what is more interesting is that whatever this function is, it seems to have a horizontal asymptote at y=\tfrac{\pi }{2}. The area is approaching a finite limit as x increases without bound.  The unbounded region has a finite area. The connection with improper integrals is obvious.

\displaystyle \underset{b\to \infty }{\mathop{\lim }}\,A\left( b \right)=\underset{b\to \infty }{\mathop{\lim }}\,\int_{0}^{b}{\frac{1}{1+{{x}^{2}}}dx=}\int_{0}^{\infty }{\frac{1}{1+{{x}^{2}}}dx}

Also, the improper integral is defined as the limit of the area function. This may give some insight as to why improper integrals are defined as they are.