Riemann Sum & Table Problems (Type 5)

AP  Questions Type 5: Riemann Sum & Table Problems

Tables may be used to test a variety of ideas in calculus including analysis of functions, accumulation, theory and theorems, and position-velocity-acceleration, among others. Numbers and working with numbers is part of the Rule of Four and table problems are one way this is tested. This question often includes an equation in a  latter part of the problem that refers to the same situation.

 What students should be able to do:

  • Find the average rate of change over an interval
  • Approximate the derivative using a difference quotient. Use the two values closest to the number at which you are approximating.  This amounts to finding the slope or rate of change. Show the quotient even if you can do the arithmetic in your head and even if  the denominator is 1.
  • Use a left-, right-, or midpoint- Riemann sums or a trapezoidal approximation to approximate the value of a definite integral using values in the table (typically with uneven subintervals). The Trapezoidal Rule, per se, is not required; it is expected that students will add the areas of a small number of trapezoids without reference to a formula.
  • Average value, average rate of change, Rolle’s theorem, the Mean Value Theorem and the Intermediate Value Theorem. (See 2007 AB 3 – four simple parts that could be multiple-choice questions; the mean on this question was 0.96 out of a possible 9.)
  • These questions are usually presented in some context and answers should be in that context. The context may be something growing (changing over time) or linear motion.
  • Use the table to find a value based on the Mean Value Theorem (2018 AB 4(b)) or Intermediate Value Theorem.
  • One of the parts of this question asks a related question based on a function given by an equation.
  • Unit analysis.

Do’s and Don’ts

Do: Remember that you do not know what happens between the values in the table unless some other information is given. For example, do not assume that the largest number in the table is the maximum value of the function, or that the function is decreasing between two values just because a value is less than the preceding value.

Do: Show what you are doing even if you can do it in your head. If you’re finding a slope, show the quotient even if the denominator is 1.

Do Not do arithmetic: A long expression consisting entirely of numbers such as you get when doing a Riemann sum, does not need to be simplified in any way. If you a simplify correct answer incorrectly, you will lose credit.

Do Not leave expression such as R(3) – pull its numerical value from the table.

Do Not: Find a regression equation and then use that to answer parts of the question. While regression is perfectly good mathematics, regression equations are not one of the four things students may do with their calculator. Regression gives only an approximation of our function. The exam is testing whether students can work with numbers.


This question typically covers topics from Unit 6 of the 2019 CED but may include topics from Units 2, 3, and 4 as well.


Free-response examples:

  • 2007 AB 3 (4 simple parts on various theorems, yet the mean score was 0.96 out of 9),
  • 2017 AB 1/BC 1, and AB 6,
  • 2016 AB 1/BC 1
  • 2018 AB 4

Multiple-choice questions from non-secure exams:

  • 2012 AB 8, 86, 91
  • 2012 BC 8, 81, 86  (81 and 86 are the same on both the AB and BC exams)

 

 

 

Revised March 12, 2021


 

 

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


Y the FTC?

So, you’ve finally proven the Fundamental Theorem of Calculus and have written on the board:

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

And the students ask, “What good is that?” and “When are we ever going to use that?” Here’s your answer.

There are two very important uses of this theorem. Show them BOTH uses right away to help your students see why the FTC is so useful and important.

First, in words the theorem says that “the integral of a rate of change is the net amount of change.” So, if you are given a rate of change (as you are every year on the AP Calculus exam) and asked to find the amount of change (as you are every year on the AP Calculus exam), this is what you use, Show an example such as 2015 AB 1/BC1 that states,

“The rate at which rainwater flows into a drain pipe is modeled by the function R, where R\left( t \right)=20\sin \left( {\frac{{{{t}^{2}}}}{{35}}} \right) cubic feet per hour….

“(a) How many cubic feet of rainwater flow into the pipe during the 8-hour time interval ?”

The answer is of course, \displaystyle \int_{0}^{8}{{20\sin \left( {\frac{{{{t}^{2}}}}{{35}}} \right)dt}}. (Which they will soon learn how to evaluate.)

Second, a more immediate use is to avoid all that work you’ve been doing setting up Riemann sums and finding their limits. No more of that! Give them this integral to evaluate:

\displaystyle \int_{2}^{7}{{2xdx}}

Draw the trapezoid representing the area between the graph of y=2x and the x-axis on the interval [2,7] and find its area =  \displaystyle \frac{1}{2}\left( 5 \right)\left( {18+4} \right)=45

Then ask, “Does anyone know of a function whose derivative is 2x?” Let them think for a minute and someone will say, “Yeah, it’s {{x}^{2}}”  And then show them

\displaystyle \int_{2}^{7}{{2xdx}}={{7}^{2}}-{{2}^{2}}=45

Then go for a harder one:  \displaystyle \int_{0}^{{\frac{\pi }{2}}}{{\cos \left( x \right)dx}}

“Does anyone know a function whose derivative is \cos \left( x \right)?”

“Why yes, it’s \sin \left( x \right)

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

That was easy!

If you want to challenge them and review some functions of the “special angles” try this one:

\displaystyle \int_{{\frac{\pi }{6}}}^{{\frac{{4\pi }}{3}}}{{\cos \left( x \right)dx}}=\sin \left( {\frac{{4\pi }}{3}} \right)-\sin \left( {\frac{\pi }{6}} \right)=\frac{{\sqrt{3}}}{2}-\frac{1}{2}

Tie the two parts together: Look at the graph of y=\sin \left( x \right). How much does it change from 0 to \frac{\pi }{2}? How much does it change from \frac{\pi }{6} to \frac{{4\pi }}{3}?

Sum up, by looking ahead:

  1. “The function whose derivative is …” is called the antiderivative.
  2. Using antiderivatives to evaluate definite integrals is easy; the hard part is finding the antiderivatives, since they are not all as straightforward as the two examples above. So, next we need to spend a few weeks learning how to find antiderivatives.[1]
  3. Given a derivative, finding its antiderivative is also the start of solving differential equations. This, too, will soon be a concern in the course.

[1] As I’ve written before, this is where it seems logical place to teach antiderivatives. Now students have a reason to find them. Teaching antidifferentiation after differentiation, before integration, seems like an intellectual exercise. Sure, it’s fun, but now we have a need for it.



Riemann Sums – the Theory

The series of post leads up to the Fundamental theorem of Calculus (FTC). Obviously, a very important destination.

  1. Working Towards Riemann Sums
  2. 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
  3. More about the FTC The derivative of a function defined by an integral – the other half of the FTC.
  4. Good Question 11 Riemann Reversed – How to find the integral, given the Riemann sum. A problem that appears on the AP Calculus exams and can be confusing for students, at first.
  5. Properties of Integrals
  6. Variation on a Theme – 2 Comparing Riemann sums
  7. Trapezoids – Ancient and Modern – some history.

 

 

 

 

Revised and updated October 22, 2018


Getting Ready to Integrate

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 sum) and the Fundamental theorem of Calculus. Theses posts help prepare students for Riemann sums.

Integration Itinerary  Some thoughts on the order of topics in your integration unit.

Some preliminary posts leading up to 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. Jobs, Jobs, Jobs Integration in real life.
  4. Working Towards Riemann Sums (12-10-2012)

While I prefer to teach antidifferentiation after students have learned about the Fundamental Theorem of Calculus, others prefer to discuss antidifferentiation firsts and the topic often precedes Riemann sums in textbooks. (See Integration Itinerary )  If you are among those, here are posts on antidifferentiation. If you teach this topic later, save this post for then.

ANTIDIFFERENTIATION

Antidifferentiation  (11-28-2012)

Why Muss with the “+C”? But still don’t forget it.

Arbitrary Ranges (2-9-2014) Integrating inverse trigonometric functions.

ANTIDIFFERENTIATION BY PARTS This is a BC topic, or you could use it after the exam in an AB course.

Integration by Parts 1 (2-2-2013) Basics

Integration by Parts 2 (2-4-2013) The Tabular Method

Modified Tabular Integration (7-24-2013) A quicker way

Parts and More Parts (8-5-2016) Reduction formulas (Not tested on the AP Calculus exams)

Good Question 12 – Parts with a Constant (12-13-2016) How come you don’t need the “+C”?


 

 

 

 

Good Question 15: 2018 BC 2(a)

My choices for the Good Question series are somewhat eclectic. Some are chosen because they are good, some because they are bad, some because I learned something from them, some because they can be extended, and some because they can illustrate some point of mathematics. This question and the next, Good Question 16, are in the latter group. They both concern units. They both are taken from this year’s AP calculus BC exam; both are suitable for AB classes. In this question 2018 BC 2(a) has some unusual units and in the next 2018 BC 2(b) the units help you figure out what to do. Part (c) concerns an improper integral and pard (d) is about parametric equation, neither of these are AB topics. 

2018 BC 2(a)

2018 BC 2 gave an equation that modeled the density p(h) of plankton in a sea in units of millions of cells per cubic meter, as a function of the depth, h, in meters.  Specifically, p\left( h \right)=0.2{{h}^{2}}{{e}^{{-0.0025{{h}^{2}}}}} for 0\le h\le 30. Part (a) asked for the value of {p}'\left( {25} \right) and also asked students “Using correct units, [to] interpret the meaning of {p}'\left( {25} \right) in the context of the problem.”

Plankton

This was a calculator active question, so the computation is easy enough: {p}'\left( {25} \right)=-1.17906

Now units of the derivative are always very easy to determine; this should be automatic. The derivative is the limit of a difference quotient, so its units are the units of the numerator divided by the units of the denominator. In this case that’s millions of cells per cubic meter per meter of depth.

While “millions of cells per meter to the fourth power” is technically correct and will probably earn credit, what is a meter to the fourth power?

It is similar to the better-known situation with velocity and acceleration. I never liked the idea of saying the acceleration is so many meters per square second. What’s a square second? Are there round seconds? Acceleration is the change in velocity in meters per second per second; that is, at a particular time the velocity is changing at the rate of so-many meters per second each second

Returning to the question, a cubic meter (volume) and a meter of depth (linear) are not things that you should combine. The notational convenience of writing meters to the fourth power hides the true meaning. So, a better interpretation is “At depth of  25 meters, the number cells is decreasing at the rate of 1.179 million cells per cubic meter per meter of depth.” or “The number of cells changing at the rate of -1.179 million cells per cubic meter per meter of depth.”

Had the model been given using volume units such as millions of cells per liter, then the units of the derivative would be millions of cells per liter per meter. That makes more sense.

But what does it mean?

Let’s look at the graph of the derivative. The window is 0 < h < 30 and –2.5 < p(x) < 2.5

It means, that as we pass thru that thin (thickness \Delta h\to 0) film of water 25 meters down, there are approximately 1.179 million cells per cubic meter less than in the thin film right above it and more than in the thin film right below it.

For reference, p\left( {25} \right)\approx 26.2014 million cells per cubic meter. Of course, that thin (thickness \Delta h\to 0) film of water has very little volume; it is kind of difficult to think of a cubic meter exactly 25 meters below the surface (maybe a cube extending from 24.5 meters to 25.5 meters?). As \Delta h\to 0 does a cubic meter approach a square meter?

The cubic meter above h = 25 has \displaystyle \int_{{24}}^{{25}}{{p\left( h \right)(1)dh}}=26.763 cells and the cubic meter below has 25.586 million cells. This is a decrease of 1.1767 million cells. So, the derivative is reasonable.

(To make the units of \displaystyle \int_{{24}}^{{25}}{{p\left( h \right)(1)dh}} correct, I included a factor of 1 square meter, this multiplied by p(h) million cells per cubic meter and by dh in meters give a result of millions of cells. More on why this is necessary in Good Question 16 on density.)


Previous Good Questions can be found under the “Thru the Year” tab on the black navigation bar at the top of the page, or here.


Units

I had a question from a reader recently asking about how to determine the units for derivatives and integrals.

Derivatives: The units of the derivative are the units of dy divided by the units of dx, or the units of the dependent variable (f(x) or y) divided by the units of the independent variable (x). The reason for this comes from the definition of the derivative:

\displaystyle {f}'\left( x \right)=\underset{{\Delta x\to 0}}{\mathop{{\lim }}}\,\frac{{f\left( {x+\Delta x} \right)-f\left( x \right)}}{{\Delta x}}

In the quotient the numerator has the units of f and in the denominator the has the same units as x.

Definite Integrals: The units of a definite integral are the units of the integrand f(x) multiplied by the units of dx. This comes directly from the definition of a definite integral:

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

The factor (ba) has the same units a x, the independent variable, and the f(x) has whatever units it has. From the Riemann sum we can see that since these factors are multiplied, that product is the units of the definite integral.

The integrand is the derivative of its antiderivative (by the FTC) and so its units are often derivative units (miles per hour, furlongs per fortnight, etc.). When multiplied by (ba)/n its units “cancel” the units of the denominator of f(x) and the result is the units of the numerator of f(x).  This is not always the case*, therefore, multiplying the units is safest.


*The definite integral \displaystyle \int_{{-2}}^{2}{{\sqrt{{4-{{x}^{2}}}}dx}} gives the area of a semi-circle of radius 2 feet. The units of the radical are feet and represent the vertical distance from the x-axis to a point in the semi-circle; the dx is the horizontal side of the Riemann sum rectangles also in feet. Both are measured in the same linear units and the area is their product: feet times feet or square feet.