Applications of Integration – Accumulation 1

Posted on December 18, 2018 by Lin McMullin

The idea that the definite integral is an “accumulator” means that integrating a rate of change over an integral gives the net amount of change over the interval.Many of the application of integration are based on this idea. Here are some past posts on this idea.

Accumulation An introductory activity to explore accumulation and the relationship between an accumulation and derivatives

Accumulation: Need an Amount? (1-21-2013) An important and always tested application.

AP Accumulation Questions (1-23-2013) Two good questions for teaching and learning accumulation.

Graphing with Accumulation 1 (1-25-2013) Everything you need to know about the graph of a function given its derivative can be found using integration techniques. Increasing and decreasing.

Graphing with Accumulation 2 (1-28-2013) Everything you need to know about the graph of a function given its derivative can be found using integration techniques. Concavity.

Next Tuesday is Christmas (already). There will be no post until Tuesday January 1, 2019 when I will there will be several more links to post on accumulation.

Happy Holidays, Merry Christmas, and Happy New Year.

Applications of Integration – Volume

Posted on December 11, 2018 by Lin McMullin

One of the major applications of integration is to find the volumes of various solid figures.

Volume of Solids with Regular Cross-sections This is where to start with volume problems. After all, solids of revolution are just a special case of solids with regular cross-sections.

Volumes of Revolution

Subtract the Hole from the Whole and Does Simplifying Make Things Simpler?

Visualizing Solid Figures

Visualizing Solid Figures 1 Making models et al.
Visualizing Solid Figures 2 Using Winplot to visualize solids with regular cross-sections.
Visualizing Solid Figures 3 Using Winplot to visualize the Disk and Washer methods.
Visu alizing Solid Figures 4 Using Winplot to visualize finding volumes by cylindrical shells. While no problem on the AP Calculus exams will require a solution by the Cylindrical Shell method, students may use the method and be eligible for full credit. So this is “enrichment.”
Challenge – the half-full problem comparing the disk and shell methods and Visualizing Solid Figures 5 the detailed solution

Why you Never Need Cylindrical Shells

Painting a Point

Revised and update October 22, 2018

Applications of Integration – Area & Average Value

Posted on December 4, 2018 by Lin McMullin

Usually the first application of integration is to find the area bounded by a function and the x-axis, followed by finding the area between two functions. We begin with these problems

First some calculator hints

Graphing Integrals using a graphing calculator to graph functions defined by integrals

Graphing Calculator Use and Definition Integrals – Exam considerations Suggestions for using a calculator efficiently in area/volume problems

Area Problems

Area Between Curves

Under is a Long Way Down How to avoid “negative area.”

Density Functions Not often asked on the AP exams, but a good application related to area, nevertheless.

Who’d a thunk it? Some more complicated area problems for CAS solution.

Improper Integrals and Proper Areas – a BC topic

Average Value

Average Value of a Function

What’s a Mean Old Average Anyway – Discusses the different “average” in calculus

Half-full and Half Empty – Average Value

Average Value Activity to help students discover the Average Value formula

Revised and updated October 22, 2018

Getting Ready to Integrate

Posted on November 13, 2018 by Lin McMullin

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

The Old Pump Where I start Integration
Flying into Integrationland Continues the investigation in the Old Pump – the airplane problem
Jobs, Jobs, Jobs Integration in real life.
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 18: 2018 BC 2(b)

Posted on May 29, 2018 by Lin McMullin

In this post we look at another part of the AP Calculus BC exam. Good Question 15 discussed the unusual units in 2018 BC 2(a). In this post we look at 2018 BC 2(b) where units help us find the correct integral to answer the question.

How do you answer a question of a type you’ve never seen before? I expect that’s what many of the students taking the 2018 AP Calculus exam were asking when they got to BC 5. If you’ve never done a density question how do you handle this one?

The question concerns density. Density gives you how much of something exists in a certain length, area, or volume. Density questions have appeared on the exam now and then, most recently 2008 AB 92 (which really isn’t recent, but then there are a lot of questions we never see). I have a blog post about the density here with several examples. In that post the alternate solution to example 3 explained how I used a unit analysis to find the answer; I used a similar approach here.

2018 BC 2 (b)

The stem of the question tells us that at a depth of h meters, 0 < h < 30, the number of plankton in a cubic meter of sea water is modeled by $p\left( h \right)=0.2{{h}^{2}}{{e}^{{-0.0025{{h}^{2}}}}}$ million cells per cubic meter. Part (b) asks for the number of million of plankton in a column of water whose horizontal cross sections have a constant area of 3 square meters.

If the density were constant, then it is just a matter of multiplying the volume of the column times the constant density. Alas, the density is not constant; it varies with the depth. What to do?

Since an amount is asked for, you usually look around for a rate to integrate. Density is a kind of rate: the units are millions of cells per cubic meter. You need to integrate something concerning the density so that you end up with millions of cells; something that will “cancel” the cubic meters.

Consider a horizontal slice thru the column at depth h meters. While I’m not sure plankton is a good topping for pizza, you could picture this as a rather large pizza box whose sides are $\sqrt{3}$ meters long and whose height is $\Delta h$ meters. This box has a volume of 3 $\Delta h$ cubic meters. For small values of $\Delta h$ the number of million plankton in the box is nearly constant, so at depth h_i , there are p(h_i) million plankton per cubic meter or ${3p\left( {{{h}_{i}}} \right)\Delta h}$ million plankton in the box.

Notice how the units of the individual quantities combine to assure you the final quantity has the correct units:

$\displaystyle (3\text{ square meters)}\cdot \left( {p\left( {{{h}_{i}}} \right)\text{ }\frac{{\text{million plankton}}}{{\text{cubic meters}}}} \right)\left( {\Delta h\text{ meters}} \right)=3p\left( {{{h}_{i}}} \right)\Delta h\text{ million plankton}$

Now to find the amount in the column of water we can add up a stack of “pizza boxes.” The sum is $\sum\limits_{{i=1}}^{n}{{3p\left( {{{h}_{i}}} \right)\Delta h}}$ . Now, if we take thinner boxes by letting $\Delta h\to 0$ , we are looking at a Riemann sum. And calculus gives us the answer.

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{3p\left( {{{h}_{i}}} \right)\Delta h}}=\int_{0}^{{30}}{{3p\left( h \right)dh}}\approx 1,675$ million plankton in the column of water (rounded to the nearest million as directed in the question.)

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.

Renumbered 3-14-24 Was Question 16.

Math vs. the “Real World”

Posted on February 2, 2018 by Lin McMullin

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.

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.
“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.
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!

Applications of Integration, part 3: Accumulation

Posted on January 2, 2018 by Lin McMullin

Integration, at its basic level, is addition. A definite integral is a sum (a Riemann sum). When you add things you get an amount of whatever you are adding: you accumulate. Here are some previous posts on this important idea that often shows up on the AP Calculus exams (usually the first free-response question!)

Accumulation: Need an Amount?

Good Question 6 – 2000 AB 4 One of my favorite questions

Painting a Point

Real “Real Life” Graph Reading

Jobs, Jobs, Jobs are real life accumulation problem based on a graph

Graphing with Accumulation 1 Increasing and decreasing.

Graphing with Accumulation 2 Concavity

Accumulation and Differential Equations Differential equations will be considered next week, but this idea relates to integration.

Good Question 8 – or Not?

Rate and Accumulation Questions (Type 1)