# Extreme Average

A recent post on the AP Calculus bulletin board observed that the maximum value of the average value of a function on an interval occurred at the point where the graph of the average value and the function intersect. I am not sure if this concept is important in and of itself, but it does make an interesting exercise.

For a function f(x), we may treat its average value as a function, A(x), defined for all x ∈ [a, b], interval [a, x] as

$\displaystyle A\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {\tfrac{1}{{x-a}}\int_{a}^{x}{{f\left( t \right)dt}}} & {x\ne a} \\ {f\left( a \right)} & {x=a} \end{array}} \right.$

Graphically, the segment drawn at y = A(x) is such that the regions between the line and the function above and below the segment have equal areas. See figure 1 in which the red curve is the function, and the blue curve is the average value function. The two shaded regions have the same area.

Regardless of the starting value, the function and its average value start at the same value. If the function is increasing the average value is less than the function and increasing. When the function starts to decrease, the average value will continue to increase for a while. When the two graphs nest intersect, the process starts over, and the average value will now start to decrease. Therefore, the intersection value is when the average value function change from increasing to decreasing and this is its (local) maximum value. See Figure 2.

This continues until the graphs intersect again after the function starts to increase: a (local) minimum value of the average value function. The process continues with the extreme values of the average value function (blue graph) occurring at its intersections with the function. Figure 3

This can be proved by finding the extreme values of the average value function by considering its derivative. Begin by finding its derivative using the product rule (or quotient rule) and the FTC.

$\displaystyle {A}'\left( x \right)=\tfrac{1}{{x-a}}f\left( x \right)+\left( {-\tfrac{1}{{{{{\left( {x-a} \right)}}^{2}}}}} \right)\int_{a}^{x}{{f\left( t \right)dt}}$

$\displaystyle {A}'\left( x \right)=\tfrac{1}{{x-a}}f\left( x \right)-\tfrac{1}{{x-a}}\left( {\tfrac{1}{{x-a}}\int_{a}^{x}{{f\left( t \right)dt}}} \right)$

$\displaystyle {A}'\left( x \right)=\tfrac{1}{{x-a}}\left( {f\left( x \right)-A\left( x \right)} \right)$

The critical points of a(x) occur when its derivative is equal to zero (or undefined). This is when $f\left( x \right)=A\left( x \right)$ (or when x = a, the endpoint). This is where the graphs intersect.

## How to use this in your class

This is not a concept that is likely to be tested on the AP Calculus Exams. Nevertheless, it is an easy enough idea to explore when teaching the average value of a function and at the same time reviewing some earlier concepts such as product (or quotient) rule, the FTC (differentiating an integral), and some non-ordinary simplification.

You could have your students use their own favorite function and show that the extreme values of its average value occur where the average value intersects the function. This is good practice in equation solving on a calculator since the points do not occur at “nice” numbers. Here’s an example.

If $\displaystyle f\left( x \right)=\sin \left( x \right)$, then its average value on the interval $\displaystyle [0,\infty )$ is

$\displaystyle A\left( x \right)=\tfrac{1}{x}\int_{0}^{x}{{\sin \left( t \right)dt}}=\frac{{-\cos \left( x \right)+1}}{x}$.

The intersections of f(x) and A(x) can be found by solving

$\displaystyle\sin \left( x \right)=\frac{{-\cos \left( x \right)+1}}{x}$

The extreme values of $\displaystyle \frac{{-\cos \left( x \right)+1}}{x}$ may also be found using a calculator.

The points are the same. the first is approximately (2.331, 0.725) and the second is (6.283, 0) or (2π, 0). This second is reasonable since at 2π the sine function has completed one period and its average value zero. (See figure 3 again.).

Other questions you could ask (for my function anyway) are what is the absolute maximum and how can you be sure? Why are all the minimums zero?

The message on the AP Calculus discussion boards that inspired this post was started by Neema Salimi an AP Calculus teacher from Georgia. He made the original observation. You can read his original post and proof, and comments by others here.

## The Rule of Four

Not much has been heard of the Rule of Four lately. The Rule of Four suggests that mathematical concepts should be looked at graphically, numerically, analytically, and verbally. It has not gone away. The Rule of Four has a new name: multiple representations. (In the latest Course and Exam Description, you will find it in Mathematical Practices (p. 14), specifically practices 2.B, 2.C, 2.D, 2.E, 3.E, 3.F, 4.A, and 4.C)

I have used the Rule of Four in this post. The post started with a verbal discussion of the concept and how the result can be seen graphically. That was followed by analytic proof. At the end is a numerical example.

Other posts on the average value of a function:

Finding the average value of a function on an interval is Topic 8.1 in the Course and Exam Description (p. 149)

Average Value of a Function – or How do you average an infinite number of numbers?

Most Triangles Are Obtuse! An obvious observation, but here’s how to figure the exact proportion of obtuse to acute triangles.

Half-full or Half-empty Visualizing the average value of a function

What’s a Mean Old Average Anyway? Be sure to distinguish between the average rate of change, the average value of a function, and the mean value theorem.

# 2019 CED Unit 8: Applications of Integration

This unit seems to fit more logically after the opening unit on integration (Unit 6). The Course and Exam Description (CED) places Unit 7 Differential Equations before Unit 8 probably because the previous unit ended with techniques of antidifferentiation. My guess is that many teachers will teach Unit 8: Applications of Integration immediately after Unit 6 and before Unit 7: Differential Equations. The order is up to you.

Unit 8 includes some standard problems solvable by integration (CED – 2019 p. 143 – 161). These topics account for about 10 – 15% of questions on the AB exam and 6 – 9% of the BC questions.

### Topics 8.1 – 8.3 Average Value and Accumulation

Topic 8.1 Finding the Average Value of a Function on an Interval Be sure to distinguish between average value of a function on an interval, average rate of change on an interval and the mean value

Topic 8.2 Connecting Position, Velocity, and Acceleration of Functions using Integrals Distinguish between displacement (= integral of velocity) and total distance traveled (= integral of speed)

Topic 8. 3 Using Accumulation Functions and Definite Integrals in Applied Contexts The integral of a rate of change equals the net amount of change. A really big idea and one that is tested on all the exams. So, if you are asked for an amount, look around for a rate to integrate.

### Topics 8.4 – 8.6 Area

Topic 8.4 Finding the Area Between Curves Expressed as Functions of x

Topic 8.5 Finding the Area Between Curves Expressed as Functions of y

Topic 8.6 Finding the Area Between Curves That Intersect at More Than Two Points Use two or more integrals or integrate the absolute value of the difference of the two functions. The latter is especially useful when do the computation of a graphing calculator.

### Topics 8.7 – 8.12 Volume

Topic 8.7 Volumes with Cross Sections: Squares and Rectangles

Topic 8.8 Volumes with Cross Sections: Triangles and Semicircles

Topic 8.9 Volume with Disk Method: Revolving around the x– or y-Axis Volumes of revolution are volumes with circular cross sections, so this continues the previous two topics.

Topic 8.10 Volume with Disk Method: Revolving Around Other Axes

Topic 8.11 Volume with Washer Method: Revolving Around the x– or y-Axis See Subtract the Hole from the Whole for an easier way to remember how to do these problems.

Topic 8.12 Volume with Washer Method: Revolving Around Other Axes. See Subtract the Hole from the Whole for an easier way to remember how to do these problems.

### Topic 8.13  Arc Length BC Only

Topic 8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled  BC ONLY

### Timing

The suggested time for Unit 8 is  19 – 20 classes for AB and 13 – 14 for BC of 40 – 50-minute class periods, this includes time for testing etc.

### Previous posts on these topics for both AB and BC include:

Average Value and Accumulation

Half-full or Half-empty

Accumulation: Need an Amount?

AP Accumulation Questions

Good Question 7 – 2009 AB 3 Accumulation, explain the meaning of an integral in context, unit analysis

Good Question 8 – or Not Unit analysis

Graphing with Accumulation 1 Seeing increasing and decreasing through integration

Graphing with Accumulation 2 Seeing concavity through integration

Area

Area Between Curves

Under is a Long Way Down  Avoiding “negative area.”

Math vs. the “Real World”  Improper integrals  BC Topic

Volume

Volumes of Solids with Regular Cross-sections

Volumes of Revolution

Why You Never Need Cylindrical Shells

Visualizing Solid Figures 1

Visualizing Solid Figures 2

Visualizing Solid Figures 3

Visualizing Solid Figures 4

Visualizing Solid Figures 5

Painting a Point

Other Applications of Integrals

Density Functions have been tested in the past, but are not specifically listed on the CED then or now.

Who’d a Thunk It? Some integration problems suitable for graphing calculator solution

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

# Applications of integrals, part 1: Areas & Average Value

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

# Good Question 8 – or not?

Today’s question is not a good question. It’s a bad question.

But sometimes a bad question can become a good one.

This one leads first to a discussion of units, then to all sorts of calculus.

Here’s the question a teacher sent me this week taken from his textbook:

The normal monthly rainfall at the Seattle-Tacoma airport can be approximated by the model $R=3.121+2.399\sin \left( 0.524t+1.377 \right)$, where R is measured in inches and t is the time in months, t = 1 being January. Use integration to approximate the normal annual rainfall.  Hint: Integrate over the interval [0,12].

Of course, with the hint it’s not difficult to know what to do and that makes it less than a good question right there. The answer is $\displaystyle \int_{0}^{12}{R(t)dt=37.4736}$ inches. You could quit here and go on to the next question, but …

Then a student asked. “If R is in inches shouldn’t be in units of the integral be inch-months, since the unit of an integral is the unit of the integrand times the units of the independent variable?”  Well, yes, they should. So, what’s up with that?

Also, the teacher figured that the integral of a rate is an amount and our answer is an amount, so why isn’t the integrand a rate?

The only answer I could come up with is that the statement “R is measured in inches” is incorrect; R should be measured in inches /month. The opening phrase “normal monthly rainfall” also seems to point to the correct units for R being inches/month.

Problem solved; or maybe does this lead to a different concern?

The teacher pointed out that R(6) = 0.7658 inches is a reasonable answer for the amount of rain in June whereas $\displaystyle \int_{0}^{6}{R(t)dt=}20.4786$ is not.

If R is a rate, then the amount of rain that falls in June (t = 6) is given by $\displaystyle \int_{5}^{6}{R(t)dt}=0.9890$.

From here on we will assume that R is a rate with units of inches/month. Here are the individual monthly rates calculated with a CAS.

The total amount of rainfall (second line above) appears be R(1) + R(2) + R(3) + … +R(12) = 37.4742. This is very close to the amount calculated by integration.

The slight difference of 0.0006 is not a round off error.

Remember, behind every definite integral there is a Riemann sum!

Again, the units are the problem. Why does the sum of the monthly rates seem to give the total amount?  The reason is that the terms of the sequence above are actually the values of a right-side Riemann sum of the rate, R(t), over the interval [0,12] with 12 equal subdivisions of width 1 (month) each with the 1’s left out as 1’s often are. Therefore, their sum should come close to the total yearly rainfall, but it is really just an approximation of it.

The actual total for any month, n, is given by $\displaystyle \int_{n-1}^{n}{r(t)}dt$. For example the amount of rain that falls in June is given by $\displaystyle \int_{5}^{6}{R(t)dt}=0.9890$ inches.

Here is the sequence of the actual monthly rainfall values in inches, and their sum.

This agrees with the integral. Why? Because one of  the properties of integrals tell us that $\displaystyle \sum\limits_{n=1}^{12}{\int_{n-1}^{n}{r(t)dt}}=\int_{0}^{12}{r(t)dt}$.

Another instructive thing with this integral is this: The function $R=3.121+2.399\sin \left( 0.524t+1.377 \right)$ is periodic with a period of  $\frac{2\pi }{0.524}\approx 11.9908\approx 12$. So the sine function takes on (almost) all its values in a year, as you would expect. Since the sine values all but cancel each other out

$\displaystyle \int_{0}^{12}{3.121+2.399\sin \left( 0.524t+1.377 \right)dt}\approx \int_{0}^{12}{3.121dt=3.121\left( 12-0 \right)=37.452}$. Close!

The total rainfall divided by 12 is $\frac{37.452}{12}=3.121$ this must be close to the average rainfall each month. The average rainfall is $\displaystyle \frac{1}{12}\int_{0}^{12}{R\left( t \right)dt}=3.1228$ inches. Close, again!

So, there you have it. Is this a good question or not? We considered all these concepts while working not just with an equation but with numbers from a poorly stated problem:

• Unit analysis
• Integration by technology
• Realizing that a pretty good approximation is not correct, due again to units.
• A Riemann sum approximation in a real situation that comes very close to the value by integration
• Using a property of a periodic function to greatly simplify an integral
• Finding average value two ways

So, it turned out to be a sunny day in Seattle.

.

# Half-full and Half-empty

A thought experiment:

Suppose you had a container with a rectangular base whose length runs from x = a to x = b, with a width of one inch. The container has four vertical rectangular sides. You put a piece of, say, plastic into the container which fits snugly along the bottom and four sides. The top of the piece is irregular and has the equation y = f(x).  If the plastic were to melt, how high up the sides would the melted plastic rise?

One way to think about this is to consider the final level, L. When melted, the plastic above the final level must fill in the part below, leaving a rectangle with the same area as that under the original function’s levels. (The one-inch width will remain the same and not affect the outcome.)

So the original area is $\displaystyle \int_{a}^{b}{f\left( x \right)dx}$ and the final area is $L\left( b-a \right)$. Since these are the same, we can write an equation and solve it for L.

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

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

But that’s the equation for the average value of a function!

What a surprise!

Well, not a surprise for you, the teacher. This might be a good way to sneak up on the average value of a function idea for your students while giving them a good visual idea of the concept.

# What’s a Mean Old Average Anyway?

Students often confuse the several concepts that have the word “average” or “mean” in their title. This may be partly because not just the names, but the formulas associated with each are very similar, but I think the main reason may be that they are keying in on the word “average” rather than the full name.

Here are the three items. We will assume that the function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b):

1.  The average rate of change of a function over the interval is simply the slope of the line from one endpoint of the graph to the other.

$\displaystyle \frac{f\left( b \right)-f\left( a \right)}{b-a}$

2. The mean (or average) value theorem say that somewhere in the open interval (a, b) there is a number c such that the derivative (slope) at x = c is equal to the average rate of change over the interval.

$\displaystyle {f}'\left( c \right)=\frac{f\left( b \right)-f\left( a \right)}{b-a}$

3. The average value of a function is literally the average of all the y-coordinates on the interval. It is the vertical side of a rectangle whose base extends on the x-axis from x = a to x =b and whose area is the same as the area between the graph and the x-axis and the function over the same interval.

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

Notice that when you evaluate the integral, the result looks very much like the ones above. This formula is also called the mean value theorem for integrals or the integral form of the mean value theorem. No wonder people get confused.

The three are closely related. Consider a position-velocity-acceleration situation. The average rate of change of position (#1 above) is the average value of the velocity (#3) and somewhere the velocity must equal this number (#2). Similarly, the average rate of change of velocity (#1) is the average acceleration (#3) and somewhere in the interval the acceleration (derivative of velocity) must equal this number (#2).

These ideas are tested on the AP calculus exams sometimes in the same question. See for example 2004 AB 1 parts c and d.

So, help your students concentrate on the entire name of the concepts, not just the “average” part.

# Most Triangles Are Obtuse!

What is the probability that a triangle picked at random will be acute? An average value problem.

The thing here is to define what you mean by picking a triangle “at random.” You could open a Geometry book and take the first triangle you come to, but are the triangles in a Geometry book really a good sample space? I doubt it.

Let’s try this: let A, B, and C be the measures, in degrees, of the angles of $\Delta ABC$. Let A be a random number between 0 and 180, and let B be a random number between 0 and 180 – A. Then, let C = 180 – AB.

Then $P(A<90)=\tfrac{1}{2}$

For any A < 90, B and C are chosen from the interval $\left( 0,180-A \right)$. In order for the triangle to be acute B and C must be within 90 of both ends of this interval. That is, B and C must both be in the interval $[90-A,90]$.

This is an interval of length A and the probability of picking numbers, B and C, at random in this interval is $\frac{A}{180-A}$. At this point you may want to stop and calculate a typical probability. For instance, if A= 30 then the probability of both B and C being acute is $\frac{30}{180-30}=0.20$.

In general $P\left( A<90\text{ and }\left( B<90\text{ and }C<90 \right) \right)=\frac{1}{2}\cdot \frac{A}{180-A}$. What we need is the average of (all) these values.

This average is $\displaystyle \frac{1}{2}\frac{1}{90-0}\int_{0}^{90}{\frac{A}{180-A}dA}\approx 0.19315$

So about 19.3% of triangles are acute, the rest are obtuse (or right). Leaving one to believe that most triangles are obtuse.

Taylor Gibson of the Greenhill School in Addison, Texas wrote this Monte Carlo simulation of the situation described above. Thank you, Taylor.
_______________
This problem was posted on the AP Calculus Electronic Discussion Group (11/22/03) by Stu Schwartz an AP Calculus teacher at Wissahickon High School in Ambler, Pennsylvania. The solution is by one of Mr. Schwartz’s students Kurt Schneider, a tenth grader at the time who completed AB calculus in eighth grade and BC calculus in ninth grade!