Improper Integrals and Proper Areas

Posted on January 25, 2014 by Lin McMullin

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. Of 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.

Modified Tabular Integration

Posted on July 24, 2013 by Lin McMullin

Several weeks ago, Dr. Qibo Jing, an AP teacher at Rancho Solano Preparatory School in Scottsdale, Arizona, posted a new way to approach tabular integration to the AP Calculus Community discussion group. He calls this the Modified Tabular Method. The algorithm makes repeated integration by parts quicker and more streamlined than the usual method. The usual method is explained here.

Here is Dr. Jing’s method outlined and illustrated with this example $\displaystyle I=\int{\left( {{x}^{3}}+7{{x}^{2}} \right)\cos \left( x \right)dx}$

Step 1: Identify u and dv in the usual way and rewrite the given integral in terms of $u={{x}^{3}}+7{{x}^{2}}$ and $dv=\cos \left( x \right)=\tfrac{d}{dx}\sin \left( x \right)$ .

$\displaystyle I=\int{\left( {{x}^{3}}+7{{x}^{2}} \right)\cos \left( x \right)dx}=\int{\left( {{x}^{3}}+7{{x}^{2}} \right)\left( \tfrac{d}{dx}\left( \sin x \right) \right)}$

Step 2: The first term of the antiderivative is the product of the two functions that appear in the revised integral:

$\displaystyle I=\int{\left( {{x}^{3}}+7{{x}^{2}} \right)\left( \tfrac{d}{dx}\left( \sin x \right) \right)}=\left( {{x}^{3}}+7{{x}^{2}} \right)\sin \left( x \right)\cdots$

Step 3: The remaining terms alternate sign. The next term has subtraction sign etc. The first factor of the next term is the derivative of the first factor of the current term. The second factor is the antiderivative of the second factor of the current term – differentiate the first factor; integrate the second factor.

$\displaystyle I =\int{\left( {{x}^{3}}+7{{x}^{2}} \right)\left( \tfrac{d}{dx}\left( \sin x \right) \right)}=\left( {{x}^{3}}+7{{x}^{2}} \right)\sin \left( x \right)-\left( 3{{x}^{2}}+14x \right)\left( -\cos \left( x \right) \right)\cdots$

Step 4: Alternate signs and repeat step 3 until the term becomes zero or until the original form appears (See next example). DO NOT simplify either factor until you are done as this may change later terms.

The simplified result is

$\displaystyle I=\left( {{x}^{3}}+7{{x}^{2}}-6x-14 \right)\sin \left( x \right)+\left( 3{{x}^{2}}+14x-6 \right)\cos \left( x \right)+C$

Here is a second example. In this example the original integral returns as the third term.

$\displaystyle I=\int{{{e}^{x}}\sin \left( x \right)dx=\int{\sin \left( x \right)\left( \tfrac{d}{dx}{{e}^{x}}dx \right)}}$

$=\sin \left( x \right)\left( {{e}^{x}} \right)-\cos \left( x \right)\left( {{e}^{x}} \right)+\left( -\sin \left( x \right) \right)\left( {{e}^{x}} \right)$

$={{e}^{x}}\sin \left( x \right)-\cos \left( x \right)\left( {{e}^{x}} \right)-I$

$2I={{e}^{x}}\sin \left( x \right)-\cos \left( x \right)\left( {{e}^{x}} \right)$

$I=\tfrac{1}{2}\left( {{e}^{x}}\sin \left( x \right)-\cos \left( x \right)\left( {{e}^{x}} \right) \right)+C$

This really is a simpler algorithm than the usual tabular method.

Variations on a Theme – 2

Posted on June 28, 2013 by Lin McMullin

This is the second in an occasional series on adapting and expanding AP calculus exam questions. Variations on multiple-choice questions may make other multiple-choice questions, or, if you do not like making up all the “wrong” choices they can be changed to short answer questions for use in your class homework or test. Other variations make for good discussion questions.

Today we will take a look at 2008 AB 10. This question gave the graph below and asked which of $\displaystyle \int_{1}^{3}{f\left( x \right)dx}$ , a left, right, or midpoint Riemann sum approximation, or a trapezoidal sum approximation of the integral all with four subintervals of equal length has the least value.

I’ll refer to these as I, L, R, M and T. The answer is R as a quick sketch will show.

In the discussion we will assume the curve does not change direction or concavity over the interval of integration. The answers to the questions in the variations are at the end of this post.

Variation 1: The question seems pretty easy so let’s beef it up a bit. How about asking the student to put I, L, R, M and T in order from smallest to largest? For a multiple-choice question answer choices are inequalities with all 5 sums in them.

The midpoint Riemann sum may be the most difficult to see. Here is the way to explain it. See the figure below.

On each subinterval draw the horizontal segment at the function value at the midpoint of the interval. At this same point draw a tangent segment from one side of the interval to the other. This forms two congruent triangles with the horizontal segment and the edges of the interval.

Therefore, this midpoint trapezoid and the midpoint rectangle have the same area. Then because of the concavity, we see that in this case the midpoint Riemann sum, M, will be larger than the integral, I. With the same concavity the trapezoidal approximation will lie below the curve and therefore T < I < M. This will be true for any curve that is concave down. The opposite is true for a curve that is concave up: M < I < T. M and T are always on opposite sides of I.

Variation 2: Change the number of subintervals, but keep the same number for each sum. Does this change anything?

Variation 3: Change the number of subintervals so that the sums over the same interval have a different number of subintervals. Does this change anything?

Variation 4: Change the shape of the curve to one of the other 4 shapes that a curve may have. How does this change the order of the sums?

Variation 5: Move the graph from above the x-axis to below. Does this change anything?

Variation 6: The reversal problem. Suppose that L < T < I < M < R; describe the shape of the curve. Other orders can be used as well. In fact this idea has appeared on an AP calculus exams. In 2003 AB 85 students were told that a trapezoidal sum over approximates and a right Riemann sum under approximates a certain integral. The question asked which of 5 graphs could be the graph of the function.

___________________

If you have other variations please click “Leave a Comment” to share them. If you have variations on a different question please share them as well.

___________________

Answers:

Variation 1: R < T < I < M < L

Variation 2: No. Since the order will be the same on each subinterval adding them will not change the order.

Variation 3: This makes a difference. In the first figure above, consider a trapezoidal sum with one subinterval compared to a right Riemann sum with many subintervals. The right Riemann sum will be very close to the integral with its rectangles ending above the trapezoidal segment between the endpoints. In this case T < R for the figure shown.

Variation 4: The left and right Riemann sums depend on whether the curve is decreasing or increasing. The Trapezoidal sum and the midpoint Riemann sum depend on the concavity. For example, an increasing concave up curve the order is L < M < I < T < R.

Variation 5: No. If the curve is entirely below the x-axis all the values will be negative, but all the results discussed in the other variations will be the same; the order will not change.

Variation 6: The curve is increasing and concave down.

Why Muss with the “+C”?

Posted on June 21, 2013 by Lin McMullin

Here is a way to find the particular solution of a separable differential equation without using the +C and it might even be a little faster. As an example consider the initial value problem

$\displaystyle \frac{dy}{dx}=\frac{y-1}{{{x}^{2}}}\text{ with }x\ne 0\text{ and }y\left( 2 \right)=0$

Separate the variables as usual and then write each side a definite integral with the lower limit of integration the number from the initial condition and the upper limit variable. I’ll replace the dummy variable in the integrand with an upper-case X or Y to avoid having x and y in two places.

$\displaystyle \int_{0}^{y}{\frac{1}{Y-1}dY}=\int_{2}^{x}{{{X}^{-2}}dX}$

Integrate

$\displaystyle \left. \ln \left| Y-1 \right| \right|_{0}^{y}=\left. -{{X}^{-1}} \right|_{2}^{x}$

$\displaystyle \ln \left| y-1 \right|-\ln \left| 0-1 \right|=-\frac{1}{x}+\frac{1}{2}$

Note that near the initial condition where y = 0, $\left( y-1 \right)<0$ so $\left| y-1 \right|=-\left( y-1 \right)=1-y$ . Continue and solve for y.

$\displaystyle {{e}^{\ln \left( 1-y \right)}}={{e}^{-\frac{1}{x}+\frac{1}{2}}}$

$\displaystyle 1-y={{e}^{-\frac{1}{x}+\frac{1}{2}}}$

$\displaystyle y=1-{{e}^{-\frac{1}{x}+\frac{1}{2}}},\quad x>0$

No muss; no fuss.

Variations on a Theme by ETS

Posted on June 14, 2013 by Lin McMullin

Experienced AP calculus teacher use as many released exam questions during the year as they can. They are good questions and using them gets the students used to the AP style and format. They can be used “as is”, but many are so rich that they can be tweaked to test other concepts and to make the students think wider and deeper.

Below is a multiple-choice question from the 2008 AB calculus exam, question 9.

The graph of the piecewise linear function f is shown in the figure above. If $\displaystyle g\left( x \right)=\int_{-2}^{x}{f\left( t \right)\,dt}$ , which of the following values is the greatest?

(A) g(-3) (B) g(-2) (C) g(0) (D) g(1) (E) g(2)

I am now going to suggest some ways to tweak this question to bring out other ideas. Here are my suggestions. Some could be multiple-choice others simple short constructed response questions. A few of these questions, such as 3 and 4, ask the same thing in different ways.

1. 1. Require students to show work or justify their answer even on multiple-choice questions. So for this question they should write, “The answer is (D) g(1) since x = 1 is the only place where ${g}'\left( x \right)=f\left( x \right)$ changes from positive to negative.”
  2. Ask, “Which of the following values is the least?” (Same choices)
  3. Find the five values listed.
  4. Put the five values in order from smallest to largest.
  5. If $\displaystyle g\left( x \right)=g\left( -2 \right)+\int_{-2}^{x}{f\left( t \right)dt}$ and the maximum value of g is 7, what is the minimum value?
  6. If $\displaystyle g\left( x \right)=g\left( -2 \right)+\int_{-2}^{x}{f\left( t \right)dt}$ and the minimum value of g is 7, what is the maximum value?
  7. Pick any number (not just an integers) in the interval [–3, 2] to be a and change the stem to read, “If $\displaystyle g\left( x \right)=g\left( a \right)+\int_{a}^{x}{f\left( t \right)dt}$ ….” And then ask any of the questions above – some answers will be different, some will be the same. Discussing which will not change and why makes a worthwhile discussion.
  8. Change the equation in the stem to $\displaystyle g\left( x \right)=3x+\int_{-2}^{x}{f\left( t \right)dt}$ and ask the questions above. Again most of the answers will change. Also this question and the next start looking like some free-response questions. Compare them with 2011 AB 4 and 2010 AB 5(c)
  9. Change the equation in the stem to $\displaystyle g\left( x \right)=-\tfrac{3}{2}x+\int_{-2}^{x}{f\left( t \right)dt}$ and ask the questions above. This time most of the answers will change.
  10. Change the graph and ask the same questions.

Not all questions offer as many variations as this one. For some about all you can do is use them “as is” or just change the numbers.

Any other adaptations you can think of?

What is your favorite question for tweaking?

Math in the News Combinatorics and UPS

Revised: August 24, 2014

Absolute Value

Posted on May 23, 2013 by Lin McMullin

The answers to the True-False quiz at the end of the last post are all false. This brings us to absolute value, another topic I want to concentrate on for my upcoming Algebra 1 class. Absolute value becomes a concern in calculus too which I will discuss as the last example below.

There a several “definitions” of absolute value that I’ve seen over the years which I mostly do not like

The number without a sign – awful: all numbers except zero have a sign
The distance from zero on the number line – true, but not too useful especially with variables
The larger of a number and its opposite – true, but not to useful with variables

So I propose to give them an algorithm: If the number is positive, then the absolute value is the same number; if the number is negative, then its absolute value is its opposite. Of course, this is really the definition.

So I’ll soon express this in symbols

$\left| a \right|=\left\{ \begin{matrix} a & \text{if }a\ge 0 \\ -a & \text{if }a<0 \\ \end{matrix} \right.$

Now interestingly this is probably the first piecewise defined function an Algebra 1 student may see, or at least the first one that’s not artificial. So this is a good place to start talking about piecewise defined function and the importance of talking about the domain. And of course, we’ll have to take a look at the graph.

Sometimes we will have to start solving equations and inequalities with absolute values. So here is the next thing I understand but do not like and will try to avoid. Solve the equation: $\left| x \right|=3$ , Answer including work: $x=\pm 3$ . But I think a longer way around is also better:

If $x<0$ then $\left| x \right|=-x=3$ so $x=-3$ or if $x>0$ , then $\left| x \right|=x=3$ Solution: $x=3$ or $x=-3$ .

Longer? Sure. I hope that by making the students write that a few times that when they get to solving $\left| x \right|>3$ that it will be natural to say

If $x<0$ then $\left| x \right|=-x>3$ so $x<-3$ or if $x>0$ , then $\left| x \right|=x>3$ Solution: $x<-3$ or $x>3$

The last case may take a little more discussion. Solve $\left| x \right|<3$ . Starting the same way

If $x<0$ then $\left| x \right|=-x<3$ so $x>-3$ which really means $-3<x<0$

if $x>0$ , then $\left| x \right|=x<3$ which really means $0\le x<3$ . Then the union of these two sets looks like an intersection. The solution is $-3<x<3$

Quite often the equation and the two types of inequalities are treated as separate problems: with = you go with $\pm$ on the other side, with > you have a union pointing away from the origin and with < you have somehow an intersection. Who needs to remember all that when this idea works all the time?

Example: Solve $\left| 4x-10 \right|<8$

If $4x-10<0$ , then $x<\tfrac{5}{2}$ and $\left| 4x-10 \right|=-\left( 4x-10 \right)=-4x+10<8$ , so $-4x<-2$ and $x>\tfrac{1}{2}$ or more precisely $\tfrac{1}{2} If$ latex 4x-10\ge 0$, then $x\ge \tfrac{5}{2}$ and $\left| 4x-10 \right|=4x-10<8$ , so $4x<18$ and $x<\tfrac{9}{2}$ or more precisely $\tfrac{5}{2}<x<\tfrac{9}{2}$ . The union again becomes an intersection and the answer is $\tfrac{1}{2}<x<\tfrac{9}{2}$

Finally an example from calculus. On the 2008 AB exam, question 5 asked student to find the particular solution of a differential equation with the initial condition $f\left( 2 \right)=0$ . After separating the variables, integrating, including the “+C” and substituting the initial condition students arrived at this equation which they now need to solve for y:

$\displaystyle \left| y-1 \right|={{e}^{\tfrac{1}{2}-\tfrac{1}{x}}}$

How can you lose the absolute value sign? Simple, near the initial condition where y = 0, $\left( y-1 \right)<0$ so replace $\left| y-1 \right|$ with $-\left( y-1 \right)$ and then go ahead and solve for y

$\displaystyle -\left( y-1 \right)={{e}^{\tfrac{1}{2}-\tfrac{1}{x}}}$

$\displaystyle y=1-{{e}^{\tfrac{1}{2}-\tfrac{1}{x}}}$

I don’t think I’ll try this one in Algebra 1, but maybe it will come in handy when they get to calculus.

What’s a Mean Old Average Anyway?

Posted on April 29, 2013 by Lin McMullin

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.

Teaching Calculus

The pleasure lies not in discovering the truth, but in searching for it.

Category Archives: Integration

Improper Integrals and Proper Areas

Modified Tabular Integration

Variations on a Theme – 2

Why Muss with the “+C”?

Variations on a Theme by ETS

Absolute Value

What’s a Mean Old Average Anyway?

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