Category Archives: Integration Applications

Most Triangles Are Obtuse!

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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 – A – B.

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.

Challenge: Try writing a calculator or computer program that picks the measures of the angles of a triangle as described above. Repeat this many times while the program counts the number of acute triangles and the total number of triangles and finds their ratio. Does it come close to 19.3%?
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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!

Average Value of a Function

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Today I want to consider a way of developing the expression for finding the average value of a function, f (x), on an interval [a, b].

Ask students how to find the average of a bunch of numbers and they will say, “add them up and divide by the number of numbers.” Then ask if they can average an infinite number of numbers. Most will say no since you cannot divide by infinity.

Well, what if the numbers were all the same?

Such as all the y-values of f (x) = 2 between x =1 and x = 5. Isn’t the average 2? So, apparently, you can sometimes average an infinite number of numbers.

Next suggest something like f (x) = x between x = 0 and x = 3. Since half the values are obviously above 1.5 and half below, can’t 1.5 be their average? Sketch this situation and draw the segment at y = 1.5 to help them see this.

Suggest another situation, say f (x) = 2 + sin(x) on the interval \left[ 0,2\pi \right] and some others. The average appears to be 2, again since half the values are above and half below 2.

When you draw the line that is the apparent average, lead the students to see that the rectangle formed by this line, the x-axis and the ends of the interval has the same area as between the function and the x-axis.

Continue with more difficult examples until someone hits on the idea of finding the “area” with an integral and then dividing the result by the width of the interval to find the height of the rectangle that is also the average value:

\displaystyle \overline{y}=\frac{\text{''area''}}{\text{length of interval}}=\frac{\int_{a}^{b}{f\left( x \right)dx}}{b-a}

Click here for an activity that you can use to develop this idea from scratch.

The next post: A fun application of average value – Most Triangles Are Obtuse.

Why You Never Need Cylindrical Shells

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Don’t get me wrong; finding volumes of solids of rotation by the method of cylindrical shells is a great method. It’s just that you can always work around it; you don’t ever need to use it. The work-around is often longer and involves more work, but it is interesting mathematically. So here’s an example of how to do it.

The region bounded by the graph of y={{x}^{3}}+x+1, and the lines y = 1 and x = 2, is revolved around the y-axis. What is the volume of the resulting solid?

Supposedly this volume must be found by the shell method. Using the washer method the volume is set up as the integral of the area of the outside circle with constant radius of 2, minus the area of the inside circle of radius x, times the “thickness” of a slice. This “thickness” is in the y-direction and so is dy. The dy also determines the limits of integration.

\displaystyle V=\int_{1}^{11}{\left( \pi {{2}^{2}}-\pi {{x}^{2}} \right)\,dy}

Since the integral contains a dy the usual way is to change the x to a function of y, which in this case involves solving a cubic polynomial. In this case that is very difficult to find x as a function of y and with a different function that may even be impossible. But do we really have to do that? In fact, what is required is to have only one variable and the variable may be x! So we find dy in terms of x and dx and substitute into the expression above. We also change the limits of integration to the corresponding x-values.

dy=\left( 3{{x}^{2}}+1 \right)dx
\displaystyle V=\int_{0}^{2}{\left( \pi {{2}^{2}}-\pi {{x}^{2}} \right)\,\left( 3{{x}^{2}}+1 \right)dx}
\displaystyle V=\pi \int_{0}^{2}{\left( 4+11{{x}^{2}}-3{{x}^{4}}\, \right)dx}

This integral is easy to evaluate and will give the same value, \tfrac{272}{15}\pi, as the shell integral.

To use this idea, the function must either be one-to-one on the interval or the solid must be broken into sections that are one-to-one. This may make the problem longer. Most problems that you want to do by shells are easier by shells. The point is that washers (or disks) may always be used; not that the washer approach is the easiest way.

Volumes of Revolution

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Applications of Integration 3 – Volumes of Rotation

In our last post we discussed volumes of figures with regular cross sections. Many common figures can be analyzed as some region rotated around a line, possibly one of its edges. For example, the region between the x-axis and the segment with equation y=\frac{r}{h}x from x = 0 to
x = h when rotated around the x-axis generates a cone with height h and a base with a radius of r. This has regular cross-sections which are circles for any value of x between 0 and h. Their areas are. So based on the idea in my previous post the volume is the integral of the cross-section area times the thickness:

\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\pi {{\left( \tfrac{r}{h}{{x}_{i}} \right)}^{2}}\Delta {{x}_{i}}}=\int_{0}^{h}{\pi {{\left( \tfrac{r}{h}x \right)}^{2}}}dx=\tfrac{1}{3}\pi {{r}^{2}}h

Since circular cross-sections are very common (at least in calculus books) this is often treated as a separate topic with its own ideas and formulas. It really is not. It’s the basic idea applied to figures with circular sections; the cross-section area as a function of x is \pi {{\left( r(x) \right)}^{2}}; the thickness is \Delta x

\displaystyle V=\int_{a}^{b}{\pi {{\left( r\left( x \right) \right)}^{2}}dx}.

If the figure has a hole through it one subtracts the inside volume (radius = r\left( x \right) ) from the outside volume (radius = R\left( x \right)). This may be done with two integrals or combined into one.

\displaystyle \int_{a}^{b}{\pi {{\left( R\left( x \right) \right)}^{2}}dx}-\int_{a}^{b}{\pi {{\left( r\left( x \right) \right)}^{2}}}dx=\int_{a}^{b}{\pi {{\left( R\left( x \right) \right)}^{2}}-\pi {{\left( r\left( x \right) \right)}^{2}}dx}

Looking at the last integral we see that this is the same as treating the cross-section as an annulus (aka a “washer”). The \pi is often factored out in front of the integral sign to make things look neater; I suggest you leave it where it is to remind students that they are subtracting the areas of two circles.

To help students see what the figures look like you can do several things. You can cut the region out of light cardboard and tape it to the end of a pencil. Then rotate the pencil quickly to see the volume. Another way is to use a good graphing program, such as Winplot, which will let you see a three-dimensional figure being formed. One of my favorites is to use paper wedding bells that start flat and then open into a three-dimensional figure.

Volume of Solids with Regular Cross-sections

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Applications of Integration 2 – Solids with Regular Cross-sections

In Vino Veritas. And not only truth, but the start of an important application of the calculus. After Johannes Kepler’s (1571–1630) first wife died he decided to marry again. Naturally, he bought wine for the festivities. There was some disagreement with the wine merchant concerning how he calculated the volume of the barrel of wine. This led Kepler, a mathematician, to study the problem of finding the volume of barrels. The barrels, then as now, had circular cross sections of different diameters. His idea was to consider the wine as a stack of thin cylinders (with height of as we would say today). He then calculated the volume of each cylinder and added them together to find the volume of the barrel. There is a nice animation of the idea and more information on Kepler here.

The basic problem is this: you have a solid object whose volume you need to compute. If you can slice the object perpendicular to the axis in such a way that you get regular cross sections (i.e. similar figures) whose area you can compute, then by multiplying the area times the thickness and adding the results, you can have the volume. This kind of addition is done nicely by integration, since the volume of the slices form a Riemann sum.

You need to set up a coordinate system so that you can slice the figure into thin slices (whose width is ) and express the cross-section’s dimensions as a function of x and then the area as A(x). The result always looks like this where A(x) is the area of the cross-section:

\displaystyle \underset{n\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{A\left( {{x}_{i}} \right)\Delta {{x}_{i}}}=\int_{a}^{b}{A\left( x \right)dx}

Think of this as the integral of the cross-section area times the thickness

Since you are probably not allowed to have wine in your calculus class, you may want to illustrate the idea some way other than using a wine barrel. Think of a loaf of sliced bread, or a deck of cards. Find some object that the students can measure and calculate the volume, such as a tree trunk, a statue, or a Coke-Cola bottle.

Another interesting exercise is to develop (prove) the formulas for volumes that students have known and used without proof since grade school: a cone, a pyramid, a sphere, the frustum of a cone.

Area Between Curves

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Applications of Integration 1 – Area Between Curves

The first thing to keep in mind when teaching the applications of integration is Riemann sums. The thing is that when you set up and solve the majority of application problems you cannot help but develop a formula for the situation. Students think formulas are handy and go about memorizing them badly. By which I mean they forget or never learn where the various things in the formula come from. A slight change in the situation and they are lost. Behind every definite integral stands a Riemann sum; each application should be approached through its Riemann sum. If students understand that, they will make fewer mistakes with the “formula.”

As I suggested in a previous post, I believe all area problems should be treated as the area between two curves. If you build the Riemann sum rectangle between the graph and the axis and calculate its vertical side as the upper function minus the lower (or right minus left if you use horizontal rectangles) you will always get the correct integral for the area. If the upper curve is the x-axis, then the vertical sides of the Riemann sums are (0 – f(x)) and you get a positive area as you should.

If both your curves are above the x-axis, then it is tempting to explain what you are doing as subtracting the area between the lower curve and the x-axis from the area between the upper curve and the x-axis. And this is not wrong. It just does not work very smoothly when one, both or parts of either are below the x-axis. Then you go into all kinds of contortions explaining things in terms of positive and negative areas.  Why go there?

Regardless of where the two curves are relative to the x-axis, the vertical distance between them is the upper value minus the lower, f(x) – g(x). It does not matter if one or both functions are negative on all or part of the interval, the difference is positive and the area between them is

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

Furthermore, this Riemann sum rectangle is used in other applications. It is the one rotated in both the washer and shell method of finding volumes. So in area and all applications be sure your students don’t just memorize formulas, but keep their eyes on the rectangle and the Riemann sum.

Finally, if the graphs cross in the interval so that the upper and lower curves change place, you may (1) either break the problem into several pieces so that your integrands are always of the form upper minus lower, or (2) if you intend to do the computation using technology, set up the integral as

\displaystyle \int_{a}^{b}{\left| f\left( x \right)-g\left( x \right) \right|dx}.

Under is a Long Way Down

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The development of the ideas and concepts related to definite integrals almost always begins with finding the area of a region between a graph in the first quadrant and the x-axis between two vertical lines. Everyone, including me in the past, refers to this as “finding the area under the curve.”

Under is a long way down. And while everyone understands what this means, I suggest that a better phrasing is “finding the area between the curve and the x-axis.” Here is why:

  • That is what you are doing.
  • You will soon be finding the area between the curve and the x-axis where the curve is below the x-axis. This often leads to something you may be tempted to call “negative area” and of course there is no such thing as a negative area, regardless of what you may find in some textbooks.

As with so many integration problems, the results is often a formula that obscures what is really going on – the Riemann sum whose value the integral gives. The first such formula is that the area is given by \int_{a}^{b}{f\left( x \right)dx}. This is correct only if  f (x) > 0. There is a natural confusion for beginning students between the facts that if f (x) < 0 the integral comes out negative, but the area is positive.

For all the applications of integration always emphasize the Riemann sum – not just the final formula. In the area problem with f (x) > 0 the integrand is the vertical length of the rectangles that make up the sum and this is the upper function’s value minus the lower function’s value, with the lower being the x-axis, y = 0. Then when f (x) < 0 the upper minus the lower is 0 – f (x) and the area is given by \int_{a}^{b}{0-f\left( x \right)dx}=-\int_{a}^{b}{f\left( x \right)dx} which is positive as it should be. And students will immediately see that \int_{a}^{b}{f\left( x \right)dx} is not automatically the area.

To help students see this you could start (very first problem) by helping them to find the area of the region between f (x) > 0 and the line y = 1 so they have to deal with the lower curve. Then consider another problem using the x-axis.

There is a fair amount of ground to cover between the first area between the curve and the x-axis problems with f (x) > 0 and other area problems. Teaching students how to set up those first Riemann sums, what a Riemann sum is, the definition of the definite integral and even the Fundamental Theorem of Calculus may all come between the first problem and when this distinction becomes important. Starting with the right words, “area between the graph and the x-axis”, will help in the long run.