25
$\begingroup$

I am trying to teach my calculus students to apply integration by thinking about what they are integrating rather than just applying formulas. Calculus books are full of formulas like "to find the volume of a solid of revolution obtained by rotating region A around axis B, write down integral C". I don't want my students to just memorize these formulas, but to understand where they come from and to be able to apply integration in contexts where these formulas don't necessarily apply. What are some good examples of applications of integration that don't fit into the textbook formulas?

Here's a simple example: find the area of a circle by dividing it into concentric rings and integrating the area $2\pi x\, dx$ of each ring from 0 to $r$. Of course you can do the same finding the volume of a sphere.

$\endgroup$
  • 1
    $\begingroup$ Are you just looking for in terms of geometry or any interesting applications of integrals? $\endgroup$ – ruler501 Mar 29 '14 at 4:40
  • 4
    $\begingroup$ @JyrkiLahtonen I think, your comment (with more details) might be better off as an answer. $\endgroup$ – Markus Klein Mar 29 '14 at 7:36
  • 1
    $\begingroup$ Give the students an $I$-$t$-graph of e.g. a capacitor loading process, the formula $I=\dot Q$ and ask them for the quantum of electrical charge in the capacitor. $\endgroup$ – Toscho Mar 29 '14 at 11:25
  • 1
    $\begingroup$ I remember there was a question about how many people at a concert given the people allowed in at any given time on the AP test a while ago. $\endgroup$ – ruler501 Mar 29 '14 at 18:49
  • 1
    $\begingroup$ There's the classic question of, cut at two randomly selected points on a unit segment; what is the probability that the resulting three pieces can be assembled to form a triangle? As I recall, among the many ways to do this is one for which an integral is carefully set up (possibly corresponding to the standard graphical approach of representing the two breaks $a$ and $b$ as the point $(a,b)$ in the unit square, and then coloring in the part of the square that admits the assembly of a triangle). $\endgroup$ – Benjamin Dickman Jun 26 '17 at 23:39
19
$\begingroup$

An example that illustrates a slightly different integral is the following.

enter image description here

An observer at point $O$ is exposed to traffic noise from a long straight highway (the line $AP$). The task is to show that the total noise level at $O$ is inversely proportional to the distance $r=|OA|$ from the highway.

Use the following approximations/simplifications.

  • The traffic density is constant along the road (a kind of an average).
  • The level of "atomic noise intensity" at $O$ follows the inverse square law (this is probably a good approximation you can easily justify).
  • The total noise intensity is the sum of the infinitesimal/atomic noise intensities (i.e. the quantity measured in Watts per square meter is additive. You only switch to the logarithmic decibel scale afterwards if ever, as things are not additive on the decibel scale).

So if you place $x$-axis along the road, and $P$ is at $x$, then the noise generated along a segment of the road of length $dx$ around $P$ has (at a unit distance from $P$) intensity $K\,dx$ for some constant $K$ proportional to traffic density (= number of passing vehicles per hour). When we take the distance $|OP|=\sqrt{r^2+x^2}$ into account we get $$ dI=K\,\frac{dx}{r^2+x^2} $$ for the contribution of this atomic noise to the total observed noise level at point $O$. The total noise intensity is thus $$ I=\int dI=\int_{-\infty}^{\infty}K\,\frac{dx}{r^2+x^2}. $$

So you get to illustrate/discuss:

  • improper integrals
  • indefinite integral of a simple rational function
  • discuss what goes wrong when $r\to0+$ (the approximations fail, and you cannot be at distance zero from a continuous source of the noise)
  • point out that a similar (but also different) integral emerges when you calculate the field generated by an electric current along a long straight wire

I have tried this both as an in class and as a homework example. As a HW problem it did create some discussion as the physics majors protested at first (assuming that I was adding decibels together). I guess the non-physics majors may have been bored. You know your customers (and if somebody can add an example from chemistry I will be very grateful).

$\endgroup$
  • 1
    $\begingroup$ This looks excellent to me, thanks! (But then I do tend to like physics examples best myself.) $\endgroup$ – Mike Shulman Mar 30 '14 at 20:38
  • $\begingroup$ Have you used any real-world data to confirm your theoretical results? For example, taking noise-meter measurements along a road that is perpendicular to such a highway? Or looking up average noise levels on a map like maps.bts.dot.gov/arcgis/apps/webappviewer/… $\endgroup$ – Jasper Jun 28 '17 at 6:07
  • $\begingroup$ Afraid not, @Jasper. It may well be that soundwaves don't propagate the way I thought. I have seen at least one exercise in some calculus book use this model. There the problem was to find the most quiet spot between two highways the other with twice as much traffic, and an inverse distance law was assumed. $\endgroup$ – Jyrki Lahtonen Jun 28 '17 at 6:57
11
$\begingroup$

Chemical reaction rates and how they relate to changes in temperature.

I use this example when teaching chemical engineering students about integrals. The Boltzmann distribution describes the velocity distribution of molecules in a gas.

$$f(v) = \sqrt{\frac{2}{\pi} \left( \frac{m}{k_B \cdot T} \right)^3} \cdot v^2 \cdot \exp\left( \frac{-m v^2}{2 k_B T} \right)$$

The integral $\int_a^b f(v) dv$ is then the fraction of the gas molecules with a velocity in the range $[a, b]$. For a certain reaction to occur there must be a minimal amount of kinetic energy present (the activation energy, $E_a$), from that the lowest possible velocity of a gas molecule with this kinetic energy can be determined by

$$v_\textrm{min} = \sqrt{\frac{2 E_a}{m}}$$

The fraction of gas molecules that can participate in the reaction is then given by

$$\int_{v_\textrm{min}}^{\infty} f(v) \, dv$$

This can be used to investigate how changes in temperature alters the fraction of gas molecules that have sufficient kinetic energy to overcome the activation energy.

One application is to test the rule of thumb, which states that an increase in temperature of 10 kelvin will roughly double the reaction rate.

Below is Mathematica code for solving this

assump = Assumptions -> {k > 0, T > 0, 
   m > 0}; MaxwellBoltzmannSpeedDistribution =
 Sqrt[2/Pi (m/(k T))^3] v^2 Exp[-m v^2/(2 k T)];
FractionOfParticlesWithEnergyAbove = 
  Integrate[
   MaxwellBoltzmannSpeedDistribution, {v, minVelocity, Infinity}, 
   assump];
TemperatureDependency = 
 FractionOfParticlesWithEnergyAbove //. {minVelocity -> 
    Sqrt[2 10^-19/m], k -> 1.3806488*10^(-23), m -> 1.6726 10^-27}
res = Table[{T, TemperatureDependency} /. {T -> Tval}, {Tval, 293, 
    343, 10}];
res // TableForm

The above code gives the following output

293   1.05132*10^-10
303   2.33909*10^-10
313   4.94258*10^-10
323   9.96656*10^-10
333   1.926*10^-9
343   3.58024*10^-9
$\endgroup$
  • 1
    $\begingroup$ Amazing! As a math educator who has never seen this equation before, what would be some good values of those constants m and k_B for a real-world example? $\endgroup$ – Chris Cunningham Apr 2 '14 at 15:05
  • 1
    $\begingroup$ The missing values are: $m = 1.67\cdot 10^{-27} kg$ (atom mass of hydrogen) and $k_B = 1.38\cdot 10^{-23} \cdot m^2 \cdot kg \cdot s^{-2} \cdot K^{-1}$ (Boltzmanns constant). $\endgroup$ – midtiby Apr 3 '14 at 6:04
10
$\begingroup$

Physics provides with a lot of examples, whose main disadvantage is that they might be considered too much physics by some, i.e., require too much thinking or knowledge about something other than mathematics. (Also engineers and physicists might rather consider them the usual applications of integration.) I kept the examples simple, and as a result they might still be solved without understanding by guessing what to integrate, but they can easily made more difficult to avoid this.

Lengthening of an elastic rope under its own weight

Q: An elastic rope of length $l$ has a weight per length of $ρ$ (e.g., $ρ=1\,\text{kg}/\text{m}$) – both if unstretched. If a total of mass $m$ is pulling on a piece of rope, it extends to $f(m)$ of its unburdened length (typically $f(m)=1+Em$). To what length does the rope extend under its own weight?

A: A infinitesimally short piece of rope of length $\mathrm{d}x$ located at point $x$ (measured from the bottom of the rope in the unstretched case) is burdened with a weight of $ρx$ and thus its burdened length is $f(ρx)\mathrm{d}x$. The burdened length of the entire rope thus is the integral $\int_0^l f(ρx) \mathrm{d}x$.

A difficulty lies in considering the rope unstretched for most purposes (but this can be given as a hint). To “spice up” this problem, one could attach weights to the ropes (at the end or other positions) or have the rope vary in thickness (i.e., make $ρ$ as well as $f$ depend on $l$).

Uniformly accelerated object

Q: An apple starts falling from a tree at $t=0$. Its speed $v(t)$ at time $t$ is $gt$ (with $g$ being the local gravity of Earth). How far has it fallen after time $T$?

A: The distance covered by the apple during an infinitesimally short time $\mathrm{d}t$ is $v(t)\mathrm{d}t$ and thus the total distance covered up to time $T$ is $\int_0^T v(t)\mathrm{d}t = \int_0^T at\mathrm{d}t = \tfrac{1}{2}aT^2$.

To “spice up” the problem, one could consider more complicated acceleration functions (e.g., considering air resistance), initial speeds and so on.

$\endgroup$
  • 1
    $\begingroup$ I think most of my students have memorized that distance is the integral of velocity, that being a common example in the book, so I'm not sure that example will be what I want. $\endgroup$ – Mike Shulman Mar 30 '14 at 14:27
  • 1
    $\begingroup$ The first one looks good, but I'm confused about what it means to talk about a rope burdened with a mass m but not with its own weight, to make sense of the function f. $\endgroup$ – Mike Shulman Mar 30 '14 at 14:32
  • 1
    $\begingroup$ @MikeShulman: I am not certain whether I understand your problem. The usual way to burden a rope is by attaching a piece of weight to it. Also, using forces may be clearer, but adds another physical concept, which is why I did not do that. Finally, I rephrased, what I thought to be the phrase in question. $\endgroup$ – Wrzlprmft Mar 30 '14 at 18:37
  • 1
    $\begingroup$ The rephrasing helps a bit. Maybe my problem is that the very phrase "If a total of mass m is pulling on a piece of rope, it extends to f(m) of its unburdened length" is only literally true when the "piece of rope" is infinitesimal. Right? $\endgroup$ – Mike Shulman Mar 30 '14 at 20:37
  • 1
    $\begingroup$ @MikeShulman: Yes. Unless the rope is not oriented vertically, but horizontally (and the force of the mass pulling is mediated by a weigth-less rope and redirected via a wheel, e.g.). $\endgroup$ – Wrzlprmft Mar 30 '14 at 20:42
6
$\begingroup$

Let them guess the mass of a mountain by showing them some pictures of it. They then can then model the mountain in different ways:

  • cone
  • rotational body
  • irregular pyramid

Additionally they can model the tip of the mountain in different ways

  • flat
  • flat with inverse semi-ball for the caldera
  • flat with inverse cone for the caldera

You can also give them elevation data and let them model the mountain as a stack of layers.

$\endgroup$
3
$\begingroup$

Newton's law of cooling: The rate at which the cup of tea cools is proportional to it's temperature difference to the environment.

Compute the areas/volumes of curved figures/solids. I.e., the area under a parabola, the area of a circle, the volume of a cone, a cylinder, a sphere. Can even introduce/prove Cavalieri's principle.

$\endgroup$
  • 1
    $\begingroup$ Doesn’t Newton’s law of cooling (or any cooling) lead to a differential equation? $\endgroup$ – Wrzlprmft Mar 29 '14 at 23:36
  • 1
    $\begingroup$ @Wrzlprmft, yes, but one easily solved (it is just $y'(t) = -a(y - y_\infty)$, with $y(0) = y_0$) $\endgroup$ – vonbrand Mar 29 '14 at 23:59
  • 1
    $\begingroup$ Your examples of areas and volumes are all easily computed with the standard textbook formulas. $\endgroup$ – Mike Shulman Mar 30 '14 at 14:17
  • 1
    $\begingroup$ @MikeShulman, I understood the question was on ways to show how integration is used, in a setting different from "compute $\int_a^b f(x) \, \mathrm{d} x$, where $a$, $b$, $f$ are given." To get the point of what it is used for across, it is best if the computation of the resulting integral is simple. You don't want a series of complex maneuvers to obscure that. $\endgroup$ – vonbrand Mar 30 '14 at 23:53
  • 2
    $\begingroup$ It has nothing to do with simplicity or complexity of the resulting integral. I don't care whether the resulting integral is even doable by elementary means or not; mainly I want them to be able to set up integrals correctly in situations that don't fit neatly into formaulas that they've memorized. $\endgroup$ – Mike Shulman Mar 31 '14 at 21:00
3
$\begingroup$

This example is actually from a textbook I am using.

You are in a submarine and want to know how long you have gone in some time (there are no km marks to read under water!). You have a speedometer, so you can just write down the speedometer writings, and integrate them numerically with respect to time.

(You can do that in a car too, but ther there is no point in doing it).

$\endgroup$
  • 1
    $\begingroup$ This sort of example is good, but for me it belongs to a different part of calculus, when we are learning the meaning of integrals and how they relate to Riemann sums. $\endgroup$ – Mike Shulman Mar 31 '14 at 21:02
2
$\begingroup$

Integration is everywhere there is superposition. One very nice class of examples comes from electrostatics. In particular, if we set the zero for voltage at $\infty$ then the tiny voltage $dV$ at $\vec{r}$ due to a tiny charge $dQ$ at $\vec{r}'$ is given by the Coulomb Potential $$ dV = \frac{dQ}{4\pi \epsilon_o \| \vec{r} - \vec{r}' \|}$$ If we're given a bunch of $dQ$'s over some reasonably nice geometric object then we can use integration to find the net voltage for such a collection.

Example: ring of charge if we have total charge $Q$ around ring of radius $R$ in the $xy$-plane (aka $z=0$ plane) then at the origin it's not bad to calculate $V$. We have $\vec{r} = (0,0,z)$ and using $s = \sqrt{x^2+y^2}$ and $\theta$ for the usual standard angle, the ring of charge is at $s=R$ and $z=0$ and $$ dV =\frac{dQ}{4\pi \epsilon_o\| \vec{r} - \vec{r}'\|} = \frac{dQ}{4\pi \epsilon_o \sqrt{R^2+z^2}} $$ since every $dQ$ is distance $\sqrt{R^2+z^2}$ from $(0,0,z)$. Thus adding up the bits of voltage due to each $dQ$ distributed over the loop which I have labeled $C$, $$V = \int dV = \int_C \frac{dQ}{4\pi \epsilon_o \sqrt{R^2+z^2}} = \frac{1}{4\pi \epsilon_o \sqrt{R^2+z^2}} \int_C dQ = \frac{Q}{4\pi \epsilon_o \sqrt{R^2+z^2}}$$ Then, by symmetry the electric field must point along the $z$-axis and we can derive its magnitude from the voltage just calculated: $E_z = -\frac{dV}{dz}$; $$ E_z = -\frac{d}{dz}\frac{Q}{4\pi \epsilon_o \sqrt{R^2+z^2}} = \frac{zQ}{4\pi \epsilon_o (R^2+z^2)^{3/2}}$$ As $z \rightarrow \infty$ we obtain $E_z \sim \frac{Q}{4\pi \epsilon_o z^2}$. Intuitively this makes sense, from very far away a ring of charge with charge $Q$ is the same as a point charge $Q$ at the origin. We find the Coulomb field in the appropriate limit.

Example: plane of charge with uniform density $\sigma = \frac{dQ}{dA}$. Building off what we just did, if we divide the plane into rings of radius $s$ with thickness $ds$ then each such ring has charge $dQ = \sigma dA = 2\pi s \sigma ds$ hence the voltage at $(0,0,z)$ due to the ring of charge is (using my first example with $R=s$ and $Q$ replaced with appropriate $dQ$): $$ dV = \frac{2\pi s \sigma ds}{4\pi \epsilon_o \sqrt{s^2+z^2}}$$ The voltage due to a disk of radius $s=R$ is thus, assuming $z>0$, \begin{align} V &= \int_{0}^R \frac{\pi s \sigma ds}{2\pi \epsilon_o \sqrt{s^2+z^2}} \\ &= \frac{ \sigma }{2 \epsilon_o}\int_{0}^R \frac{ s ds}{\sqrt{s^2+z^2}} \\ &= \frac{ \sigma }{2 \epsilon_o}\sqrt{s^2+z^2} \bigg{|}_0^R \\ &= \frac{ \sigma }{2 \epsilon_o} \left( \sqrt{R^2+z^2} - z \right) \end{align} Thus, again by symmetry the electric field must be of the form $\vec{E} = \langle 0,0, E_z \rangle$ at $(0,0,z)$ and we calculate: for $z >0$, $$ E_z = -\frac{dV}{dz} = -\frac{ \sigma }{2 \epsilon_o}\left( \frac{z}{(R^2+z^2)^{3/2}} - 1\right) $$ As $R \rightarrow \infty$ for fixed $z>0$ the expression above simply yields $E_z \sim \frac{ \sigma }{2 \epsilon_o}$. Now, we usually derive this result using a little cylinder and Gauss' Law, but, it's nice to see it can be derived directly by potential methods. Also, we can stop before taking $R \rightarrow \infty$ and we have formulas for the potential due to a disk of charge.

Typically we teach how to find electric fields by direct superposition of the Coulomb fields due to the point charges in the source distribution. However, that is harder than what I show here in that the integration is over vectors. In contrast, the potential is a scalar function. In both calculation schemes, if we diverge from standard geometries then the actual integrals get really tough. For example, if we get away from the axis in my example then I think Bessel functions are needed. Much of the study of electrostatics is centered around how to set up these sort of integrals.

$\endgroup$
  • 1
    $\begingroup$ These are okay examples, but they really fit much better into a multivariable calculus course, since you're really doing double and triple integrals. $\endgroup$ – Mike Shulman Jun 26 '17 at 6:51
  • 1
    $\begingroup$ But, we cover area and volume integrals in calculus I (or maybe II depending on the curriculum). Those take $dA$ or $dV$ and relate them a single integral depending on the coordinate. Here I do the same with $dV$. Volumes by method of washers or shells really fit your question, but, only as much as you free yourself from a blackbox calculus book. The same could be said about the chain rule in some regard, you've probably noticed calculus texts with 10 different chain rules, replete with their boxes for each function. Instead, we ought to teach one chain rule and illustrate those 10 as cases.. $\endgroup$ – James S. Cook Jun 26 '17 at 12:17
  • 1
    $\begingroup$ Ha, now that I look at my comment, maybe I've made the case for using $\varphi$ instead of $V$ for voltage. $\endgroup$ – James S. Cook Jun 26 '17 at 12:18
  • $\begingroup$ Seems pretty advanced. $\endgroup$ – guest Jun 28 '17 at 2:12
1
$\begingroup$

Answer:

You can do something graphical where they use "box counting" to determine the integral. For instance an electrical current that has a set of triangles on top of a base current. (for every t unit increment, I goes from 1 to 2 in a line and then drops immediately to 1. [the average current is 1.5 per t.] Obviously they won't know a formula for that weird sawtooth graph.

If you want something "different" to the normal applications, do some box counting with a peak oil bell curve shape. See the 1955 or maybe 1952 paper by Hubbard for a nice discussion. (And FWIW, I'm not pushing any peak oil slant. If you want to show failure of peak oil predictions give them the US natural gas prediction and then production to date. But the nice thing about the Hubbard paper is the explication of the math. Even someone who does not know calculus can follow it super easy. And just the boxes are super well drawn...I know it sounds silly, but it makes it seem simple.) The one thing to watch out for with this sort of answer though is you need to get an estimate versus the triangle graph where it can be exact. Maybe tell the kids "to the closest TCF" (or whatever the unit is). Also knowing the range can help your grading later! Also, this is a little more sophisticated as you ask them to estimate, show a Gaussian and they think a little more about needing a formula. Than the sawtooth, where it is a quicker insight to count boxes.

Comment:

I do think having the kids learn formulas, bag of tricks, etc. is also useful. I would not replace the "toolkit" approach with an emphasis on integration fundamental. Instead make the sort of thing that you want additive and enriching. But the bag of tricks and knowing how to find the hammer for the nail and screwdriver for the screw and even to combine tools when needed is also powerful. And important for engineers and physicists.

$\endgroup$
  • 1
    $\begingroup$ Once you have a graph and are looking for the area under it, then that's what I call a textbook formula. Whether you compute the integral by counting boxes or antidifferentiating is irrelevant for the question I'm asking. $\endgroup$ – Mike Shulman Jun 26 '17 at 6:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.