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# 4.5: Applications of Definite Integrals

Difficulty Level: At Grade Created by: CK-12

## Area Between Two Curves

CONTENT

This content follows naturally from the definition of an integral. The factor that determines a student’s success will be his/her ability to accurately draw the curves that are being evaluated. Therefore, it is a good idea to include some review of plotting.

PROCESS

Teaching this lesson should begin with a careful review of plotting. From this, students should be asked to shade particular regions of plots. For example, students could be asked to draw a plot in which the region bounded by \begin{align*}f(x)= x \end{align*}and \begin{align*}g(x)=x^2\end{align*} is shaded:

With practice, students will begin to get the hang of finding regions whose area they are calculating.

The next step is to recognize that this area is given by the area beneath the higher function minus the area beneath the lower function. So in the example above students should be able to add to their picture the equation:

PRODUCTS

Understanding and becoming good at this is really just a matter of practice. It is best to start with diagrams that are provided, and then later to give word problems where the students are asked to make their own diagrams.

## Volumes

CONTENT

There is a great deal of content covered in this lesson and it can be differentiated in a number of ways. The goal with every problem is to find some volume in three-dimensional space, and it should be made clear that there are always a variety of ways to carry this out. Usually there is one way that is simpler than others; however, students should be encouraged to try whatever occurs to them and see if it works.

Now, the biggest hurdle to using integration for finding volumes is drawing an accurate diagram or being able to visualize the exact volume that is of interest. Some students are naturally very good at visualizing three-dimensional shapes and manipulating them mentally. These students can usually rotate the objects in their minds to tell whether to use the disc method or shell method and which function should go where in the integral. However, for the students without this natural ability it is important to not frustrate them. For this reason the content of drawing accurate figures should be stressed and students who possess a unique ability for this should be asked to explain their reasoning to the class and present the problems.

PROCESS

To teach students how to find volumes it is best to carry out a detailed problem and then have students repeat. Have students put away their books and pencils and go over a volume of rotation problem in detail. Make sure everyone understands each step, and then have them do the problem on their own after erasing the board.

If there are a decent number of students that are very good at visualizing the volumes, it is a great idea to form groups where each group has at least one of these students. The groups can work a number of problems and the teacher should focus on making sure that everyone is participating and learning. Students who are ahead of the curve should be told that they are not done with a problem until everyone in the group understands it as well as they do.

PRODUCTS

Naturally, students can be tested on this material by having them calculate the volume of a number of different shapes. It is a good idea to start with problems where the shape is provided and then move into problems which are more abstract.

A great problem to start with is to have students prove that the volume of a sphere is \begin{align*}V = \frac{4}{3} \pi R ^ 3\end{align*} . Tell them to think about taking the curve \begin{align*}f(x) = \sqrt { R ^ 2 - x ^ 2}\end{align*} for \begin{align*}-R \le x \le R \end{align*} :

and rotating it around the x-axis:

To calculate the volume we can use the disc method as \begin{align*}x\end{align*} runs from \begin{align*}- R\end{align*} to \begin{align*}+R\end{align*} and the radius of each disc is \begin{align*}\sqrt {R ^ 2 - x ^ 2}\end{align*} . Then the area of the discs’ faces are \begin{align*}\pi ( R ^ 2 - x^ 2 )\end{align*} and their thickness’ are \begin{align*}dx\end{align*} so they each have volume \begin{align*}dV = \pi ( R ^ 2 - x ^ 2)\end{align*} . Putting this in we find the total volume to be:

\begin{align*}V = \int\limits_{-R}^{R} \pi ( R ^ 2 - x ^ 2 ) dx = \left [ \pi R ^ 2 x - \pi \frac{x ^ 3}{3} \right ] _ {x = -R} ^ {x = + R} = 2 \pi R ^ 3 - 2 \pi \frac{R ^3}{3} = \frac{4}{3} \pi R ^ 3 \end{align*}

Drawing careful diagrams of the discs that are being used will be very helpful:

## The Length of a Plane Curve

CONTENT

The formula for arclength:

\begin{align*} L = \int\limits_{a}^{b}dx \sqrt { 1 + \left ( \frac{dy}{dx} \right )^ 2 }\end{align*}

will seem awkward to many students at first, but it can be explained on geometric grounds. But before doing this, it is a good idea to connect the concepts of integral and summation once more.

To this end, consider a general integral of any function \begin{align*}f(x)\end{align*}:

\begin{align*}\int\limits_{a}^{b} f(x) dx\end{align*}

This can certainly be thought of as the “area under the curve \begin{align*}y=f(x)\end{align*} between \begin{align*}x=a\end{align*} and \begin{align*}x=b\end{align*}.” Or equivalently, we can understand this as the sum of the values of \begin{align*}f(x)\end{align*} at each and every point \begin{align*}x\end{align*} between \begin{align*}a\end{align*} and \begin{align*}b\end{align*}. That is, qualitatively we have:

\begin{align*}\int\limits_{a}^{b}f(x)dx \approx \sum_{x \in [a,b]}f(x)\end{align*}

Since the set \begin{align*}[a,b]\end{align*} is an interval with an uncountably infinite number of elements the summation on the right here cannot be expanded in a way that each “term” of the sum could be written individually. On the other hand, we could divide up the interval \begin{align*}[a,b]\end{align*} into smaller and smaller pieces and write these as the terms. Then the limit of any given “piece” or subinterval would be a single point and eventually we’d be summing the value of \begin{align*}f\end{align*} at every point.

Now back to the arc-length of a curve. Intuitively, the total length of any curve is equal to the sum of tiny “pieces” along the curve. So if we zoom way in on one of these pieces it might look like the following:

The height is equal to \begin{align*}\frac{dy}{dx} dx\end{align*} since the rise over run (or slope) is given by the derivative \begin{align*}\frac{dy}{dx}\end{align*} and if we multiply this by the run \begin{align*}dx\end{align*} we obtain the rise. Using the Pythagorean Theorem we can see that the length \begin{align*}dl\end{align*} of a piece along the curve is then equal to the hypotenuse of the right triangle:

\begin{align*}dl = \sqrt{dx^2 + \left ( \frac{dy}{dx} dx \right )^2 } = dx \sqrt{1 + \left (\frac{dy}{dx} \right )^2 }\end{align*}

So summing up these tiny segments means taking the integral of this and we are left with the given formula for arc-length.

PROCESS/PRODUCTS

In teaching this material it is enough to have students try some problems. A fun project may be to have them derive the formula for the circumference of a half-circle. Here we have that \begin{align*}y = \sqrt{R^2-x^2}\end{align*} and so

\begin{align*}dl = dx \sqrt{1+ \left ( \frac{dy}{dx} \right )^2} = \frac{R \ dx}{\sqrt{R^2-x^2} }\end{align*}

Then students can practice integrating:

\begin{align*}\int \limits_{-R}^R \frac{R \ dx}{\sqrt{R^2-x^2}} = \pi R\end{align*}

which can be done with a trig substitution or simply by recognizing that \begin{align*}\frac{1}{\sqrt{1 - x^2}} = \frac{d}{dx} \text{Arc Sin}(x)\end{align*}.

## Area of Surface of Revolution

CONTENT

The formulas used here relate closely to the formula for the circumference of a circle. In teaching this material the intuitive nature of the concepts should not be disregarded. The area of a surface of revolution is given essentially by summing up the circumferences of all the circles that make it up. Think about a high stack of thin tires. The surface that they form can be thought of as a combination of all of their circumferences, and so we expect to just integrate over these values in order to find its area.

PROCESS/PRODUCTS

As with the volume calculations, the biggest hurdle to overcome in mastering this material is being able to draw a very accurate picture. Students should be begin with practice problems where they are told this is now an art class. They should then be given three positive functions like:

1. \begin{align*}f(x)=4-x & \ 0 \le x \le 4 \end{align*}

2. \begin{align*}f(x)=x^4+5 & \ -1 \le x \le 1 \end{align*}

3.\begin{align*}f(n) = \begin{cases} x^2, & 0 \le x \le 4 \\ 2-x^2, & 1 \le x \le \sqrt{2} \end{cases}\end{align*}

and asked to draw nice pictures of what the surfaces will look like when these are rotated around the \begin{align*}x\end{align*}-axis.

Once they have completed this, students can be paired and each partner can be given a curve to be rotated and they should guess which have the larger area. Then each should calculate the area for his/her curve and they should compare to see if their guess was correct.

## Applications from Physics, Engineering and Statistics

CONTENT

The content herein is really quite simple if it is introduced in a fashion that follows naturally from earlier concepts. The concept of work can be described simply and fundamentally as the area under a particular curve. In this case, the function is the force on an object and we think of this as depending upon the location of that object along the \begin{align*}x\end{align*}-axis. So the force \begin{align*}F\end{align*} is a function of \begin{align*}x\end{align*}:

Then the work as the object moves from \begin{align*}x=a\end{align*} to \begin{align*}x=b\end{align*} is simply the shaded area above, or:

\begin{align*}W_{ab}=\int\limits_a^b F(x)dx\end{align*}

The concept of pressure is a little more slippery (no pun intended). In the simplest terms, its description pertains to a force that is exerted over a surface. Then the pressure is just the amount of force exerted upon each unit area. Since this is not the same for every piece of area in general, the pressure is most generally a function of position along the surface. Therefore, we must sum up the pressure over the entire \begin{align*}2\end{align*}-dimensional surface in order to obtain the force in general:

\begin{align*}F=\int\limits dx \int\limits dy P(x,y)\end{align*}

When the pressure is a constant along one of the two directions (like in the example) then this becomes a single integral.

Probability densities will also seem very natural if introduced properly. For example, it’s a good idea to consider the following examples:

1. Suppose a fair \begin{align*}6\end{align*}-sided die is rolled.

a. What is the probability of obtaining any given number in particular?

b. If we were to plot the probability as a function of the possible outcome, what would it look like?

c. Now suppose we have a dartboard and somebody who throws a dart randomly. What would the plot look like for the probabilities here:

d. Now suppose that the person is aiming specifically for the number \begin{align*}10\end{align*}. Then the probability might look like:

where the values for \begin{align*}6\end{align*} and \begin{align*}15\end{align*} are larger since they are next to \begin{align*}10\end{align*} on a dartboard.

With this kind of example in mind, it may be easier to then introduce a probability density as a plot like this where the points along the \begin{align*}x\end{align*}-axis are spaced even closer together. If you like, we could plot the probability of hitting the dartboard at a particular angle as measured clockwise from \begin{align*}12\end{align*} o-clock. Then the distribution would be a continuous plot between \begin{align*}0\end{align*} and \begin{align*}360\end{align*} or \begin{align*}0\end{align*} and \begin{align*}2\pi\end{align*}.

PROCESS/PRODUCTS

Problems given in purely physical terms here are likely to confuse students that have not learned physics in detail. Therefore it is preferable to teach this material in fairly abstract terms. For example, suppose a force of \begin{align*}10\;\mathrm{Newtons}\end{align*} acts on an object as it moves for \begin{align*}20\;\mathrm{feet}\end{align*}, how much work is done? Or, suppose a spring force of \begin{align*}F(x) = -k \ x\end{align*} acts on an object as it moves from \begin{align*}x=0\end{align*} to \begin{align*}x=A\end{align*}. How much work is done?

To teach probability density functions it is a good idea to have students imagine what they may look like for different situations. For example, have students consider a shark in the ocean off of a popular swimming beach. For each distance from the shore \begin{align*}x\end{align*}, what is the probability \begin{align*}f\end{align*} that this will be the shark’s distance of closest approach? This should be a function that begins at zero for \begin{align*}x=0\end{align*} and rises for a little before sinking back to zero. Once students have thought through an example like this, they should model their plot with some function. In this case, for example, a good function might be something like \begin{align*}f(x)=e^{-(x-100)^2}\end{align*}. The first step in making this a good function is normalizing so that

\begin{align*}\int \limits_{0}^{\infty} f(x)dx = 1\end{align*}

In this case we would have to modify the function so that \begin{align*}f(x)\cong \frac{1}{1.77} e^{-(x-100)^2}\end{align*}.

Then students could be asked specific questions about their function, like in this case they could be asked: 1. What is the probability that a shark will venture inside of \begin{align*}99\;\mathrm{meters}\end{align*}? a. ANSWER:

\begin{align*}\int \limits_0^{99}f(x)dx \cong 7.9 \%\end{align*}

2. What is the probability that a shark will venture inside of \begin{align*}100\;\mathrm{meters}\end{align*}?

\begin{align*}\int \limits_0^{100}f(x)dx \cong 50 \%\end{align*}

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