]> Applications

### Chapter 2: Integration; Section 2: Definite Integrals; page 2

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# Applications

## Enclosed area

We have so far learned how to calculate the area under a curve by using the definite integral. We'll now see how the definite integral can be applied, for calculating an area enclosed by a loop.

We'll begin with the case of an enclosed area between two single valued continuous functions that intersect at two points, defining in this way an interval of the independent variable  x . In order to use definite integrals for calculating the enclosed area, we also require that the functions are integrable in the interval. In order to use the notion of an area under a function, we'll start by also requiring that both functions are non-negative in this interval.

For any point in this interval (except the end points), one of the functions, say  y1(x) , should be greater than the second  y2(x; otherwise we'll have intersection points inside the interval. From the definition of area under a curve, it follows that the enclosed area  S  is the difference of the integrals of the functions. Formally this can be written as

 $\begin{array}{l}\text{for}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\text{interval}\text{ }\left[a,b\right]\\ \left\{\begin{array}{l}{y}_{1}\left(a\right)={y}_{2}\left(a\right)\\ {y}_{1}\left(b\right)={y}_{2}\left(b\right)\end{array}\\ \text{for}\text{\hspace{0.17em}}x\ne a\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}x\ne b:\\ {y}_{1}\left(x\right)>{y}_{2}\left(x\right)\ge 0\\ S=\underset{a}{\overset{b}{\int }}{y}_{1}\text{d}x-\underset{a}{\overset{b}{\int }}{y}_{2}\text{d}x=\\ =\underset{a}{\overset{b}{\int }}{y}_{1}\text{d}x+\underset{b}{\overset{a}{\int }}{y}_{2}\text{d}x\end{array}\right\}$ (2.2.2.1)

The last row of (2.2.2.1) can be interpreted as the integral, whose path is clockwise along the loop, defined by the appropriate functions.

Let's take the following example

 $\begin{array}{l}\left\{\begin{array}{l}{y}_{1}={x}^{2}+1\text{ }\text{and}\text{ }{y}_{2}=2{x}^{2}\\ a=-1\text{ }\text{and}\text{ }b=1\end{array}\\ S=\underset{-1}{\overset{1}{\int }}\left({x}^{2}+1\right)\text{d}x+2\underset{1}{\overset{-1}{\int }}{x}^{2}\text{d}x=\\ =\left(\frac{{x}^{3}}{3}+x\right)\underset{-1}{\overset{1}{|}}+2\frac{{x}^{3}}{3}\underset{1}{\overset{-1}{|}}=\\ =\left(\frac{1}{3}+1\right)-\left(-\frac{1}{3}-1\right)+2\left(-\frac{1}{3}-\frac{1}{3}\right)=\\ =-\frac{2}{3}+2=\frac{4}{3}\end{array}\right\}$ (2.2.2.2)

which is shown in Fig. Enclosed area of parabolas(1) .

This example uses functions  y  that are non-negative for all the points of the loop enclosing the area. The notion of the clockwise sense of the integration was graphically emphasized in the figure. The question that should now be asked is, what happens if there are points of the loop with negative  y  values? As we'll see it does not matter: the result of the integration gives the enclosed area with a positive sign, as far as the clockwise sense is observed.

In order to prove this, let's use (2.2.2.1) and translate both functions in  y  by the same constant value  Δy . By doing so, the enclosed area remains unchanged, but on the other hand the values of  y  change and can also be negative. The clockwise sense of integration becomes

 $\begin{array}{l}\underset{a}{\overset{b}{\int }}\left({y}_{1}+\Delta y\right)\text{d}x+\underset{b}{\overset{a}{\int }}\left({y}_{2}+\Delta y\right)\text{d}x=\\ =\underset{a}{\overset{b}{\int }}{y}_{1}\text{d}x+\underset{b}{\overset{a}{\int }}{y}_{2}\text{d}x+\Delta y\underset{a}{\overset{b}{\int }}\text{d}x+\Delta y\underset{b}{\overset{a}{\int }}\text{d}x=\\ =S+\Delta y\left[\left(b-a\right)+\left(a-b\right)\right]=S\end{array}\right\}$ (2.2.2.3)

After the proof, we can modify the example (2.2.2.2) by adding Δy=−1 and using

 $\begin{array}{l}{y}_{1}={x}^{2}\\ {y}_{2}=2{x}^{2}-1\end{array}\right\}$ (2.2.2.4)

The graphical presentation this time can be seen in Fig. Enclosed area of parabolas(2)

We can now talk about any area enclosed in a loop not intersecting itself. It does not matter if this loop is formed from two functions, or it is defined by many segments of different curves, as long as the integration is done in a clockwise sense, it will result in a positive value. The result of a counter-clockwise sense of integration will give also the area but with a negative sign.

The enclosing loop does not have to be defined by explicit functions, but could be given also by a parametric representation. There are cases, where the whole loop can be defined by just one parametric presentation, whose area can be obtained by one single integration. In Exercise 2 from the previous page, the area of  ¼  ellipse was obtained, by the use of the angular parameter  β , integrated over the interval  [π/2,0] .

In the following example we'll use the area enclosed by the loop formed by the Folium of Descartes, which was defined implicitly by (1.3.3.1) and reproduced here:

Its parametric presentation was given in an example of the previous chapter (1.3.4.23) and is reproduced here:

It was shown there that the function has a minimum at the origin  (x,y) = (0,0) , and it was illustrated in the Fig Folium of Descartes that the point  x = y = 3a ⁄2  belongs to the curve. From the parametric presentation one immediately obtains that this point corresponds to  t = 1 . This information is sufficient for calculating the area of the loop, as shown in Fig. Folium of Descartes loop.

As explained in this figure, the area S of the loop can be calculated from

 $S=2\left(\underset{0}{\overset{\frac{3a}{2}}{\int }}x\text{d}x+\underset{\frac{3a}{2}}{\overset{0}{\int }}y\text{d}x\right)$ (2.2.2.5)

where  y  corresponds to the curve (the Folium of Descartes) for the interval of the parameter $0\le t\le 1$ and the integral can be solved by substituting  u = 1+t³

 $\begin{array}{l}\underset{\frac{3a}{2}}{\overset{0}{\int }}y\text{d}x=\underset{1}{\overset{0}{\int }}y\frac{\text{d}x}{\text{d}t}\text{d}t=\\ =\underset{1}{\overset{0}{\int }}\frac{3a{t}^{2}}{1+{t}^{3}}\frac{3a\left(1+{t}^{3}\right)-3{t}^{2}3at}{{\left(1+{t}^{3}\right)}^{2}}\text{d}t=\\ ={\left(3a\right)}^{2}\underset{1}{\overset{0}{\int }}{t}^{2}\frac{1-2{t}^{3}}{{\left(1+{t}^{3}\right)}^{3}}\text{d}t=\\ =3{a}^{2}\underset{2}{\overset{1}{\int }}\frac{3-2u}{{u}^{3}}\text{d}u=\\ =3{a}^{2}\left(-\frac{3}{2{u}^{2}}+\frac{2}{u}\right)\underset{2}{\overset{1}{|}}=-\frac{3{a}^{2}}{8}\end{array}\right\}$ (2.2.2.6)

and finally, from (2.2.2.5) and (2.2.26) one obtains the area  S

 $S=2\left(\frac{9{a}^{2}}{8}-\frac{3{a}^{2}}{8}\right)=\frac{3{a}^{2}}{2}$ (2.2.2.7)

## Mean value

We have already seen applications of definite integrals connected with areas bounded by functions. In addition to that the definite integrals have many applications in science, and one of them is the calculation of mean values. We are going to use the very common notation  <Q>  for the mean value of the quantity  Q .

Let's start with a simple example: there are  N  boxes with marbles in each one of them, but the distribution is uneven. If the boxes are ordered by index  i , we denote the number of marbles at the  ith  box as  ni . The mean number of marbles per box will be

 $〈n〉=\frac{\sum _{i=1}^{N}{n}_{i}}{N}$ (2.2.2.8)

We will now deal with a similar, but a bit more complicated example. A rod of length  l  made from an insulator has a given temperature  T  as a function of its coordinate  x . We want to know what is the mean temperature of the rod. In order to do this, we can use  N  values of  T  measured on small equal segments of  l  and use (2.2.2.8). On the other hand since  l = NΔx , where  Δx  is the length of one segment, we obtain

 $〈T〉=\frac{\sum _{i=1}^{N}{T}_{i}}{N}=\frac{\sum _{i=1}^{N}{T}_{i}\Delta x}{l}=\frac{\sum _{i=1}^{N}{T}_{i}\Delta x}{\sum \Delta x}$ (2.2.2.9)

but the result should be more accurate if  N  is bigger (or in other words  Δx  is smaller). This yields the best result, when definite integrals replace the sums:

 $〈T〉=\frac{\underset{a}{\overset{b}{\int }}T\text{d}x}{\underset{a}{\overset{b}{\int }}\text{d}x}\text{ }\text{ }\text{where}\text{ }l=b-a$ (2.2.2.10)

The expression (2.2.210) is called the mean value of  T  with respect to  x . The mean value of a quantity conserves its dimensions, as can be expected.

In the case of a continuous function, the mean value should coincide with a value of the function at least at one point of the interval. But this is not necessary for a discontinuous function.

The mean value of a continuous function is presented graphically in Fig. Mean value of a continuous function.

The case of a discontinuous function is shown graphically in Fig. Mean value of a discontinuous function.

We'll now see an example which looks straight-forward at first, but there is a small complication. Let's assume that a car drove exactly half of a distance at a constant velocity  v1  and the other half at the velocity  v2 . Relating to the question what was the mean velocity  <v>  of all the journey, it looks almost certain that it should be  $〈v〉=\frac{{v}_{1}+{v}_{2}}{2}$ . If we want to be more accurate we have to use (2.2.2.10), and obtain the same result:

 $〈v〉=\frac{{v}_{1}\Delta x+{v}_{2}\Delta x}{2\Delta x}=\frac{{v}_{1}+{v}_{2}}{2}$ (2.2.2.11)

where  Δx  denotes half of the total distance. For instance if  v1 = 60 km/hr  and  v2 = 100 km/hr , then  <v> = 80 km/hr .

So far so good, but let's try another approach: the mean velocity should be equal to the total distance divided by the total time. If we denote by  Δt1  and  Δt2  the travel time of the first and the second half of the distance accordingly, then we should have

 $〈v〉=\frac{2\Delta x}{\Delta {t}_{1}+\Delta {t}_{2}}=\frac{2\Delta x}{\frac{\Delta x}{{v}_{1}}+\frac{\Delta x}{{v}_{2}}}=\frac{2{v}_{1}{v}_{2}}{{v}_{1}+{v}_{2}}$ (2.2.2.12)

and by substituting the numerical values of the velocities we obtain  <v> = 75 km/hr . Where has the poor driver lost the  5 km/hr  and arrived late? The answer to this puzzle is that we should take the mean value of the velocity with respect to time and not - distance. Indeed

 $〈v〉=\frac{\underset{a}{\overset{b}{\int }}v\text{d}t}{\underset{a}{\overset{b}{\int }}\text{d}t}=\frac{{v}_{1}\Delta {t}_{1}+{v}_{2}\Delta {t}_{2}}{\Delta {t}_{1}+\Delta {t}_{2}}=\frac{2\Delta x}{\Delta {t}_{1}+\Delta {t}_{2}}$ (2.2.2.13)

gives identical result as (2.2.2.12). This example shows the importance of chosing the right variable for calculating the mean value with respect to it.

Since this example uses a discontinuous function  v , it is not surprising that the calculated mean did not correspond to any actual value of the function in the interval.

## Energy conservation

Energy conservation is believed to be a basic law of nature, but although there is no experimental evidence for its violation, it cannot be proven theoretically so far. However there are particular cases in which this law can be deduced from basic principles, as in the following case.

According to (2.1.4.2) the second law of Newton for the motion of a mass point  m  in one dimension with a force  F  dependent on the coordinate  x  only is

 $m\frac{\text{d}v}{\text{d}t}=F\left(x\right)$ (2.2.2.14)

where  v  is the velocity and  t  - the time. We already integrated this equation in an exercise of section 1. For the sake of completeness, part of it is repeated here and may look familiar.

The indefinite integral of the left-hand side with respect to x is

 $\begin{array}{l}\int m\frac{\text{d}v}{\text{d}t}\text{d}x=m\int \frac{\text{d}v}{\text{d}t}\frac{\text{d}x}{\text{d}t}\text{d}t=m\int v\frac{\text{d}v}{\text{d}t}\text{d}t=\\ =m\int v\text{d}v=\frac{m{v}^{2}}{2}+C=K+C\end{array}\right\}$ (2.2.2.15)

where  K  denotes kinetic energy. The corresponding indefinite integral of the right-hand side of (2.2.2.14), in the most general form is a function of  x  that we'll denote by  −U(x) , where we'll call  U  the potential energy.

 $\int F\left(x\right)\text{d}x=-U\left(x\right)$ (2.2.2.16)

By applying definite integration on both sides of (2.2.2.14) between the limits a and b and in view of (2.2.2.15-16) we obtain

 $\begin{array}{l}{K}_{b}-{K}_{a}={U}_{a}-{U}_{b}\text{ }\text{or}\\ {K}_{b}+{U}_{b}={K}_{a}+{U}_{a}=E\end{array}\right\}$ (2.2.2.17)

where  E  is a constant of the motion called total energy. By using the definite integral (2.2.2.17), any additive constant of the indefinite integrals (2.2.2.15-16) becomes immaterial. According to the definition the kinetic energy cannot be negative, while there is no limitation on the sign of the potential energy.

One can obtain the total energy from the initial conditions and rewrite the expression (2.2.2.17) as

 $\frac{m{v}^{2}}{2}+U\left(x\right)=E$ (2.2.2.18)

which represents a very useful relation between  v²  and the coordinate  x .

Let's return to the example of tossing up vertically (positive direction of  x) a mass point  m  with an initial velocity of  vo>0  under the constant gravitational force  F = −mg  from a starting height of  xo = 0 . Then the equation (2.2.2.18) becomes

 $\frac{m{v}^{2}}{2}+mgx=\frac{m{v}_{0}^{2}}{2}$ (2.2.2.19)

from which we obtain that the maximal height  h  (corresponding to  v = 0) is

 $h=\frac{{v}_{0}^{2}}{2g}$ (2.2.2.20)

From the energy conservation and the direction of the force alone, one can obtain a qualitative description of the motion. One must keep in mind that at the point of vanishing velocity the motion will continue in the direction of the force (which is also the direction of the acceleration).

Let's apply this to the example we are dealing with. The motion should look as follows

• From the initial position, the motion is upwards due to the initial velocity.
• The velocity decreases due to the energy conservation until rest.
• At rest the height is (2.2.2.20), due to the energy conservation.
• After rest the motion continues downwards (negative velocity) due to the direction of the force.
• The absolute value of the velocity increases but it remains negative.
• At  x = 0  the velocity is  v = −vo .
• If the mass does not hit the ground it will continue to move downwards and accelerate due to the energy conservation (the potential energy will become negative).

By continuing the same example, the relations (2.2.2.19-20) allow an easy calculation of the mean kinetic energy with respect to x. By denoting the kinetic energy by  K , (2.2.2.19) becomes

 $K={K}_{0}-mgx$ (2.2.2.21)

and the mean value of  K  for the vanishing initial  x and final  x  corresponding to  h  (2.2.2.20), we obtain

 $\begin{array}{l}{〈K〉}_{x}={K}_{0}-mg〈x〉=K{}_{0}-mg\frac{\underset{0}{\overset{\frac{{v}_{0}^{2}}{2g}}{\int }}x\text{d}x}{\underset{0}{\overset{\frac{{v}_{0}^{2}}{2g}}{\int }}\text{d}x}=\\ ={K}_{0}-\frac{2m{g}^{2}}{{v}_{0}^{2}}\frac{{x}^{2}}{2}\underset{0}{\overset{\frac{{v}_{0}^{2}}{2g}}{|}}=K{}_{0}-\frac{m{g}^{2}}{{v}_{0}^{2}}\frac{{v}_{0}^{4}}{4{g}^{2}}=\\ ={K}_{0}-\frac{m{v}_{0}^{2}}{4}={K}_{0}\left(1-\frac{1}{2}\right)=\frac{1}{2}{K}_{0}\end{array}\right\}$ (2.2.2.22)

In the same case, in order to obtain the mean value of the kinetic energy with respect to the time  t  we need the time dependence, which is already known from (2.1.4.7) and is reproduced here:

 $\begin{array}{l}v=-gt+{v}_{o}\\ x=-\frac{g{t}^{2}}{2}+{v}_{o}t\end{array}\right\}$ (2.2.2.23)

The initial time is  to = 0  and the time corresponding to the mass reaching its maximal height is  t1 = vo/g  according to (2.2.2.23). The required mean value of the kinetic energy becomes

 $\begin{array}{l}{〈K〉}_{t}=K{}_{0}-mg\frac{\underset{0}{\overset{\frac{{v}_{0}}{g}}{\int }}\left({v}_{0}t-\frac{1}{2}g{t}^{2}\right)\text{d}t}{\underset{0}{\overset{\frac{{v}_{0}}{g}}{\int }}\text{d}t}=\\ ={K}_{0}-\frac{m{g}^{2}}{{v}_{0}}\left(\frac{1}{2}{v}_{0}{t}^{2}-\frac{1}{6}g{t}^{3}\right)\underset{0}{\overset{\frac{{v}_{0}}{g}}{|}}=\\ =K{}_{0}-\frac{m{g}^{2}}{2{v}_{0}}\frac{{v}_{0}^{2}}{{g}^{2}}\left({v}_{0}-\frac{1}{3}g\frac{{v}_{0}}{g}\right)=\\ ={K}_{0}-\frac{m{v}_{0}^{2}}{2}\left(1-\frac{1}{3}\right)={K}_{0}\left(1-\frac{2}{3}\right)=\frac{1}{3}{K}_{0}\end{array}\right\}$ (2.2.2.24)

It is not unexpected that the mean values (2.2.2.22) and (2.2.2.24) are different.

## Exercises

Exercise 1. The two curves

$y={x}^{4}\text{ }\text{and}\text{ }{y}^{2}=x$

enclose an area.

1. Make a sketch and mark the enclosed area.
2. Find the intersection points.
3. Calculate the area.

Exercise 2. The two ellipses with a>b>0 are

$\begin{array}{l}\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\text{ }\left(\text{horizontal}\right)\\ \frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1\text{ }\left(\text{vertical}\right)\end{array}\right\}\text{and}$

We are interested in the central enclosed area.

1. Make a sketch and mark this area.
2. Find the intersection points.
3. Calculate the area.

Exercise 3. For each of the following forces F which depend on the x coordinate, and are applied to a mass point m, use the given initial conditions in order to

• obtain the expression of energy conservation
• describe qualitatively the motion
1. F=−ax³ where a is a positive constant and the initial conditions are xo>0 and vo=0.
2. F=kx where k is a positive constant and the initial conditions are xo>0 and vo=0.
3. F=bx−2 where b is a positive constant and the initial conditions are xo>0 and vo<0.

Exercise 4. The simple harmonic motion of a mass point m is obtained by the force F=−kx where k is a positive constant, and was already discussed in (2.1.4.18-24). In the present case the initial conditions are ${x}_{0}\ne 0\text{\hspace{0.28em}}\text{and}\text{\hspace{0.28em}}{v}_{0}=0$, where x is the coordinate and v the velocity.

1. Obtain the expression of energy conservation.
2. What is the maximal possible kinetic energy  Kmax ?
3. Obtain the mean value of the kinetic energy with respect to  x  in terms of the maximal kinetic energy! Use the interval $\left(0,{x}_{0}\right)$ .
4. Use the already known solution of the motion: $\begin{array}{l}x={x}_{0}\mathrm{cos}\left(\omega t\right)\\ v=-\omega {x}_{0}\mathrm{sin}\left(\omega t\right)\\ \omega =\sqrt{\frac{k}{m}}\end{array}\right\}$ in order to obtain the mean value of the kinetic energy with respect to time, integrated over one time period!

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