Definite Integrals and Some of Their Applications

One use of the integral calculus is to enable us to ascertain the values of areas bounded by curves.

Let us try to get at the subject bit by bit.

Let (the following figure) be a curve, the equation to which is known. That is,  in this curve is some known function of . Think of a piece of the curve from the point  to the point .

Let a perpendicular  be dropped from , and another  from the point . Then call and , and the ordinates and . We have thus marked out the area  that lies beneath the piece . The problem is, how can we calculate the value of this area?

The secret of solving this problem is to conceive the area as being divided up into a lot of narrow strips, each of them being of the width . The smaller we take , the more of them there will be between and . Now, the whole area is clearly equal to the sum of the areas of all such strips. Our business will then be to discover an expression for the area of any one narrow strip, and to integrate it so as to add together all the strips. Now think of any one of the strips. It will be like this: being bounded between two vertical sides, with a flat bottom , and with a slightly curved sloping top. Suppose we take its average height as being ; then, as its width is , its area will be  (the next figure). And seeing that we may take the width as narrow as we please, if we only take it narrow enough its average height will be the same as the height at the middle of it. Now let us call the unknown value of the whole area , meaning surface. The area of one strip will be simply a bit of the whole area, and may therefore be called . So we may write

If then we add up all the strips, we get

So then our finding depends on whether we can integrate for the particular case, when we know what the value of  is as a function of .

For instance, if you were told that for the particular curve in question , no doubt you could put that value into the expression and say: then I must find .

That is all very well; but a little thought will show you that something more must be done. Because the area we are trying to find is not the area under the whole length of the curve, but only the area limited on the left by , and on the right by , it follows that we must do something to define our area between those ‘limits.’

This introduces us to a new notion, namely that of integrating between limits. We suppose to vary, and for the present purpose we do not require any value of  below  (that is ), nor any value of  above  (that is ). When an integral is to be thus defined between two limits, we call the lower of the two values the lower limit, and the upper value the upper limit. Any integral so limited we designate as a definite integral, by way of distinguishing it from an indefinite integral (which is the reverse of differentiation) to which no limits are assigned.

In the symbols which give instructions to integrate, the limits are marked by putting them at the top and bottom respectively of the sign of integration. Thus the instruction will be read: find the integral of  between the lower limit  and the upper limit .

Sometimes the thing is written more simply Well, but how do you find an integral between limits, when you have got these instructions?

Look again at the first figure of this chapter. Suppose we could find the area under the larger piece of curve from to , that is from to , naming the area . Then, suppose we could find the area under the smaller piece from to , that is from to , namely the area . If then we were to subtract the smaller area from the larger, we should have left as a remainder the area , which is what we want. Here we have the clue as to what to do; the definite integral between the two limits is the difference between the integral worked out for the upper limit and the integral worked out for the lower limit.

Let us then go ahead. First, find the general integral thus: and, as is the equation to the curve (Fig. 19.1), is the general integral which we must find.

Doing the integration in question by the rule, we get and this will be the whole area from  up to any value of  that we may assign.

Therefore, the larger area up to the upper limit  will be and the smaller area up to the lower limit  will be

Now, subtract the smaller from the larger, and we get for the area  the value,

This is the answer we wanted. Let us give some numerical values. Suppose , , and and . Then the area  is equal to

Let us here put down a symbolic way of stating what we have ascertained about limits: where is the integrated value of  corresponding to , and  that corresponding to .

All integration between limits requires the difference between two values to be thus found. Also note that, in making the subtraction the added constant  has disappeared.

In summary, The symbol means to evaluate for and for and subtract the latter from the former.

Examples

Example 19.1. To familiarize ourselves with the process, let us take a case of which we know the answer beforehand. Let us find the area of the triangle (the following figure), which has base and height . We know beforehand, from obvious mensuration, that the answer will come .

Now, here we have as the “curve” a sloping line for which the equation is

The area in question will be

Integrating and putting down the value of the indefinite integral in square brackets with the limits marked above and below, we get

Let us satisfy ourselves about this rather surprising dodge of calculation, by testing it on a simple example. Get some squared paper, preferably some that is ruled in little squares of one-eighth inch or one-tenth inch each way. On this squared paper plot out the graph of the equation,

The values to be plotted will be:

The plot is given below.

Now reckon out the area beneath the curve by counting the little squares below the line, from as far as on the right. There are  whole squares and four triangles, each of which has an area equal to  squares; or, in total,  squares. Hence  is the numerical value of the integral of between the lower limit of and the higher limit of .

As a further exercise, show that the value of the same integral between the limits of and is .

Example 19.2. Find the area, between limits and , of the curve (the following figure).

Solution.

Note—Notice that in dealing with definite integrals the constant  always disappears by subtraction.

Let it be noted that this process of subtracting one part from a larger to find the difference is really a common practice. How do you find the area of a plane ring (next figure), the outer radius of which is  and the inner radius is ? You know from mensuration that the area of the outer circle is ; then you find the area of the inner circle, ; then you subtract the latter from the former, and find area of ring ; which may be written .

Example 19.3. Here’s another case—that of the die-away curve. Find the area between and , of the curve (the following figure) whose equation is

Solution. The integration (see here) gives

Example 19.4. Another example is afforded by the adiabatic curve of a perfect gas, the equation to which is , where  stands for pressure,  for volume, and  is of the value  (see below).

Find the area under the curve (which is proportional to the work done in suddenly compressing the gas) from volume  to volume .

Solution. Here we have

Area of a Disk

Example 19.5. Prove the ordinary mensuration formula, that the area  of a circle whose radius is , is equal to .

Solution. Consider an elementary zone or annulus of the surface (the next figure), of breadth , situated at a distance  from the centre. We may consider the entire surface as consisting of such narrow zones, and the whole area  will simply be the integral of all such elementary zones from centre to margin, that is, integrated from to .

We have therefore to find an expression for the elementary area  of the narrow zone. Think of it as a strip of breadth , and of a length that is the periphery of the circle of radius , that is, a length of . Then we have, as the area of the narrow zone,

Hence the area of the whole circle will be:

Now, the general integral of is . Therefore, or whence

Mean (or Average) Value of a Function

Example 19.6. Let us find the mean ordinate (the average -value) of the positive part of the curve , which is shown below.

Solution. To find the mean ordinate, we shall have to find the area of the piece , and then divide it by the length of the base . But before we can find the area we must ascertain the length of the base, so as to know up to what limit we are to integrate. At the ordinate  has zero value; therefore, we must look at the equation and see what value of  will make . Now, clearly, if is ,  will also be , the curve passing through the origin ; but also, if , ; so that gives us the position of the point .

Then the area wanted is

But the base length is .

Therefore, the average ordinate of the curve .

[Note—It will be a pretty and simple exercise in maxima and minima to find by differentiation what is the height of the maximum ordinate. It must be greater than the average.]

The mean (or the average) ordinate of any curve, over a range from to , is given by the expression,

Areas in Polar Coordinates

When the equation of the boundary of an area is given as a function of the distance  of a point of it from a fixed point  (see the next figure) called the pole, and of the angle which  makes with the positive direction of the -axis, the process just explained can be applied just as easily, with a small modification. Instead of a strip of area, we consider a small triangle , the angle at  being , and we find the sum of all the little triangles making up the required area.

The area of such a small triangle is approximately or ; hence the portion of the area included between the curve and two positions of  corresponding to the angles  and  is given by

Note— In the above formula must be expressed in radian measure.

Examples

Example 19.7. Find the area of the sector of  radian in a circumference of radius  inches (the following figure).

Solution. The polar equation of the circumference is evidently . The area is

Example 19.8. Find the area of the first quadrant of the curve (known as “Pascal’s Snail”), the polar equation of which is (the following figure).

Solution.

Volumes by Integration

What we have done with the area of a little strip of a surface, we can, of course, just as easily do with the volume of a little strip of a solid. We can add up all the little strips that make up the total solid, and find its volume, just as we have added up all the small little bits that made up an area to find the final area of the figure operated upon.

Examples.

Example 19.9. Find the volume of a sphere of radius .

Solution. Method (a). A thin spherical shell has for volume  (see Fig. 19.9); summing up all the concentric shells which make up the sphere, we have

Method (b). We can also proceed as follows: a slice of the sphere, of thickness , has for volume . This thin disk is generated by rotating the strip of thickness shown in Fig. 19.14 around the -axis (see Fig 19.15). The length of this strip, , is related to the distance of this strip from the origin, , through Hence

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                Interactive Fig. 19.15
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Interactive Fig. 19.16

On Quadratic Means

In certain branches of physics, particularly in the study of alternating electric currents, it is necessary to be able to calculate the quadratic mean of a variable quantity. By “quadratic mean” is denoted the square root of the mean of the squares of all the values between the limits considered. Other names for the quadratic mean of any quantity are its “virtual” value, or its “r.m.s.” (meaning root-mean-square) value. If  is the function under consideration, and the quadratic mean is to be taken between the limits of and ; then the quadratic mean is expressed as

Examples.

Example 19.11. Find the quadratic mean of the function (next figure).

Solution. Here the integral is , which is .

Dividing by  and taking the square root, we have

Here the arithmetical mean is ; and the ratio of quadratic to arithmetical mean (this ratio is called the form-factor) is .

Exercises

Exercise 19.2. Find the area of the parabola between and . Show that it is two-thirds of the rectangle of the limiting ordinate and of its abscissa.

 

Answer

of .

 

 

 

 

Solution

 

 

Exercise 19.3. Find the area of the positive portion of a sine curve and the mean ordinate.

 

Answer

; .

 

 

 

 

Solution

 

 

Exercise 19.4. Find the area of the positive portion of the curve (), and find the mean ordinate.

 

Answer

; .

 

 

 

 

Solution

 

The problem asks for the hatched area

 

 

Exercise 19.5. Find the area included between the two branches of the curve from to , also the area of the positive portion of the lower branch of the curve (see Fig. 11.12).

 

Answer

, .

 

 

 

 

Solution

 

 

Exercise 19.6. Find the volume of a cone of radius of base , and of height .

 

Answer

.

 

 

 

 

Solution

 

If line rotates around the -axis, it creates a cone of radius of base , and of height .

We just need to calculate the volume of this solid of revolution.

The volume of this thin disk is

Therefore, if we add the volumes of such thin disks, we get the volume of the solid (cone)