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 \(AB\) (Fig. 52) be a curve, the equation to which is known. That is, \(y\) in this curve is some known function of \(x\). Think of a piece of the curve from the point \(P\) to the point \(Q\).

Let a perpendicular \(PM\) be dropped from \(P\), and another \(QN\) from the point \(Q\). Then call \(OM = x_1\) and \(ON = x_2\), and the ordinates \(PM = y_1\) and \(QN = y_2\). We have thus marked out the area \(PQNM\) that lies beneath the piece \(PQ\). 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 \(dx\). The smaller we take \(dx\), the more of them there will be between \(x_1\) and \(x_2\). 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 \(dx\), and with a slightly curved sloping top. Suppose we take its *average* height as being \(y\); then, as its width is \(dx\), its area will be \(y\, dx\). 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 \(S\), meaning surface. The area of one strip will be simply a bit of the whole area, and may therefore be called \(dS\). So we may write \[\text{area of $1$ strip} = dS = y \cdot dx.\] If then we add up all the strips, we get \[\text{total area $S$} = \int dS = \int y\, dx.\]

So then our finding \(S\) depends on whether we can integrate \(y \cdot dx\) for the particular case, when we know what the value of \(y\) is as a function of \(x\).

For instance, if you were told that for the particular curve in question \(y = b + ax^2\), no doubt you could put that value into the expression and say: then I must find \(\int (b + ax^2)\, dx\).

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 \(PM\), and on the right by \(QN\), 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 \(x\) to vary, and for the present purpose we do not require any value of \(x\) below \(x_1\) (that is \(OM\)), nor any value of \(x\) above \(x_2\) (that is \(ON\)). When an integral is to be thus defined between two limits, we call the lower of the two values *the inferior limit*, and the upper value *the superior limit*. Any integral so limited we designate as a *definite integral*, by way of distinguishing it from a *general integral* 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 \[\int_{x=x_1}^{x=x_2} y \cdot dx\] will be read: find the integral of \(y \cdot dx\) between the inferior limit \(x_1\) and the superior limit \(x_2\).

Sometimes the thing is written more simply \[\int^{x_2}_{x_1} y \cdot dx.\] Well, but *how* do you find an integral between limits, when you have got these instructions?

Look again at Fig. 52. Suppose we could find the area under the larger piece of curve from \(A\) to \(Q\), that is from \(x = 0\) to \(x = x_2\), naming the area \(AQNO\). Then, suppose we could find the area under the smaller piece from \(A\) to \(P\), that is from \(x = 0\) to \(x = x_1\), namely the area \(APMO\). If then we were to subtract the smaller area from the larger, we should have left as a remainder the area \(PQNM\), 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 superior limit and the integral worked out for the lower limit.

Let us then go ahead. First, find the general integral thus: \[\int y\, dx,\] and, as \(y = b + ax^2\) is the equation to the curve (Fig. 52), \[\int (b + ax^2)\, dx\] is the general integral which we must find.

Doing the integration in question by the rule (Chapter 18), we get \[bx + \frac{a}{3} x^3 + C;\] and this will be the whole area from \(0\) up to any value of \(x\) that we may assign.

Therefore, the larger area up to the superior limit \(x_2\) will be \[bx_2 + \frac{a}{3} x_2^3 + C;\] and the smaller area up to the inferior limit \(x_1\) will be \[bx_1 + \frac{a}{3} x_1^3 + C.\]

Now, subtract the smaller from the larger, and we get for the area \(S\) the value, \[\text{area $S$} = b(x_2 – x_1) + \frac{a}{3}(x_2^3 – x_1^3).\]

This is the answer we wanted. Let us give some numerical values. Suppose \(b = 10\), \(a = 0.06\), and \(x_2 = 8\) and \(x_1 = 6\). Then the area \(S\) is equal to \[\begin{gathered} 10(8 – 6) + \frac{0.06}{3} (8^3 – 6^3) \\ \begin{aligned} &= 20 + 0.02(512 – 216) \\ &= 20 + 0.02 \times 296 \\ &= 20 + 5.92 \\ &= 25.92. \end{aligned}\end{gathered}\]

Let us here put down a symbolic way of stating what we have ascertained about limits: \[\int^{x=x_2}_{x=x_1} y\, dx = y_2 – y_1,\] where \(y_2\) is the integrated value of \(y\, dx\) corresponding to \(x_2\), and \(y_1\) that corresponding to \(x_1\).

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

*Examples.*

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, \[y = \frac{x}{3}.\]

The values to be plotted will be: \[\begin{array} {|c|| *{5}{c|}} \hline {x} & {0} & {3} & {6} & {9} & {12} \\ \hline {y} & {0} & {1} & {2} & {3} & {4} \\ \hline \end{array}\]

The plot is given in Fig. 54.

Now reckon out the area beneath the curve *by counting the little squares* below the line, from \(x = 0\) as far as \(x = 12\) on the right. There are \(18\) whole squares and four triangles, each of which has an area equal to \(1\frac{1}{2}\) squares; or, in total, \(24\) squares. Hence \(24\) is the numerical value of the integral of \(\dfrac{x}{3}\, dx\) between the lower limit of \(x = 0\) and the higher limit of \(x = 12\).

As a further exercise, show that the value of the same integral between the limits of \(x = 3\) and \(x = 15\) is \(36\).

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 (Fig. 56), the outer radius of which is \(r_2\) and the inner radius is \(r_1\)?

You know from mensuration that the area of the outer circle is \(\pi r_2^2\); then you find the area of the inner circle, \(\pi r_1^2\); then you subtract the latter from the former, and find area of ring \(= \pi(r_2^2 – r_1^2)\); which may be written \[\pi(r_2 + r_1)(r_2 – r_1)\] \(= \text{mean circumference of ring} \times \text{width of ring}\).

*An Exercise.*

*Another Exercise.*

The mean ordinate of any curve, over a range from \(x= 0\) to \(x = x_1\), is given by the expression,

\[\bbox[#F2F2F2,5px,border:2px solid black]{\text{mean $y$} = \frac{1}{x_1} \int^{x=x_1}_{x=0} y \cdot dx.}\]

One can also find in the same way the surface area of a solid of revolution.

*Example.*

### Areas in Polar Coordinates.

When the equation of the boundary of an area is given as a function of the distance \(r\) of a point of it from a fixed point \(O\) (see Fig. 61) called the *pole*, and of the angle which \(r\) makes with the positive horizontal direction \(OX\), 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 \(OAB\), the angle at \(O\) being \(d\theta\), and we find the sum of all the little triangles making up the required area.

The area of such a small triangle is approximately \(\dfrac{AB}{2}\times r\) or \(\dfrac{r\, d\theta}{2}\times r\); hence the portion of the area included between the curve and two positions of \(r\) corresponding to the angles \(\theta_1\) and \(\theta_2\) is given by \[\tfrac{1}{2} \int^{\theta=\theta_2}_{\theta=\theta_1} r^2\, d\theta.\]

*Examples.*

### 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.*

### 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. The French term is *valeur efficace*. If \(y\) is the function under consideration, and the quadratic mean is to be taken between the limits of \(x=0\) and \(x=l\); then the quadratic mean is expressed as \[\sqrt[2] {\frac{1}{l} \int^l_0 y^2\, dx}.\]

*Examples.*