One of the most important applications of the processes of integration which have been explained in the preceding sections is to the calculation of areas of plane curves. Suppose that $$P_{0}PP’$$ (Fig. 44) is the graph of a continuous curve $$y = \phi(x)$$ which lies wholly above the axis of $$x$$, $$P$$ being the point $$(x, y)$$ and $$P’$$ the point $$(x + h, y + k)$$, and $$h$$ being either positive or negative (positive in the figure).

The reader is of course familiar with the idea of an ‘area’, and in particular with that of an area such as $$ONPP_{0}$$. This idea we shall at present take for granted. It is indeed one which needs and has received the most careful mathematical analysis: later on we shall return to it and explain precisely what is meant by ascribing an ‘area’ to such a region of space as $$ONPP_{0}$$. For the present we shall simply assume that any such region has associated with it a definite positive number $$(ONPP_{0})$$ which we call its area, and that these areas possess the obvious properties indicated by common sense, e.g. that $(PRP’) + (NN’RP) = (NN’P’P),\quad (N_{1}NPP_{1}) < (ONPP_{0}),$ and so on.

Taking all this for granted it is obvious that the area $$ONPP_{0}$$ is a function of $$x$$; we denote it by $$\Phi(x)$$. Also $$\Phi(x)$$ is a continuous function. For \begin{aligned} \Phi(x + h) – \Phi(x) &= (NN’P’P)\\ &= (NN’RP) + (PRP’) = h\phi(x) + (PRP’).\end{aligned}

As the figure is drawn, the area $$PRP’$$ is less than $$hk$$. This is not however necessarily true in general, because it is not necessarily the case (see for example Fig 44a) that the arc $$PP’$$ should rise or fall steadily from $$P$$ to $$P’$$. But the area $$PRP’$$ is always less than $$|h|\lambda(h)$$, where $$\lambda(h)$$ is the greatest distance of any point of the arc $$PP’$$ from $$PR$$. Moreover, since $$\phi(x)$$ is a continuous function, $$\lambda(h) \to 0$$ as $$h \to 0$$. Thus we have $\Phi(x + h) – \Phi(x) = h\{\phi(x) + \mu(h)\},$ where $$|\mu(h)| < \lambda(h)$$ and $$\lambda(h) \to 0$$ as $$h \to 0$$. From this it follows at once that $$\Phi(x)$$ is continuous. Moreover $\Phi'(x) = \lim_{h \to 0} \frac{\Phi(x + h) – \Phi(x)}{h} = \lim_{h \to 0} \{\phi(x) + \mu(h)\} = \phi(x).$ Thus the ordinate of the curve is the derivative of the area, and the area is the integral of the ordinate.

We are thus able to formulate a rule for determining the area $$ONPP_{0}$$. Calculate $$\Phi(x)$$, the integral of $$\phi(x)$$. This involves an arbitrary constant, which we suppose so chosen that $$\Phi(0) = 0$$. Then the area required is $$\Phi(x)$$.

If it were the area $$N_{1}NPP_{1}$$ which was wanted, we should of course determine the constant so that $$\Phi(x_{1}) = 0$$, where $$x_{1}$$ is the abscissa of $$P_{1}$$. If the curve lay below the axis of $$x$$, $$\Phi(x)$$ would be negative, and the area would be the absolute value of $$\Phi(x)$$.