7.1. THE DEFINITION OF THE INTEGRAL FROM ANTIDERIVATIVES 173

a

b

y

V (y)

A(y)

dy

and so the total volume of the solid between a and y,V (y) , satisfies the initial value problemdVdy = A(y) , V (a) = 0. The volume of the solid is V (b) .

Example 7.0.7 Suppose a solid is obtained as a circular disk of radius r and center at(0,0) is revolved about the y axis. Find the volume of the solid ball of radius r.

x2 + y2 = r2

yx

r x

Here the line would be the y axis from −r to r. For given y, you have x2 + y2 = r2,so x =

√r2 − y2 and this is the radius of a cross section located at y. Then by the above,

V ′ (y) = π(r2 − y2

),V (−r) = 0. The volume is

∫ r−r π

(r2 − y2

)dy. An antiderivative is

πr2y− πy3

3 so the volume is(

πr2 · r−πr3

3

)−(

πr2 · (−r)+πr3

3

)= 4

3 πr3. This is thevolume of a ball of radius r.

The above procedure from the 1700’s is how we find integrals. Thus we are finding so-lutions to an initial value problem and this will include many important applied problemsin geometry and physics. However, there are significant theoretical questions, the most im-portant being the existence of an antiderivative for a continuous function. These questionsare resolved in the next section. After this, I will give a careful treatment of Darboux’sformulation of the Riemannn integral which will bring the understanding of the integral upto the 1850’s. Expressing the integral as a limit of sums is the precise way of avoiding thefuzzy notion of infinitesimals which was the original idea of Leibniz. However, Leibniz’soriginal idea is still very useful in formulating problems to be solved in terms of integrals.

At this point, the reader can do all of the do-able examples involving integrals exceptfor finding antiderivatives, techniques for which are presented later. The following sectionsare theoretical in nature.

7.1 The Definition of the Integral from AntiderivativesNext is the definition of what is meant by an oriented interval.

Definition 7.1.1 For the rest of this section, [a,b] will denote the closed intervalhaving end points a and b but a could be larger than b or smaller than b. It is written thisway to indicate that there is a direction of motion from a to b which will be reflected by thedefinition of the integral given below. It is an “oriented interval”.

Definition 7.1.2 The integral of a continuous function defined on an oriented inter-val [a,b] is defined by Procedure 7.0.1.