If is the set of points in
such that and
, then we showed in (2.6) that

By (2.9)

Also

so

for all .

The right side of (2.13) is greater than and the left side is less than for all , but by taking large enough, both sides can be made as close to as we please. Hence we conclude that the ratio is equal to . Thus, we have proved the following theorem:

i.e.

**Remark:** The last paragraph of the proof of theorem 2.14
is a little bit vague. How large is ``large enough'' and what does
``as close as we please'' mean? Archimedes and Euclid would not have
considered such an argument to be a proof. We will reconsider the end of this
proof after we have developed the language to complete it more carefully.
(Cf Example 6.54.)

The first person to calculate the area of a parabolic segment was

Archimedes
(287-212 B.C.). The parabolic segment considered by Archimedes corresponds to
the set bounded by the parabola and the line joining
to
where .

The result of this exercise shows that the area of a parabolic segment depends only on its width. Thus the segment determined by the points and has the same area as the segment determined by the points and , even though the second segment is 400 times as tall as the first, and both segments have the same width. Does this seem reasonable?

**Remark:** Archimedes
stated his result about the area of a parabolic segment
as follows. The area of the parabolic segment cut off by the
line is four thirds of the area of the inscribed triangle , where
is the point on the parabola at which the tangent line is parallel
to . We cannot verify Archimedes formula at this time, because we do
not know how to find the point .

The following definition is introduced as a hint for exercise 2.18

If denotes the reflection of about the line , then and have the same area.