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