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Mathematics Magazine, v. 64, no. 1, Feb. 1991.
A Euclidean Approach to the FTC
Gregory's Proof of the FTC
This Euclidean definition of the tangent is pivotal to Gregory's contradiction. For if KC is not tangent to OEK, it means that some point a on KC must lie above the curve OEK. In particular, if DF is the line segment perpendicular to OA which intersects OA at D, OEK at E, KC at a, and OFL at F, then DE < Da by choice of a. This is shown in the figure below (adapted from Gregory's proof-see the Appendix ).
Actually, the figure tells only half of the story. It's also possible that a could lie on the other side of K on the line CK. However, the argument for that case is similar, and the details make a good exercise (or see the Appendix ).
By the definition of the curve OEK, if area(OFD) denotes the area enclosed by the curve OF and the line segments OD and DF, then DE = area(OFD). Likewise, IK = area(OFLI). Thus, IK/DE = area(OFLI)/area(OFD). Since DE < Da, it follows that IK/Da < IK/DE and so
This last step may seem mysterious, but it serves the purpose of converting the original inequality into an inequality about areas which will lead Gregory to an obvious geometrical contradiction. Specifically, DC·IL is the area of the rectangle with sides IL and DC, and IL·IC is the area of the rectangle with sides IL and IC. Inverting the ratios in the last inequality yields:
But recall the defining equality for C-namely, IK/IC = IL. This implies that IK = IC·IL. Since IK = area(OFLI) by definition of the curve OEK, the denominators in the last inequality are the same. Therefore, the numerators must satisfy DI·IL < area(DFLI). But remember that OFL is increasing by assumption. Thus, the rectangle with sides IL and DI must circumscribe the region DFLI.
Hence area(DFLI) < DI ·IL as well. This contradiction shows that KC must in fact be tangent to OEK. Hence the fundamental theorem of calculus follows.