Barrow surely discovered that the tangent to the area curve of a function at a point equals the value of the function at that point. Also, I’ve seen geometric proofs of this.

But did he also discover that the area function of the tangent function (or derivative, in modern terms) of a function at a point also evaluated to the function value at that point plus some constant for all the points in the domain?


1 Answer 1


Insofar as Barrow realised the first theorem, he also realised the second. Barrow's theorem X.11 can be interpreted as FTC1 $$\frac{d}{dt}\int_{a}^t y(x)dx = y(t)$$ while his theorem XI.19 can be interpreted as FTC2 $$\int_{a}^b y'(x) dx = y(b)-y(a)$$ But according to most historians it is more accurate to say that he did not realise either, and that this anachronistic interpretation of what he said does not reflect his own way of thinking. As Katz says in his history of mathematics, "Barrow presented all of his work in a classic geometric form. It does not appear that he was aware of the fundamental nature of the two theorems [equivalent to FTC1 and FTC2] presented in the text. Barrow did not mention that they are particularly important; he just presented them as two among many geometrical results dealing with tangents and areas."

The work by Barrow in question has twice been translated into English: https://archive.org/details/geometricallect00newtgoog https://archive.org/details/geometricallectu00barruoft


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