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which is undefined due to division by zero. The logarithm function, which is the actual antiderivative of 1/x, must be introduced and examined separately.


The derivative
(x^n)'=nx^{n-1} can be geometrized as the infinitesimal change in volume of the n-cube, which is the area of n faces, each of dimension n − 1.
Integrating this picture – stacking the faces – geometrizes the fundamental theorem of calculus, yielding a decomposition of the n-cube into n pyramids, which is a geometric proof of Cavalieri's quadrature formula.
For positive integers, this proof can be geometrized: if one considers the quantity xn as the volume of the n-cube (the hypercube in n dimensions), then the derivative is the change in the volume as the side length is changed – this is xn−1, which
can be interpreted as the area of n faces, each of dimension n − 1 (fixing one vertex at the origin, these are the n faces not touching the vertex), corresponding to the cube increasing in size by growing in the direction of these faces – in the 3-
dimensional case, adding 3 infinitesimally thin squares, one to each of these faces. Conversely, geometrizing the fundamental theorem of calculus, stacking up these infinitesimal (n − 1) cubes yields a (hyper)-pyramid, and n of these pyramids form the
n-cube, which yields the formula. Further, there is an n-fold cyclic symmetry of the n-cube around the diagonal cycling these pyramids (for which a pyramid is a fundamental domain). In the case of the cube (3-cube), this is how the volume of a pyramid
was originally rigorously established: the cube has 3-fold symmetry, with fundamental domain a pyramids, dividing the cube into 3 pyramids, corresponding to the fact that the volume of a pyramid is one third of the base times the height. This illustrates
geometrically the equivalence between the quadrature of the parabola and the volume of a pyramid, which were computed classically by different means.

Alternative proofs exist – for example, Fermat computed the area via an algebraic trick of dividing the domain into certain intervals of unequal length; alternatively, one can prove this by recognizing a symmetry of the graph y = xn under inhomogeneous
dilation (by d in the x direction and dn in the y direction, algebraicizing the n dimensions of the y direction), or deriving the formula for all integer values by expand
--- end quoting Wikipedia on Cavalieri's quadrature formula ---

--- quoting Google Search hits ---

A New Proof of Cavalieri's Quadrature Formula

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by NJ Wildberger · 2002 · Cited by 5 — Theorem of Calculus. Here is a proof of Cavalieri's formula that uses the (hidden) symmetry of the func- tion x" and the Binomial ...

A New Proof of Cavalieri's Quadrature Formula

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PDF | On Nov 1, 2002, N. J. Wildberger published A New Proof of Cavalieri's Quadrature Formula | Find, read and cite all the research you need on ...

(PDF) A New Proof of Cavalieri's Quadrature Formula

Academia.edu
https://www.academia.edu › A_New_Proof_of_Cavali...
We use the contemporary mathematical technologies to prove the fundamental assumptions of the Euclidean Goemetry with indivisibles and we develop a model- ...

12.A. The proof of Cavalieri's Principle

University of California,