How Fermat integrated xt dx

Theorem. (Fermat) For all rational t other than -1, and any positive a,b, the integral of xt dx from x=a to x=b equals (bt+1-at+1)/(t+1).

Proof: Without loss of generality assume b>a. Suppose b=rna for some integer n and positive r. Use the partition {xi=ria | 0<=i<=n} of [a,b]. Then a Riemann sum for the integral is

(x1-x0) xt0 + (x2-x1) xt1 + ... + (xn-xn-1) xtn-1 .
This is the sum of a geometric series with initial term (r-1)at+1, common ratio rt+1, and post-final term (r-1)bt+1. Hence the Riemann sum for this partition is
((r-1) / (rt+1-1))  ·  (bt+1-at+1).
As n ``approaches infinity'', r approaches 1, so the mesh (1-r-1)b of our partition approaches zero. Therefore the Riemann sum approaches the integral. So, it remains to prove that as r approaches 1 the factor ((r-1) / (rt+1-1)) tends to 1/(t+1). This is true for all nonzero t+1, but it's particularly easy to prove when t is rational. Let t+1=c/d, and let z be the dth root of r. Then we want the limit as z approaches 1 of (zd-1)/(zc-1). Simply remove the common factor z-1 from both numerator and denominator of this fraction; then the numerator approaches d, and the denominator approaches c. Hence the limit is d/c=1/(t+1) as claimed. QED

P.S. For irrational t we can now integrate xt dx over [a,b] by writing xt as the uniform limit over that interval of xt' with t' ranging over a sequence of rational numbers that approach t.