**
How Fermat integrated ***x*^{t} dx
**Theorem.** (Fermat) For all rational *t*
other than -1, and any positive *a,b*, the integral of
*x*^{t} dx from
*x*=*a* to *x*=*b* equals
(*b*^{t+1}-*a*^{t+1})/(*t*+1).

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

(*x*_{1}-*x*_{0})
*x*^{t}_{0}
+
(*x*_{2}-*x*_{1})
*x*^{t}_{1}
+ ... +
(*x*_{n}-*x*_{n-1})
*x*^{t}_{n-1} .
This is the sum of a geometric series with initial term
(*r*-1)*a*^{t+1},
common ratio *r*^{t+1},
and post-final term
(*r*-1)*b*^{t+1}.
Hence the Riemann sum for this partition is
((r-1) / (*r*^{t+1}-1))
·
(*b*^{t+1}-*a*^{t+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) / (*r*^{t+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 *d*^{th} root of *r*.
Then we want the limit as *z* approaches 1 of
(*z*^{d}-1)/(*z*^{c}-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
*x*^{t} dx over [*a*,*b*]
by writing *x*^{t}
as the uniform limit over that interval of *x*^{t'}
with *t'* ranging over a sequence of rational numbers
that approach *t*.