where the interval [a,b] is partitioned into [x₀=a, x₁, x₂,...,x_n=b]
and Δ_ix=x_i-x_i-1

The limiting process is over n and partitions of the interval such that the maximum increment goes to zero; i.e., max (Δ_i) → 0.

It would be extremely cumbersome and tedious to evaluate integrals by this definition. Fortunately there is an easier way.

If F(x) is a function such that its derivative is equal to f(x);

D(F(x)) = dF/dx = f(x),
 

then (The Fundamental Theorem of Calculus)

∫_a^b>f(x)dx = F(b) - F(a)
 

The function F(x) is called the anti-derivative of f(x).

Ronald L. Graham, Donald E. Knuth and Oren Patashnik prsent an interest extension of this system for the summation of series. They call their system Finite Calculus.

Analogies exist in Finite Calculus for most concepts, operations and relations in the Differential Calculus. (Differential Calculus is a misnomer; the field should be called Derivative Calculus.)

In Finite Calculus the differencing operation

Δf = f(x+1)−f(x)
 

takes the place of the derivative. Then if F(x) is such that ΔF=f(x) then

Σ_a^b>f(x)dx = F(b) - F(a)
 

F(x) is called the anti-difference of f(x).

While it may be easy to find the anti-difference of some functions, such as b^x, in general it is not easy to do so without some special tools. In the case of b^x

Δb^x = b^x+1−b^x = (b-1)b^x.
 

Therefore the anti-difference of b^x is b^x/(b-1). This means that for b=2

Δ2^x = 2^x.
 

The function 2^x serves the same role in the Finite Calculus as e^x serves in the Differential Calculus.

The power functions xⁿ do not have obvious anti-differences, but there is another type of function that provides a convenient basis for computing the anti-differences. This function is called the falling factorial power function. It is defined as

x^m = x(x-1)(x-2)··(x-m+1)
 

Note that there are m factors in this product.

Consider now

Δx^m = (x+1)x(x-1)(x-2)··(x+1-m+1)−x(x-1)(x-2)··(x-m+1)
x(x-1)(x-2)··(x-m+2)[(x+1)−(x-m+1)]
= x(x-1)(x-2)··(x-m+2)m
= mx^m-1.
 

This is analogous to d(x^m)/dx =mx^m-1.

Since Δx^m = mx^m-1 the anti-diffence of x^m is x^m+1/(m+1).

The definition of x^m may be extended beyond the positive integers. The appropriate definition of negative integers is

x^−m = 1/(x+1)(x+2)··(x+m)
 

This leves x⁰ and x⁻⁰ to be defined. They are equal to each other and equal to 1. To justify this definition note that

x¹ = xx⁰
 

but x¹=x so x⁰ must be equal to 1. Likewise,

x⁻¹ = x⁻⁰/(x+1).
 

But x⁻¹=1/(x+1) so x⁻⁰=1.

For negative integers then

Δ(x^−m) = 1/(x+2)··(x+m+1) − 1/(x+1)··(x+m)
= 1/(x+2)··(x+m)[1/(x+m+1) − 1/(x+1)]
= 1/(x+2)··(x+m)[(x+1)-(x+m+1)]/(x+1)(x+m+1)
= - m[1/(x+1)(x+2)··(x+m+1)]
= - mx^m+1
 

Therefore the anti-difference of x^−m is x^−(m-1)/(m-1) for m≠1.

For the special case of m equal to 1 consider the harmonic function defined for positive integers as:

H(x) = 1 + 1/2 + 1/3 + ·· 1/x.

The difference ΔH is

H(x+1)−H(x) = 1/(x+1).
 

Therefore the anti-difference of x⁻¹ is H(x). The harmonic function is the analogue in the Finite Calculus of the logarithm function in the Differential Calculus.