∫

Imagine that someone has plotted a
curve y = x^{n} on a standard Cartesian grid. Between two given values
of x (x_{0} and x_{1}) and between the curve and the abscissa of
the coordinate grid lies an area that you want to calculate. You would set up
that calculation by dividing the area into vertical strips, each strip x^{n}
long by dx wide. You would then add up the areas of the strips.

But if you make dx too wide, you will
have an error in your calculation. That error comes from the fact that the area
that you calculate for each strip equals the product of the length of the strip
and its width, the area of a long, thin rectangle, while the actual strip under
the curve has one end that does not meet the sides at a right angle. That
difference creates an error in the calculation of the total area of the patch
under the curve between x_{0} and x_{1}. However, if dx
approaches zero, the error will also approach zero and we may thus ignore it.
Now we want to add up the areas of the little strips.

We take x_{0} = 0 to start and
then set x = kdx and substitute it into the integrand. Thus we get for our
integrand

(Eq'n 1)

Now we progress along the abscissa,
taking successive values of k from k = 0 to k = K such that Kdx = x_{1}.
We have now turned our problem into one of calculating the sum of the n-th power
of k for all integer values of k from 1 to K. That looks like a kind of Gaussian
sum, which we can analyze through the theorem of finite differences. As it
applies to the Gaussian sums of powers, that theorem gives us

(Eq'n 2)

in which "etcetera" represents a sequence of terms of progressively smaller powers of k. If we substitute that expression into the Riemannian sum of Equation 1, we get

(Eq'n 3)

If we now let dx tend toward zero, all of the terms on the right side of the equality sign, except the first, also tend toward zero: they become vanishingly small, so we can ignore them. We then substitute Kdx = x into the result and get

(Eq'n 4)

I must point out that because of the way in which we deduced that equation, we can only apply it for positive integer values of n. Our next step, after working through an example, will extend the range of applicable values of n for this integration.

**Example
Application:**

**The kinetic
energy gained by an accelerating body.**

We know from classical physics that the energy that a body of mass m gains equals the force exerted upon the body multiplied by the distance the body moves in the direction of the force. Newton's second law of motion also tells us that the body's response to the force equals the product of the body's mass and its acceleration (F = ma = mdv/dt). If the force acts in the x-direction and the body moves in that direction, then it moves the pseudo-infinitesimal distance dx at the speed v = dx/dt. The force then does the increment of work

(Eq'n 5)

If we integrate that expression indeterminately, we get

(Eq'n 6)

which expresses the kinetic energy that the body possesses by virtue of the velocity it gains from the force.

eeefff

Now I want to carry out an integration by parts. We have the standard form of the equation as

(Eq'n 7)

Let f = x^{n}, so that df = nx^{n-1}dx,
and let dg = x^{-n}dx. Equation 7 then becomes

(Eq'n 8)

The formula on the right side of the
equals sign must ultimately be proportional to the first power of x, so we must
have as true to mathematics that g = Kx^{1-n}. Equation 8 thus becomes

(Eq'n 9)

From that we infer that K = 1/(1-n) and that tells us, in turn, that

(Eq'n 10)

for all values of n except 1. We will have to integrate dx/x by a different process.

This gives us a good example of how we parlay our discoveries into extending the range of our knowledge.

**Example
Application:**

**To calculate the
work done upon a body that falls freely in a Newtonian gravitational field.**

A small body of mass m falls straight toward a much larger body of mass M. Again we know that the work done upon the smaller body equals the product of multiplying the force exerted upon it by the larger body's gravitational field and the distance the smaller body falls. By Newton's formula we have for the force exerted upon the smaller body

(Eq'n 11)

Because the falling body goes toward ever smaller radial distances, the increment of distance over which the gravitational force does work must be negative (-dr), so we have for the energy gain

(Eq'n 12)

eeefff

So far we have integrated x^{n}dx
for integer values of n. We may now ask whether the formula in Equation 4 also
holds true for fractional exponents, for the roots of x. We want to know the
form of F(x) in the equation

(Eq'n 13)

In this case let's apply the chain
rule. We define x = z^{n} and have dx = nz^{n-1}dz, which gives
us

(Eq'n 14)

Substituting back to recover our function of the variable x gives us then

(Eq'n 15)

which has the same form as does
Equation 4; that is, we increment the exponent by one and divide the result by
the new exponent.

**Example
Application:**

I haven't found a good example to use to illustrate the integration of a fractional power of a variable, but when I do find one, I'll post it here.

eeefff