Back to Contents
Imagine that someone has plotted a curve y = xn on a standard Cartesian grid. Between two given values of x (x0 and x1) 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 xn 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 x0 and x1. 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 x0 = 0 to start and then set x = kdx and substitute it into the integrand. Thus we get for our integrand
Now we progress along the abscissa, taking successive values of k from k = 0 to k = K such that Kdx = x1. 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
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
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
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.
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
If we integrate that expression indeterminately, we get
which expresses the kinetic energy that the body possesses by virtue of the velocity it gains from the force.
Now I want to carry out an integration by parts. We have the standard form of the equation as
Let f = xn, so that df = nxn-1dx, and let dg = x-ndx. Equation 7 then becomes
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 = Kx1-n. Equation 8 thus becomes
From that we infer that K = 1/(1-n) and that tells us, in turn, that
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.
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
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
So far we have integrated xndx 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
In this case let's apply the chain rule. We define x = zn and have dx = nzn-1dz, which gives us
Substituting back to recover our function of the variable x gives us then
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.
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.
Back to Contents