1—Basic Stuff 17To improve the sum, keep adding more and more points to the partition so that in the limit all the intervals
xk−xk− 1 → 0. This limit is called the Riemann-Stieljes integral,
∫
f dα (22)What’s the big deal? Can’t I just say thatdα=α′dxand then I have just the ordinary integral
∫
f(x)α′(x)dx?Sometimes you can, but what ifαisn’t differentiable? Suppose that it has a step or several steps? The derivative
isn’t defined, but this Riemann-Stieljes integral still makes perfectly good sense.
An example. A very thin rod of lengthLis placed on thex-axis with one end at the origin. It has a uniform
linear mass densityλandan added point massm 0 atx= 3L/ 4. (a piece of chewing gum?) Letm(x)be the
function defined as
m(x) =
(
the amount of mass at coordinates≤x)
=
{
λx ( 0 ≤x < 3 L/ 4 )
λx+m 0 ( 3 L/ 4 ≤x≤L)This is of course discontinuous.
m(x)
x
The coordinate of the center of mass is∫
xdm/∫
dm. The total mass in the denominator ism 0 +λL,
and I’ll go through the details to evaluate the numerator, attempting to solidify the ideas that form this integral.
Suppose you divide the lengthLinto 10 equal pieces, then
xk=kL/ 10 , (k= 0, 1 ,...,10) and ∆mk={
λL/ 10 (k 6 = 8)
λL/10 +m 0 (k= 8)