Recall that when we set up definite integrals with Rieman sums, we
ended up with sums like, e.g., to do
int12 (
x2)
dx,
(1)(.25) + (1.56)(.25) + (2.25)(.25) + (3.06)(.25)
=(1 + 1.56 + 2.25 + 3.06)(.25)
In this case I did a left-hand sum with four intervals. Each term
gets a factor of (.25) -- the delta-
x for the sum. When
we go to an actual definite integral, we are letting the number of
rectangles get infinitely large, so that the delta-
x gets
infinitely small -- but in some way it's still there. The
dx in the definite integral is this infintesimal
delta-
x, which gives a sense of the width of each infinitely
small slice of the interval being integrated.
Now, what happens when we do a substitution? Consider
int (3x - 1) dx. We could subsitute
w=3x - 1. Then
dw/dx = 3, so
dw = 3dx, or dx = (1/3) dw.
What does this tell us? It tells us that
w changes 3 times as
fast as
x does -- in some sense, the little slice
dw of the interval is three times as wide as the little slice
of represented by
dx. Thus when we change the integral from
being in terms of
x to being in terms of
w, we inherit a
factor of (1/3):
int (3x - 1) dx =
int w (1/3)dw.
When the substitution isn't linear we get other factors of
x
and so forth, but this idea that we have to account for the fact that
the way
w changes is different than the way
x changes
remains.