Of all the techniques of integration that we have learned, the most
powerful, and the simplest, is u-substitution. If u=g(x), then
∫baf(g(x))g′(x)dx=∫g(b)g(a)f(u)du.
You probably learned this method in terms of anti-derivatives, as a
way to turn the chain rule inside out. That sort of
reasoning doesn't generalize very well to multiple dimensions. Instead,
let's analyze u-substitution from the definition of an integral.
By definition, ∫baf(g(x))g′(x)dx is the limit of a sum
N∑i=1f(g(x∗i))g′(x∗i)Δix,
where
we have broken the interval from a to b into N pieces, x∗i is a
point in the i-th piece, and Δix
is the length of the i-th piece. If we
let u∗i=g(x∗i), then Δiu≈g′(x∗i)Δix, and
our sum is approximately
N∑i=1f(u∗i)Δiu.
The limit of this sum is, by definition, the definite integral of f(u)du
from the starting value of u (namely g(a)) to the ending value of u
(g(b)).
We have converted an integral in x-space into an integral in u-space,
but the functions being integrated are not the same. In one case we integrate
f(u). In the other case we integrate f(g(x)) times a distortion factor
g′(x). This factor corrects for Δu being a different size
than Δx.
We also learned about inverse u-substitution, especially in the
context of trig substitutions. If x=g(u) (rather than the other way
around), then
∫baf(x)dx=∫βαf(g(u))g′(u)du,
where a=g(α) and b=g(β). That's the same rule we had before,
only with the roles of x and u reversed. We say that g is a
mapping from u-space to x-space, sending the interval
[α,β] to the interval [a,b],
and the distortion factor g′(u) is called the Jacobian of this
mapping. When we write the equation
dx=g′(u)du,
we are saying that the length of a short interval in x-space is
g′(u) longer than the interval of the corresponding interval in u-space.
Example: Compute ∫100e−x/5dx.
Solution: We let x=g(u)=5u, so e−x/5=e−u and dx=5du.
The map g sends the interval [0,2] in u-space to the interval
[0,10] in x-space, |
and g′(u)=5 tells us that the x-interval is
5 times bigger than the u-interval, so
∫100e−x/5dx=∫205e−udu=5(1−e−2).
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