## Change of Variables

Just like we did for integration by parts we’re going to use the FToC as a mirror, but this time we’ll reflect the chain rule.

Remember that this rule tells us how to take the derivative of the composite in terms of the two derivatives and . Basically, the derivative of the product is the product of the derivatives, but we have to be careful where to evaluate each derivative. In Newton’s notation we write .

So now let’s take an integral of each side:

So if we’ve got an integrand that involves one function acting on an expression , it may be worth our while to see if we also see as a factor in the integrand, because we might then be able to reduce to integrating itself. We’re intentionally glossing over questions of where and must exist, have their ranges, be integrable or differentiable, and so on.

But let’s look a little closer at what’s really going on here. We say that the function is taking the interval and sending it to the interval . Actually, might send some points outside this image interval. Consider, for example, on the interval . We’re saying this goes to the interval , but clearly . What’s going on here.

We have to use the sign convention for intervals to understand this. First, let’s break up the domain interval into regions where is nonincreasing and where it’s nondecreasing. In this example, is nonincreasing on and nondecreasing on . The images of these monotonous intervals are exactly what we expect: and — how boring (sorry).

But now when we use the sign convention we see that our image interval is along with and . The first two of these two cancel out! In fact, anything outside the interval must be traversed an even number of times in opposite directions that will cancel each other out when we’re thinking about integrals.

So in this sense we’re using to tell us how to get our from to as we integrate . And as we change our variables from to we have to multiply by . Why? Because the derivative of is what tells us how to translate displacements in the domain of into displacements in the range of . That is, we think of a tiny little sliver of a rectangle in the integral over as having width . We need to multiply this by to get the width of a corresponding rectangle in the integral over .

Next time we’ll try to put this intuition onto a firmer ground not usually seen in calculus classes, and not often in advanced calculus classes either, these days.

[…] Riemann-Stieltjes Integral III Last Friday we explained the change of variables formula for Riemann integrals by using Riemann-Stieltjes integrals. Today let’s push it a little […]

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[…] which is just our change of variables formula. […]

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[…] The Riemann-Stieltjes Integral I Today I want to give a modification of the Riemann integral which helps give insight into the change of variables formula. […]

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[…] of Variables in Multiple Integrals I In the one-variable Riemann and Riemann-Stieltjes integrals, we had a “change of variables” formula. This let us […]

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[…] will proceed by induction on the dimension . For , this is exactly the one-dimensional change of variables formula, which we already know to be true. And so we’ll assume that it holds in all -dimensional […]

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[…] up, we have an analogue of the change of variables formula, which was closely tied to the chain rule in the first place. If and are totally -finite […]

Pingback by Corollaries of the Chain Rule « The Unapologetic Mathematician | July 13, 2010 |

The order of f and g is changed to g \circ f in [g \circ f]'(x) in the first paragraph and in the middle integral. The way it is written at the moment reminds me of a transpose (or of an adjoint for that matter), i.e. of an involution in general, but I guess that is just my imagination. Going back to your Radon-Nikodym posts now…

Comment by Evert De Nolf | June 21, 2011 |