The Unapologetic Mathematician

Mathematics for the interested outsider

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 z=f(g(x)) in terms of the two derivatives z=f(y) and y=g(x). 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 \left[g\circ f\right]'(x)=f'(g(x))g'(x).

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

\displaystyle\int\limits_a^bf'(g(x))g'(x)dx=\int\limits_a^b\left[g\circ f\right]'(x)dx=f(g(b))-f(g(a))=\int\limits_{g(a)}^{g(b)}f'(y)dy

So if we’ve got an integrand that involves one function f acting on an expression g(x), it may be worth our while to see if we also see g'(x) as a factor in the integrand, because we might then be able to reduce to integrating f itself. We’re intentionally glossing over questions of where f and g 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 g is taking the interval x\in\left[a,b\right] and sending it to the interval y\in\left[g(a),g(b)\right]. Actually, g might send some points outside this image interval. Consider, for example, g(x)=x^2 on the interval x\in\left[-1,2\right]. We’re saying this goes to the interval y\in\left[1,4\right], but clearly g(0)=0. 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 g is nonincreasing and where it’s nondecreasing. In this example, g is nonincreasing on \left[-1,0\right] and nondecreasing on \left[0,2\right]. The images of these monotonous intervals are exactly what we expect: \left[1,0\right] and \left[0,4\right] — how boring (sorry).

But now when we use the sign convention we see that our image interval is \left[0,1\right]^- along with \left[0,1\right]^+ and \left[1,4\right]. The first two of these two cancel out! In fact, anything outside the interval \left[g(a),g(b)\right] 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 y=g(x) to tell us how to get our y from g(a) to g(b) as we integrate f(y). And as we change our variables from y to x we have to multiply by g'(x). Why? Because the derivative of g is what tells us how to translate displacements in the domain of g into displacements in the range of g. That is, we think of a tiny little sliver of a rectangle in the integral over x as having width dx. We need to multiply this dx by g'(x) to get the width dy of a corresponding rectangle in the integral over y.

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.

February 27, 2008 - Posted by John Armstrong | Analysis, Calculus | | 3 Comments

3 Comments »

  1. [...] 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 [...]

    Pingback by The Riemann-Stieltjes Integral III « The Unapologetic Mathematician | March 6, 2008

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

    Pingback by The Riemann-Stieltjes Integral II « The Unapologetic Mathematician | March 6, 2008

  3. [...] 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. [...]

    Pingback by The Riemann-Stieltjes Integral I « The Unapologetic Mathematician | March 6, 2008

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