# The Unapologetic Mathematician

## 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 | Analysis, Calculus | 7 Comments