Video Lecture

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Recall from our derivative rules for a composition of functions, say $f(g(x))$, we had to employ the chainrule. In particular:

$$\frac {d}{\dx } \left [ f(g(x))\right ] = f'(g(x)) \cdot g'(x)$$

Our goal is to figure out how to reverse this process. In particular, we want to be able to recognize, and then undo, the end result of the chain rule. Consider, for example, the following derivative:

$$\frac {d}{\dx }e^{x^2}= e^{x^2}\cdot \frac {d}{\dx }\left [x^2\right ] = 2xe^{x^2}$$

Computing this derivative is straightforward with the chain rule, but reversing this process becomes a bit challenging. Just looking at $2xe^{x^2}$, it may not be obvious what the antiderivative is (at least, if we hadn’t just seen the derivative that generated it). Even if we suspect that it is something close to $e^{x^2}$, it’s not clear if that is exactly right or if it needs modification, at least without trying it out. This is where it helps to introduce a technique that, although technically from precalculus, probably hasn’t been used (or even seen) much prior to this - the $u$-substitution.

Before we (re)-introduce the $u$-substitution idea, it’s helpful to revisit the chain rule notation for our specific setting - the exponential. In essence, we want to think of “the derivative of $e^{\text {stuff}}$” as $e^{\text {stuff}} \cdot \frac {d}{\dx } \left (\text {stuff}\right )$, meaning that we want to think of the derivative as itself, times the derivative of the exponent. The whole idea of the $u$-substitution, is to use $u$ to represent that “stuff” - the part that the chain rule is differentiating.

Let’s return to our example - computing the indefinite integral $\displaystyle \int 2xe^{x^2}\dx$. We may suspect that we need to reverse a chain rule to get the antiderivative. Ideally we would notice a link between the exponent, $x^2$, and the $2x$ out front. Once we suspect that we need a u-substitution, we can try it and see what happens. Specifically, we want to let $u$ be the “inside function”, in this case the $x^2$. But remember, the whole point of the chain rule is that we have to multiply the entire term by the derivative of the inside function - which means we need to account for this as part of our $u$-sub process. This means we will also need to calculate the derivative of $u$ - in this case, $\du = 2x\dx$.

The “$\dx$” here is deceptively important. Remember that the integral uses $\dx$ at the end to tell you what variable you are trying to integrate. But we are changing that variable to $u$, which means we can’t try to integrate until all the variables are converted to the same variable! This includes the $\dx$. A good rule of thumb is that you don’t want to compute the integral (at least in this class) until you only have one variable present - if you have more than one letter in the integral, you need to fix that. By way of demonstration, let’s see what happens if we only do the $u$ part of the substitution:

$$\int 2xe^{x^2}\dx = \int 2xe^u\dx$$

Since we have a mix of $x$ and $u$, and the integral is still a $\dx$ integral, we can’t do the antiderivative (yet). But this is actually a good thing because we need to account for the $g'(x)$ part of the $f'(g(x))g'(x)$ chain rule form. In other words, we need to figure out what to do with the $u'$ we calculated earlier.

The easiest way to avoid errors is to solve for “$\dx$” in that $u'$ calculation, then substitute out the “$\dx$” in the integral. This is technically using the differential form of the derivative and integral, but we don’t need to worry about that. We calculated that $\du = 2x\dx$, so solving for $\dx$ we get $\dx = \frac {\du }{2x}$. Substituting this in for the $\dx$ in our integral gets us:

$$\int 2xe^{x^2}\dx = \int 2xe^u\dx = \int 2xe^u\frac {\du }{2x} = \int e^u \du$$

Now that we have performed all the substitutions, we have an integral where the only variable is $u$, and it’s a $\du$ integral, so we can calculate the indefinite integral, which is just $e^u + C$. Notice here that once all the letters are the same (in our case they are all $u$) we can treat it like any other integral - the fact that it’s a $u$ doesn’t really matter to the integration process (again, as long as all the letters in the integrand are the same letter!) But then we need to put it back in terms of $x$ using the definition of $u$ that we recorded earlier - which gives us $e^{x^2} + C$.

There are a couple notes to make about this method. First, by far the most common error students make is to forget to substitute out $\dx$. This is exacerbated by the fact that a lot of students don’t write the $\dx$ in the first place as it seems redundant, but it turns out to be incredibly important here as you need to keep track of whether you have done all the substitutions necessary. Moreover, forgetting to substitute out the $\dx$ tends to actually make the problem much harder, not easier, so it’s a double shot in the foot when students make the error as it makes the problem much harder to solve, and even if they manage to solve it, it’s wrong!

The second thing to keep in mind is that we call this technique “$u$-substitution” but there is nothing special about the letter $u$ here. Indeed, there are times when you can, and should, use multiple substitutions in the same problem (we have another segment on just this idea!) and if you used the same letter every time it would go... very badly. The general rule is to make sure you use a new letter every time you do a substitution - a letter that has not yet shown up at any stage of the problem. And never, ever, use the same letter for a substitution that is already in the integral at the time you are doing the substitution. This is almost certainly going to lead to confusion and errors (not to mention it’s mathematically incorrect) if you do!

We’ve introduced the idea of the $u$-substitution and common errors when trying to apply it, as a way to reverse the chain rule. This is another technique which requires extensive practice to get better at, so students are advised to do extensive practice for this technique to improve their speed, skill, and ability to recognize when and where to use $u$-substitutions. Much like the chain rule, $u$-substitutions become fairly ubiquitous moving forward, and for the same reason - function composition is pretty much everywhere.