Why does substitution work in antiderivatives?

I'm not entirely sure what I want to ask here, so please bear with me!

I think the explanation they give us in school for how finding the antiderivative by substitution works is: $$\int f(g(t))g'(t)dt=(Fg)(t)=F(g(t))=F(u)=\int f(u)du$$

But I never really understood the equality $F(u)=\int f(u)du$. Why can we behave as if the identity $u=g(t)$ doesn't exist, compute the integral $\int f(u)du$, and then 'substitute' $u$ in the result yielding the correct answer? In a sense, it feels like the variable $u$ switches from being 'meaningful' in $F(u)$ (where it stands for $g(t)$) and being insignificant in $\int f(u)du$, since as I perceive it the "variable name" here is meaningless and we could say $\int f(u)du = \int f(k)dk = \int f(x)dx = ...$ all the same. Another way to ask the same thing: why $(Fg)(t)=\int f(u)du$ and not, say, $(Fg)(t)=\int f(x)dx$? Where am I getting confused?

Maybe someone can explain this better than my high school teacher? Is there a 'formal' explanation for this? Thank you a lot!

I will try to deal with indefinite integrals. Justifications for the definite integral are more complicated.

Here is the standard argument, somewhat extended. We want to find $$\int f(g(x)) g'(x)\;dx.$$

Note that if we can find an antiderivative $F(x)$ of $f(x)$, then one antiderivative of $f(g(x)) g'(x)$ is $F(g(x))$. We can check this by differentiating. For $(F(g(x)))'= g'(x)F'(g(x))$ by the Chain Rule, since $F'(x)=f(x)$. It follows that $$\int f(g(x)) g'(x)\;dx=F(g(x))+C.$$

There was a deliberately unfortunate choice of variable when I wrote "an antiderivative $F(x)$ of $f(x)$." The $x$ here is playing a different role than the $x$ in the original integration question. It would have been better to choose a letter different than $x$, say $u$ for the sake of tradition, or $w$, or $z$, and then to write that we want to find an antiderivative $F(u)$ of $f(u)$. Then our integral is $F(g(x))+C$.

As a shortcut to this, note that the collection of all antiderivatives of $f(u)$ is by definition $\int f(u)\;du$. This is $F(u)+C$. Now substitute $g(x)$ for $u$.

As a shortcut to the shortcut, imagine like before that $u$ is a symbol, but at the same time it is, mysteriously, $g(x)$. Then the substitution step of the previous paragraph is unnecessary, and we simply get $$\int f(g(x)) g'(x)\;dx =\int f(u)\,du.$$

How can $u$ be treated as a variable for the sake of the formal manipulation by which we find $\int f(u)\;du$, and simultaneously as the function $g(x)$? One might, as you do, legitimately wonder about this. One point to be made in favour of it is that the above argument proves that the procedure will always give the right answer.

In the integration by substitution process, instead of saying that we will find an antiderivative of $f(u)$ and then substitute $g(x)$ for $u$, we write instead at the beginning "Let $u=g(x)$," then find $\int f(u)\;du$. But it all amounts to the same thing.

It gets more complicated. Soon we will treating $du$ (whatever that means) as an abbreviation for $g'(x)\;dx$. But we can check that the symbolic manipulations that we do are still the Chain Rule in disguise, and we can certainly always check whether our symbolic manipulations give the right answer.

Comment: The following general idea is useful. Differentiation is ordinarily easy, integration not so much. If after some work one has calculated an indefinite integral, one can quickly check whether the answer is right, by differentiating. This can save us from errors both major and minor. As a simple example, suppose we want $\int e^{-3x}\,dx$. I will write down a wrong answer, $3e^{-3x}+C$. Let's check whether this is right. Differentiate, using the Chain Rule. We get $-9e^{-3x}$. Oops, wrong answer! But we can see how to fix our wrong answer, by dividing that answer by $-9$.

• Just enough (not "more than enough") :) – The Chaz 2.0 Oct 31 '11 at 5:28

The «substitution rule» works precisely because the chain rule is true for derivatives.

Lets say you know that $F$ is an antiderivative of $f$.

You want to find an antiderivative of $f(g(t)) g'(t)$.

Well, the substitution rule sais that $F(g(t))$ is the answer. Why? Simply because of the chain rule:

$$[F(g(t))]'=F'(g(t)) g'(t)=f(g(t))g'(t) \,.$$

What we see is that if we know an antiderivative of $f$ we can find an antiderivative of $f(g(t)) g'(t)$.

This is exactly what the substitution rule sais, in different words: to find an antiderivative of $f(g(t)) g'(t)$ try to find an antiderivative of $f$ and use the above formula.

• My confusion is maybe more about notation than the actual idea here. In a stage of the equality chain we say $F(u)=\int f(u)du$, but why can we not just as well say $F(u)=\int f(k)dk$? What role does the equality $u=g(t)$ play in $\int f(u)du$? – david Oct 30 '11 at 21:11
• Minor typo, you probably mean $F(k)=\int f(k)\,dk$. If so, that's perfectly fine. Now we need to evaluate $F(k)$ at $k=g(t)$. The "let $u=g(t)$" and calculation of $F(u)$, using $u$ and not some other symbol, is to make sure we remember to substitute $g(t)$ at the end. – André Nicolas Oct 30 '11 at 21:20
• Hmm. I actually meant $F(u)=\int f(k)dk$. In a sense, I don't understand why $F(a)-F(b)=\int f(u)du=\int f(k)dk$ would be true if instead of an antiderivative we had an integral from a to b there, but at the same time $F(u)=\int f(k)dk$ isn't true... I guess? – david Oct 30 '11 at 21:28
• @david: There is some confusion here, between definite integrals $\int_a^b f(s)\;ds$, where the name of the dummy variable is irrelevant, and indefinite integrals (antiderivatives). Your question looked as if it was about antiderivatives. And $F(u)=\int f(k)\;dk$ is definitely not correct. – André Nicolas Oct 30 '11 at 21:40
• The question is definitely about antiderivatives, I just don't know what I'm trying to ask! Can you explain 'why' is $F(u)=\int f(k)dk$ not correct? Or I guess why $F(u)=F(k)$ is not correct (but is incorrect) in general, given that 'u' and 'k' don't strictly stand for a number? – david Oct 30 '11 at 21:43

If I understand correctly, the issue is due to unfortunately sloppy (standard) notation. Let's try tracing substitution through for a definite integral. Suppose we're looking at $$\int_a^b f(g(t)) g'(t) dt.$$ If we make the substitution $u=g(t)$, then $du=g'(t)dt$. However, there's more to do before plugging everything into the integral. The current limits of integration ($a$ and $b$) are the limits for $t$ and not $u$. But if $t$ goes from $a$ to $b$, then $u$ goes from $g(a)$ to $g(b)$. Thus our integral becomes $$\int_a^b f(g(t))dt = \int_{g(a)}^{g(b)} f(u) du,$$ which isn't as absurd as claiming that the indefinite integral $\int f(g(t))g'(t)dt$, which is a function of t, equals the indefinite integral $\int f(u)du$, which is a function of $u$. (To turn these definite integrals into something that looks more like a function, replace the upper limit $b$ by $x$)

If you're ever trying to prove an identity/rule for integration, if you look at definite integrals (e.g. $\int_a^x f(t)dt$, which is a function of $x$) instead of indefinite integrals (e.g. \int f(x)dx) then much of the confusion should go away.

• That's why I always write the substitution formula for indefinite integrals in the form $$\int f\bigl(g(t)\bigr)\ g'(t)\ dt\ =\ \int f(u)\ du\Biggr|_{u:=g(t)}\ .$$ – Christian Blatter Oct 31 '11 at 10:08

What you are looking for is an intuitive interpretation of the formula for integration by substitution. I will try to provide it in rather simple words. Consider the example $$\displaystyle F(x)=\int 2xe^{x^2}\, dx$$.

You can easily infer by the inverse chain rule $$\displaystyle F(x)=\int 2xe^{x^2}\, dx=e^{x^2}+c$$. However, substituting $$u=x^2$$ (or more specifically $$x=\sqrt{u}$$) gives $$\displaystyle F(u)=\int 2\sqrt{u}e^{u}\, du$$ which leads not to the correct result. So substituting alone is not enough.

Actually, the difference between the function in terms of $$x$$ and the substituted (but invalid) one in terms of $$u$$ is as follows: It can be seen from the plot that the areas under the curves differ a lot and, hence, the antiderivatives will differ as well. If $$x$$ moves $$\Delta x$$ then $$u$$ will move $$\Delta u$$ where for most corresponding intervals $$\Delta x<>\Delta u$$. In the plot, the density reflects this fact: For example, some intervals of the function on the top map to smaller intervals on the substituted function indicating "a higher density". Let's think of the density as $$\Delta x/ \Delta u$$.

A high density here means also that some area below the graph gets lost, because the graph becomes "squeezed". And it gets "stretched" (resulting in a too great area below the graph) for low densities. You can fix this behaviour by multiplying by the density and therefore virtually add area below the graph for high densities (density > 1) and virtually remove space at low densities (density < 1). In calculus, we need to define the infinitesimal density as $$dx/du$$ in order to represent it at a single point. But $$dx/du$$ is a differential quotient:

$$dx/du=x'(u)=\frac{1}{2\sqrt{u}}$$

And so,

$$\displaystyle \int 2xe^{x^2}\, dx=\int 2\sqrt{u}e^{u} \left ( \frac{dx}{du}\right ) \, du$$

$$\displaystyle \int 2\sqrt{u}e^{u} \left ( \frac{dx}{du}\right ) \, du=\int 2\sqrt{u}e^{u} \frac{1}{2\sqrt{u}}\, du=\int e^{u} \, du$$

$$\displaystyle \int e^{u} \, du=e^{u}+c=e^{x^2}+c$$

I also struggled with this topic when I was younger and hope that the explanation helps a lot of people. You might also want to have a look into the full article on Insight Things.