TITLE: Why does substitution work in antiderivatives?
QUESTION [12 upvotes]: 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!

REPLY [6 votes]: 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.