Why are the extremes of integration considered values of $x$ when computing an integral by substitution?

Question

My Calculus instructor showed in a video (a screenshot of the calculation is shown below) how to compute an integral by substitution, and midway through the solution, he claimed that the extremes of integration $1$ and $2$ are also values of $x$, and must be substituted as follows:

$$x = 1 \to u = 2 \qquad x = 2 \to u = 5$$

I understand how he got the solution, but he didn't explain why they are considered values of $x$, and I can't figure it out on my own.

Why are extremes of integration considered values of $x$?

Answer 1

In an integration, such as $\int_{1}^{2} x(x^2 + 1)^{100}dx$ in your case, the "$dx$" states that it's summing all increments of the function, i.e., $x(x^2+1)^{100}$, for $x$ from the lower limit, i.e., $1$, to the upper limit, i.e., $2$. As such, the lower and upper limits are considered to be the extreme values of $x$ in the integration, and thus are the corresponding extreme values of any new variable determined by using a substitution.

Answer 2

The integral is made with respect to $ x $, so when introducing a new variable, $ u $, this new function that depends on $ u $ must have its own integration interval.

The general statement is the following : Substitution formula. If $f$ and $g'$ are continuous functions, then $$\int_a^b f(g(x))g'(x)dx= \int_{g(a)}^{g(b)} f(u)du$$ (here, take $u=g(x)$)

Answer 3

You consider the diffeomorphism $u:[1,2]\rightarrow[2,5],x\mapsto x^2+1$ and use the transformation formula $$\int_1^2f(u(x))u^{'}(x)dx =\int_2^5 f(u)du$$ where You think of $u^{'}$ as a deformation factor over the "flattened" set $[1,2]$. It is important for this that the mapping $u$ has a differentiable inverse.