what is the interval of integration of $\int^a_b \frac {dx}{dt}dt $ for $\int^a_b dt$ the interval of integration is [a,b] so how about
$$\int^a_b \frac {dx}{dt}dt = \int^a_b dx $$
Does the interval of integral still [a,b] or it's changed. If it's, in what way it's changed?
 A: There seem to be multiple issues:


*

*If $a < b$, then $\displaystyle \int_{a}^{b} \dots$ denotes an integral over $[a, b]$. (The left-hand endpoint is the lower limit.)

*If $b < a$, then $\displaystyle \int_{a}^{b} \dots = -\int_{b}^{a} \dots$, and the interval of integration would be $[b, a]$.

*If $x$ is a continuously-differentiable function on $[a, b]$, the "substitution rule" or "change of variables formula" reads
$$
\int_{a}^{b} \frac{dx}{dt}\, dt = \int_{x(a)}^{x(b)} dx.
$$
(By the fundamental theorem, each side is equal to $x(b) - x(a)$.) Thus, the interval of integration on the right is $\bigl[x(a), x(b)\bigr]$ if $x(a) < x(b)$, or is $\bigl[x(b), x(a)\bigr]$ if $x(b) < x(a)$.
A: Using the Change of Variables Theorem:

Let $f:[a,b]\rightarrow \mathbb{R}$ a continuous function and $g:[c,d]\rightarrow \mathbb{R}$ with continuous derivative and $g([c,d])\subseteq [a,b]$. Then
  \begin{equation}
\int_{g(c)}^{g(d)}f(x)\,dx=\int_c^d f(g(t))g'(t)\,dt.
\end{equation}
  We say that the variable $x$ changed to $g(t)$.

In your case, we clearly have $f(x) = 1$. So you must know the function $g(t)$ to compute $\frac{dx}{dt}$, which is$\frac{dg(t)}{dt}$. Thus, if $g$ is a bijection,
\begin{equation}
\int_b^a\,dx=\int_{g^{-1}(b)}^{g^{-1}(a)}\frac{dx}{dt}\,dt.
\end{equation}
So, for example, if you defined $x=t$, then $g(t) = t$ and $\frac{dx}{dt} = 1$. Therefore
\begin{equation}
\int_{g^{-1}(b)}^{g^{-1}(a)}\frac{dx}{dt}\,dt= \int_b^a \,dt
\end{equation}
and your interval would not change. For other functions $g$, your interval may change.
A: The differentials $dt$ cancel out. So you are referring to variable $x$ only and its interval is $(a,b)$ and its integral is $(a-b)$.
