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?
There seem to be multiple issues:


If $a &lt; b$, then $\displaystyle \int_{a}^{b} \dots$ denotes an integral over $[a, b]$. (The left-hand endpoint is the lower limit.)
If $b &lt; a$, 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 $x(a) &lt; x(b)$, or is $\bigl[x(b), x(a)\bigr]$ if $x(b) &lt; x(a)$.
The differentials $dt$ cancel out. So you are referring to variable $x$ only and its interval is $(a,b)$ and its integral is $(a-b)$.