Integration confusion: $\int_0^t\frac{dx}{dt}\,dt=\int_0^t\,dx=[x]_0^t=\cdots$? What is $\displaystyle\int_0^t \frac{dx}{dt}$ $dt$ ?
Where $x=x(t)$
I thought that the $dt$'s cancelled to give $\displaystyle\int_0^t dx$=$[x]^t_0$
Now here is the answer $t-0=t$ or
$x(t)-x(0)$
Appparently it is the latter but I do not understand why.
 A: When you "cancel" the $dt$'s, you are actually performing a substitution.
And when substituting, you have to change the integration bounds accordingly:
$$\int_0^T \frac{dx}{dt} dt = \int_{x(0)}^{x(T)} dx = x(T) - x(0)$$
A: First, since we want to avoid confusion, let's not abuse notation by having $t$ simultaneously a dummy variable and a fixed limit of integration.  Instead let's integrate
$$\int_a^b \dfrac{dx}{dt}\,dt,$$
as the variability of the integration limits should be irrelevant for your question.  The "cancelling" you refer to does not strictly make sense, in that you are not cancelling by division of real numbers. In order to justify saying that the integral above is equal to $\int_a^b dx,$ for one thing we have to know what that integral means.  As written, it is misleadingly ambiguous, because $x$ is a function of $t$, and the $a$ and $b$ are $t$ values, not $x$ values.  We might remedy this by writing $\int_{t=a}^{t=b}dx$, or $\int_a^b dx(t)$.  If $x(t)$ is of bounded variation and right continuous, then the Lebesgue–Stieltjes integral $\int_a^b dx(t)$ is indeed defined and equal to $x(b) - x(a)$.  
On the other hand, if $x$ is continuously differentiable, it follows immediately from the fundamental theorem of calculus that $\int_a^b\dfrac{dx}{dt}\,dt =x(b) - x(a)$, so they do agree, and the "cancellation" works.   
