Dummy Variables and variables that change a funtion I am reading notes for my Calculus 2 class and am confused why these two functions are not equivalent
"
$x$ is not a dummy variable, for example,
$$\int 2xdx = x^2+C$$
and 
$$\int 2tdt = t^2 + C$$
are functions of different variables, so
they are not equal. 
"
Is it because the arbitrary C's can be different?
 A: You're right on both counts; the arbitrary $C$'s can be different, and $x$ and $t$ can be different. The author is referring only to $x$ and $t$, from what I can tell. I would say that the way the book puts this is a bit confusing. (At least, it confused me). Consider
$$
\begin{aligned}
f(x) &= x^2 + 1\\
g(t) &= t^2 + 1\\
\end{aligned}
$$
These two functions are equal, because the $x$'s and $t$'s are what your book would call "dummy variables" -- variables that exist nowhere except inside the function.
When you use definite integration, the variable you're integrating over goes away, and you get a number. When you use indefinite integration, the variable doesn't go away. So the author says "they are not equal" because their values still depend on $x$ and $t$, which might not be equal. But they are equal (or "equivalent") as functions.
For example, $f$ and $g$ are equal as functions, but their values are different if into $f$ I insert $x = 1$ and into $g$ I insert $t = 2$.
A separate issue, which you are right to bring up, is that the $C$ in the first equation is not the same $C$ in the second.
Here's how I think about indefinite vs. definite integration. When viewed as functions in their own right, machines with inputs and outputs, they differ totally. Indefinite integration takes in a function and spits out a lot of functions (all of its antiderivatives, differing from each other by choice of constant $C$):
$$\int:: \textbf{a function} \rightarrow \textbf{all of its antiderivatives}$$
while definite integration takes in a function, a start point, and an end point, and gives you a number:
$$\int_a^b:: (\textbf{a function}, a, b) \rightarrow \textbf{a number}$$
What the author is trying to express is that indefinite integration gives you back a function, while definite integration gives you back a number.
A: As discussed in the comments, the two expressions are likely not stated to be equal because the initial conditions are not know. For example, take the trig identity $\tan^2 x + 1 = \sec^2 x$. If we were to take the derivative of these functions we would get the same answer; however, once we take the antiderivative again we could get that $\tan^2 x + C_1 = \sec^2 x + C_2$. Information concerning 
constants is lost during integration.
A: Either $x$ or $t$ is just a token representing numbers, so they are equivalent.
Let's take the expression $x+1$ (or $t+1$) for instance: Letting $x=1.5$ is equivalent to letting $t=1.5$, because both of them make the expression evaluate to $1.5+1=2.5$.
A: What we see here is an abuse of notation ingrained in our way of doing calculus since centuries. If ${\rm sq}:t\mapsto t^2$ is the squaring function on ${\mathbb R}$ nobody would denote the set of all functions differing from ${\rm sq}$ by an additive constant by $\{{\rm sq}+c\>|\>c\in{\mathbb R}\}$. Instead we all write $t^2+C$, or $x^2+C$, for this set of functions.
Similarly, given any function  $x\mapsto f(x)$ $(x\in I)$, e.g. $f(x):=2x$ $(x\in{\mathbb R})$, the typographical picture $\int f(x)\>dx$ by definition denotes the set of all functions $$t\mapsto F(t)\quad(t\in I)\tag{1}$$ satisfying $F'(x)=f(x)$ for all $x\in I$. We might as well have written $u\mapsto F(u)$ $(u\in I)$ in $(1)$. Indeed it is customary in integral tables to denote the independent variable on the right hand side again by $x$.
