Integral notation confusion - $x$ vs. $t$ One of our worksheets contains the following integrals (areas) to find:
$$\int_0^x 1 \, dx$$
$$\int_0^x x \, dx$$
I am skeptical. Although this is our first day learning about integrals in class, I already have a sense that perhaps the variable in the bound of integration shouldn't be the same as that appearing in the function being integrated. 
If someone wrote... $$\int_0^x 1 \, dt$$ ... I know perfectly well that this would be equal to $x$ -- I could figure that out presumably by drawing $y=f(t)=1$ in the $y,t$ coordinate plane and using basic geometry. But it seems to me that the integral I noted at the very top of my post is different -- perhaps even nonsensical. Am I correct about that, i.e. did my teacher mess up when writing the questions? Any explanation is greatly appreciated. 
 A: They most likely meant, as you correctly guess,
$$
\int_0^xt\,dt.
$$
Notice the completely different 
$$
\int_0^x x\,dt=x\int_0^x1\,dt=x^2. 
$$
The use of $dx$ in an integral tells you the "variable of integration". It plays the same role as the index in a sum. So writing $\int_0^x x\,dx$ is akin to write 
$$\tag1
\sum_{n=0}^n n.
$$
One can possibly abstract the two separate roles from $n$ and say that, since $\sum_{n=0}^Mn=\frac{M(M+1)}2$, then $(1)$ should be $\frac{n(n+1)}2$. But consider what happens if you now write 
$$\tag2
n\,\sum_{n=0}^nn.
$$
Should $(2)$ be interpreted as $\sum_{n=0}^n n^2=\frac{n(n+1)(2n+1)}6$, or $n\left(\sum_{n=0}^n n\right)=n(\frac{n(n+1)}2)$? The same problem occurs with writing $\int_0^xx\,dx$. It make the expressions like $x\int_0^x\,x\,dx$ unnecessarily ambiguous.
A: Technically, the $dx$ in the first integral "binds" the variable name $x$ to the dummy variable of the integral.
That is, it makes a local definition of the symbol $x$.
This definition applies only within the integrand of the integral, nowhere else--certainly not outside the integral, and not even to the boundary values of the integral.
If you program, you might compare $dx$ to a declaration of a local variable $x$ inside the definition of a function.
No code outside the function can set the value of $x,$ and the value of $x$ inside the function does not directly control the value of any other variable named $x$ outside the function definition.
Since $x$ is a dummy variable, defined only within the integrand by the notation $dx,$ we can substitute any other name of any other variable as long as that name doesn't already appear in the integrand.
So for example, when you see
$$\int_0^x x \, dx $$
you can think of
$$\int_0^x t \, dt. $$
Technically, they're the same.
I say "technically", because in more practical terms, writing
$\int_0^x x\,dx$ is bad style.
If you know it is the same thing as $\int_0^x t\,dt$,
why not write the $dt$ version and avoid the inevitable confusion that results from using the same variable name for two different things in two different places?
Human beings are not good at that sort of distinction; it gets in the way of doing real math.
And if you don't know $\int_0^x x\,dx$ is the same as $\int_0^x t\,dt$,  you have no business writing it.
In short, I think you have a legitimate complaint that notation like this shouldn't be on your worksheet. In the meantime, 
solve $\int_0^x 1\,dx$ as $\int_0^x 1\,dt$
and solve $\int_0^x x\,dx$ as $\int_0^x t\,dt$.
A: There is a distinction between definite and indefinite integrals.  Indefinite integrals are integrals that have no values for lower and upper bounds of integration.   Alternatively, in your example the indefinite integral is an integral having 0 as lower bound and x as upper bound. 
\begin{align}
\int_0^x t dt = \int_0^x x dx = \int x dx = \frac{x^2}{2}  
\end{align}
note that $t$ is a dummy variable.  It can be replaced by any other variable, including $x$. 
I'm assuming that the constant of integration is 0
http://mathworld.wolfram.com/IndefiniteIntegral.html
More generally
\begin{align}
\int_C^x t dt = \int_C^x x dx = \int x dx  = \frac{x^2}{2} - D  
\end{align}
where
\begin{align}
D =  \frac{C^2}{2}
\end{align}
