Integral without differential It is clear to me what the meaning of this is:
$$\int_a^b f(x)dx$$
Also, I know that
$$\int_a^b \frac{df}{dx}dx=f(b)-f(a)$$
However, I don't quite understand what the meaning of this is, namely if we treat $dx$ as a variable and simplify the fraction:
$$\int_a^b df= f(b)-f(a) ??$$
And generally, what is the meaning of this? (without the $dx$)
$$\int_a^bf(x)$$
 A: This question deserves a long answer discussing "differentials" and whether you can think of $dx$ as the denominator of a fraction and "cancel it". There are several such discussions elsewhere on this site. I'll try a short intuitive answer here.
Leibniz thought of 
$$
\int_a^b f(x)dx
$$
as adding up infinitely many infinitely thin rectangles, one for each $x$ with height $f(x)$ and base the infinitely small number $dx$.
Then he observed that if you had a function $F$ whose derivative was $f$ then you could approximate the change $F(x+dx) - F(x)$  by  $f(x)dx$ . (In modern terms that's one way to think of the derivative as the limit of the difference quotient.) For him it was natural to write that change as $dF(x) = f(x)dx$ or just $dF = fdx$ and the difference quotient as $f(x) = dF/dx$, which is what we now call Leibniz notation for the derivative.
Now if you add up all the changes in $F$ you get the change from the beginning to the end of the interval, so
$$
\int_a^b dF = 
\int_a^b \frac{dF}{dx}dx=F(b)-F(a).
$$
This is of course one way to look at the fundamental theorem of calculus, relating derivates and integrals.
This posting may provide more help:
Why can't the second fundamental theorem of calculus be proved in just two lines?
A: Generally speaking, integral is the analog of sum for a continuous variable i.e. the variable that can take infinitely many values over the specified interval. Let's say you have an infinitely dense grid of all these values over the interval [a, b] with an infinitesimal step size $dx$. Then this expression
$$\int_a^b f(x)dx$$
basically says: "evaluate function at each point in your grid, multiply it by the step size and add up all these values".
Even though you have infinitely many terms, this sum actually converges to a specific value. Why? 
Think about having a finite grid with finite $dx$. Each $f(x_i)dx$ term is the area of a rectangle with height $f(x_i)$ and width $dx$. Now, if you start decreasing step size $dx$ you will eventually hit the point when $f(x_i)\approx f(x_{i+1})$ and so $f(x_i)\frac{dx}{2}+f(x_i+\frac{dx}{2})\frac{dx}{2}\approx f(x_i)dx$ so further decrease in step size $dx$ will not change the value of the integral.
Now, your question is what does this thing mean:
$$\int_a^bf(x)$$
Well, it's still a sum over all possible values in between a and b but without $dx$ term it is unbounded and so the denser the grid the higher the value of this integral is. So infinitely dense grid would yield infinitely high value $\pm\infty$ (or 0, if the function is antisymmetric with respect to the midpoint of the interval).
A: Although $\int_{a}^{b} f(x)$ may be poor notation, one may encounter $$ \int_{a}^{b} f $$
where the real interval $[a,b]$ is in the domain of $f$. This notation might be preferable since it's more concise, for one. A deeper reason is that it doesn't arbitrarily prefer a variable to represent $f$. $f$ can be defined as $f(x)$, or $f(t)$, or in many other ways, but the choice of $x$, $t$, etc. doesn't matter, except to suggest what the inputs to $f$ represent in your particular application of the integral.
The integral of $f$ over $[a,b]$ may also be written $$ \int_{[a,b]} f $$
which is a specific case of $$ \int_{\Omega} f $$
where $\Omega$ is again in the domain of $f$, but is also "sufficiently nice" to integrate over. An example is $\Omega = [0, 1] \times [0, 1]$, a unit square, in which case we might have $f : \Omega \rightarrow \mathbb{R}$ and the integral $\int_{\Omega} f$ is the signed volume "under" the surface $(\Omega, f(\Omega))$.
