You are correct to be thinking in terms of areas. The comments
about $(t)$ and $dt$ are mathematically correct, but perhaps
not useful at your mathematical level. (This is going to be a long
answer, stop when you have your answer or it starts to get too
mathematical.)
The total area under the density function of a random variable $X$ is $1.$ The probability that $X$ lies in a particular interval $(a, b]$,
is written as $P(a < X \le b)$. It is the area beneath the
density curve $f_X$ above the interval $(a,b].$ The notation
$\int_a^b f_X(t)\,dt$ is the way mathematicians write that area.
The probability that the random variable $X$ is smaller than the
number $x$ is written:
$$P(X \le x) = P(-\infty < X \le x) = \int_{-\infty}^x f_X(t)\,dt.$$
[The 'variable of integration' $t$ is part of the process of
numerical evaluation of the probability, not a part of the answer.
The integral could just as well be written $\int_{-\infty}^x f_X(\xi)\, d\xi$ or $\int_{-\infty}^x f_X(Q)\, dQ,$ or with any other symbol. Hence the term "dummy variable."]
Once you have the CDF, you can use it to find probabilities of
various intervals. For example,
$$P(0 < X \le 1/2) = F_X(1/2) - F_X(0) = \int_0^{1/2} f(t)\,dt.$$
Here is a specific example: Suppose the density function of $X$
is $f_X(x) = 2x,$ for $x$ between 0 and 1, and $f_X(x) = 0$ for
other values of $x.$ You can draw a sketch of it: mainly it looks
like a right triangle with vertices at $(0,1), (1,2),$ and $(1,0).$
You can check that it encloses total area $1$--as a density function must.
If you want to find $P(0 < X \le 1/2)$ for this simple case,
you can see that it is equal to 1/4. The area of the small
triangle under $f_X(x)$ and above $(0, 1/2)$ is half its base
times its height: $(1/2)(1/2)(1) = 1/4.$
If you know some calculus, you can find that the CDF of this
random variable $X$ is $F_X(x) = x^2,$ for $0 < x \le 1.$
Then $$P(0 < X \le 1/2) = F_X(1/2) - F_X(0) = (1/2)^2 - 0 = 1/4.$$
Notes: (1) In this simple example, calculus isn't necessary
because you can find areas under $f_X$ using elementary geometry.
(2) Using $<$ on one side of inequalities and $\le$ on the other
is just a 'convention' (habit). The $\le$ could just as well be $<$
because there is zero probability at any individual point.
(3) In many cases (such as the simple example above) you never have to deal with $-\infty$ because
all of the probability is in some finite interval.
(4) In some examples, using calculus is impossible and other
methods need to be used to find areas under density curves. The
famous normal distribution ("bell-shaped curve") is an example of
this. Instead of finding $F$, you use tables of probabilities, a
calculator, or statistical software.
Yes. That's correct. A PDF is a probability density function. It is stating the probability of a particular value coming out. Taking this analogy to a discrete distribution, the PDF of a 6-sided die is: $[x<1:0,x=1:\frac{1}{6},x=2:\frac{1}{6},x=3:\frac{1}{6},x=4:\frac{1}{6},x=5:\frac{1}{6},x=6:\frac{1}{6},x>6:0]$. For a continuous probability distribution, you can't really use the PDF directly, since the probability of an infinitesimally thin slice of the PDF being selected is intuitively zero.
That's where the cumulative density function, or CDF, comes it. It is a measure of how likely the value is to be less than some arbitrary value (which we pick). For a discrete case, you start with the first possible value, and add all the entries in the PDF up to the value of interest:
$$CDF=\sum PDF \rightarrow [x<1:0,x<2:\frac{1}{6},x<3:\frac{2}{6},x<4:\frac{3}{6},x<5:\frac{4}{6},x<6:\frac{5}{6},x\geq 6:\frac{6}{6}]$$
Notice how the final value of the CDF is $1$. This is expected, since every possible outcome of rolling a 6-sided die is less than or equal to 6.
Now let's go back to the continuous probability distribution. In this case, we don't have a finite set of options for the answer to be, so we can't constrain $X$. Thus, we start from $-\infty$, since that encompasses everything to the left of the chosen $x$. As you should be aware from calculus, the integral is to continuous functions what a sum is to discrete functions - loosely. The value of a CDF is that you can use it to determine the probability of the number falling within a specific range as follows:
$$F(a\leq X \leq b) = F(X \leq b) - F(X \leq a) = \int_{-\infty}^{b} f(x)dx - \int_{-\infty}^{a} f(x)dx = \int_{a}^{b} f(x)dx$$
Best Answer
The former integral is a Stieltjes integral. See this, for example, in particular the section "Application to probability theory".
In brief, the integral $I_1 = \int {xdF(x)} $ is a generalization of $I_2 = \int {xf(x)dx} $. $I_2$ can be used only when the distribution is absolutely continuous, that is, has a probability density function. $I_1$ can be used for any distribution, even if it is discrete or continuous singular (provided the expectation is well-defined, that is $\int {|x|dF(x)} < \infty $). Note that if $F'(x) = f(x)$, then $\frac{d}{{dx}}F(x) = f(x)$ gives rise to $dF(x) = f(x)dx$.
For a thorough account of this topic, see, for example, this.
EDIT:
An example. Suppose that a random variable $X$ has CDF $F$ such that $F'(x) = f_1 (x)$ for $x < a$, $F'(x) = f_2 (x)$ for $x > a$, and $F(a) - F(a-) = p$, where $f_1$ and $f_2$ are nonnegative continuous functions, and $0 < p \leq 1$. Note that $F$ is everywhere differentiable except at $x=a$ where it has a jump discontinuity of size $p$. In particular, $X$ does not have a PDF $f$, hence you cannot compute ${\rm E}(X)$ as the integral $I_2$. However, ${\rm E}(X)$ can be computed as follows: $$ {\rm E}(X) = \int_{ - \infty }^\infty {xdF(x)} = \int_{ - \infty }^a {xf_1 (x)dx} + a[F(a) - F(a - )] + \int_a^\infty {xf_2 (x)dx} . $$ In case $f_1$ and $f_2$ are identically zero and, hence, $p=1$, this gives $$ {\rm E}(X) = \int_{ - \infty }^\infty {xdF(x)} = a[F(a) - F(a - )] = a, $$ which is obvious since $X$ has the $\delta_a$ distribution (that is, ${\rm P}(X=a)=1$).
Exercise 1. Suppose that $X$ takes finitely many values $x_1,\ldots,x_n$, with probabilities $p_1,\ldots,p_n$, respectively. Conclude that ${\rm E}(X) = \int_{ - \infty }^\infty {xdF(x)} = \sum\nolimits_{k = 1}^n {x_k p_k } $.
Exercise 2. Suppose that with probability $0 < p < 1$ a random variable $X$ takes the value $a$, and with probability $1-p$ it is uniform$[0,1]$. Find the CDF $F$ of $X$, and compute its expectation (note that $X$ is neither discrete nor continuous random variable).
Note: The integral $I_1$ is, of course, a special case of $\int {h(x)dF(x)} $ (say, for $h$ a continuous function). In particular, letting $h=1$, we have $\int {dF(x)} = 1$ (which is a generalization of $\int {f(x)dx} = 1$).
EDIT: Further details.
Note that if $h$ is continuous, then ${\rm E}[h(X)]$, if exists, is given by $$ {\rm E}[h(X)] = \int {h(x)dF(x)}. $$ In particular, if the $n$-th moment exists, it is given by $$ {\rm E}[X^n] = \int {x^n dF(x)}. $$
In principle, the limits of integration range from $-\infty$ to $\infty$. In this context, consider the following important example. Suppose that $X$ is any nonnegative random variable. Then its Laplace transform is $$ {\rm E}[e^{ - sX} ] = \int {e^{ - sx} dF(x)} = \int_{ 0^- }^\infty {e^{ - sx} dF(x)}, \;\; s \geq 0, $$ where $0^-$ can be replaced by $-\varepsilon$ for any $\varepsilon > 0$. While the $n$-th moment of (the nonnegative) $X$ is given, for any $n \geq 1$, by $$ {\rm E}[X^n] = \int_{ 0^- }^\infty {x^n dF(x)} = \int_0^\infty {x^n dF(x)}, $$ it is not true in general that also $$ {\rm E}[e^{ - sX} ] = \int_{0 }^\infty {e^{ - sx} dF(x)}. $$ Indeed, following the definition of the integral, $$ {\rm E}[e^{ - sX} ] = \int_{ 0^- }^\infty {e^{ - sx} dF(x)} = e^{-s0}[F(0)-F(0-)] + \int_{ 0 }^\infty {e^{ - sx} dF(x)}, $$ hence the jump of $F$ at zero should be added (if positive, of course). In the $n$-th moment case, on the other hand, the corresponding term is $0^n[F(0)-F(0-)] = 0$, hence a jump of $F$ at zero does not affect the overall integral.