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:
$$F(a\leq X \leq b) = F(X \leq b) - F(X \leq a) = \int_{-\infty}^{b} f(x)dx - \int_{-\infty}^{a} f(x)dx = \int_{a}^{b} f(x)dx$$
The fisher information only has a precise meaning when you are dealing with a normally distributed value. In that case, the log likelihood function will be parabolic, and the fisher information will equal the curvature at the MLE. It turns out mathematically that the curvature of the log likelihood is the inverse of the variance of the associated normal random variable.
This is what guides the intuition surrounding fisher information, even though it will only hold approximately for non-normal variables (although, subject to some tehnical conditions, it will usually be asymptotically true.) Also, it serves as a good lower bound on the variance, see here.
Demonstration of the relationship between $I$ and $\sigma$ for gaussian likelihood for the mean
Let $f(X;\theta):= f(x;\mu,\sigma) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$
Take the logarithm of this:
$$\log(f) = -\log\sqrt{2\pi} - \log\sigma -\frac{(x-\mu)^2}{2\sigma^2}$$
We are looking at the likelihood for the mean ($\mu$) given a sample of data ($\mathbf{x}$) so we treat the above as a function of $\mu$. Given a sample $\mathbf{x}$, we can formulate the likelihood function for $\mu$.
$$L(\mu|\mathbf{x},\sigma) =-n\log\sqrt{2\pi} - n\log\sigma -\frac{1}{2\sigma^2}\sum\limits_{x_i \in \mathbf{x}}(x_i-\mu)^2$$
This function will be quadratic in $\mu$. Therefore, when we take $\frac{d^2L}{d\mu^2}$ we will get a constant value (due to it being quadratic). Specifically:
$$\frac{dL}{d\mu} =\frac{1}{\sigma^2}\sum\limits_{x_i \in \mathbf{x}}(x_i-\mu) \implies \frac{d^2L}{d\mu^2} = \frac{-n}{\sigma^2} = \textrm{constant} $$
Therefore,
$$-E_{\mu}\left(\frac{-n}{\sigma^2}\right) = \frac{n}{\sigma^2} = I(\mu)$$
Which is the Fisher Information about $\mu$, but:
$$ se(\hat \mu) = \frac{\sigma}{\sqrt{n}} \implies I(\mu) = \frac{1}{se(\mu)^2} = \frac{1}{\sigma^2_{\hat \mu}}$$
Therefore, the Fisher Information is the inverse of the variance of the MLE.
Best Answer
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.