Elaborating a bit on Trisoloriansunscreen's answer: it's very much true that you only got a probability density function. I'd like to draw an analogy for you. Imagine you have a 3D object, say some complex spaceship, and you know the mass density at every point.
For example, some parts of the spaceship might contain water, which has a mass density of $997 \frac{\text{g}}{\text{l}}$. Does this already tell you anything about the mass of the whole spaceship? No, it does not! Precisely because you only know this value at a specific point. You've got no information on how much water there actually is. It might be $1\ \text{ml}$ or $1\ \text{l}$.
Now suppose you know the amount of water, let's say $2\ \text{l}$. By simple multiplication $997 \frac{\text{g}}{\text{l}} \cdot 2\ \text{l}$, you get roughly $1994\ \text{g}$. I would like to make the point that you just did integration in disguise! Consider the following picture:

The mass you calculated is just the greenly shaded rectangular area. This was only doable as a simple multiplication because the mass density was constant for the amount of water considered and thus yielded a rectangular area.
What if you had mixed forms of water, e.g. some gaseous, some liquid, some in varying temperatures and so on? It could look like this:

Now for computing the mass you would need to integrate that mass density function over the amount of water. Do you see the parallel to probability density functions now? To get an actual probability (cf. mass) you need to integrate the probability density (cf. mass density) over some domain.