Why does integrating a probability density function give probability? It is well know that integrating a probability density function gives probability, that is, 
$$P(X\geq a) = \int_a^\infty f_X(x)\, dx$$
where $X$ is a continuous random variable, $a$ is a scalar and $f_X(x)$ is the probability density function of $X.$

Question: Why is this the case?

When I learn this, I just memorize this without understanding it. 
But recently I started pondering about this.
 A: The probability density function is actually defined by this requirement.  In introductory probability courses, you will see the density function defined as the (Riemann) derivative of the cumulative distribution function:
$$f_X(x) \equiv \frac{d F_X}{dx}(x) = \frac{d}{dx} \mathbb{P}(X \leqslant x) 
= \lim_{\Delta \downarrow 0} \frac{\mathbb{P}(x < X \leqslant x + \Delta)}{\Delta}.$$
Using the fundamental theorem of calculus, you then have:
$$\begin{equation} \begin{aligned}
\mathbb{P}(X \leqslant a) 
= F_x(a) 
= \int \limits_{-\infty}^a \frac{d F_X}{dx}(x) dx
= \int \limits_{-\infty}^a f_X(x) dx.
\end{aligned} \end{equation}$$
The corresponding equation you give in your question is then obtained using the norming axiom of probability.  You can see from this reasoning that the integral equations for probabilities of continuous random variables follow from the definition of the density function, combined with the fundamental theorem of calculus.
If you go on to do graduate-level probability theory (based on measure theory), the relationship is even more direct.  In the measure-theoretic definitions, the density function is defined as a broader type of derivative that is called the Radon-Nikodym derivative, which is a type of "anti-integral" defined directly by the stated integral requirement.  In this case, the integral requirement is essentially the definition of the density ---i.e., any function that integrates to give the probability of any stipulated event (integrating over the space of that event) is a valid density function.
