# What does it mean that the probability is an integral in continuous probability?

I have started to study continuous probability and probability density functions. I understand very basic concepts like the probability of an object twisted on a wheel of fortune to form an angle between 240° (excluded) and 260° (included), which turns out to be:

$$P(X \text{ angle to be between 240° and 260°}) = \frac{260 - 240}{360} = \frac{20}{360} = \frac{1}{18} = 0.0\overline{5}$$

What I do not understand is why the probability of $X$ being between the interval is expressed as an area (integral) when it comes to probability density functions.

Is it because if $f(x)$ gives a higher value (which should mean higher frequency, a data which is more frequent) when $X = x$, then the probability of X being within an infinitesimal interval containing $x$ is higher because the frequency $f(x)$ itself is higher?

In the above example, that would mean (to me) to express the previous probability indicatively as:

$$\int_{260}^{360} f(x)dx$$

Where $f(x) = \frac{1}{360}$ for any $x$ between $0$ and $360$ and $dx = 20$. However, this does not make much sense to me yet.

Could someone make a demonstration of why do we use integrals and how they relate to probability?

Note that the two method results in the same number.

$$\int_{260}^{360} f(x)dx=$$

$$\int_{260}^{360} \frac {1}{360}dx =$$

$${1}{360}x|_{260}^{360}=$$

$$\frac {1}{360}(360-260) =$$

$$\frac {20}{360}= \frac {1}{12}$$

• I know, the thing I do not understand is how to understand conceptually that the probability is the result of the integration... – user3019105 Mar 19 '18 at 8:32
• Anyway, it's 1 over 18, not 12. – user3019105 Mar 19 '18 at 11:05