### Why does the probability density function integrate over the domain of 1?

A probability density function, which might as well be set up as a one-dimensional f(x), means that the probability that you take the value of x (or call it an event) is f(x), and yet the probability of all the events occurring sums up to 1, and so the probability density function integrates over the domain of 1.

In mathematics, a continuous random variable’s probability density function (which can be shortened to density function when it is not confusing) is a function that describes the likelihood that the output value of this random variable will be in the vicinity of some definite point of value.

And the probability that the value of the random variable falls within a certain region is the integral of the probability density function over that region. When a probability density function exists, the cumulative distribution function is the integral of the probability density function. The probability density function is usually labeled in lower case.

Extension:

The probability density function of a continuous random variable has the following properties:

If the probability density function fX(x) is continuous at a point x, then the cumulative distribution function is derivable, and its derivatives:

Since the value of the random variable X depends only on the integral of the probability density function. So the value of the probability density function at individual points does not affect how the random variable behaves. More precisely, a function can also be a probability density function of X if there are only a finite, countably infinite number of points at which the function and the probability density function of X take different values, or if it measures 0 (is a zero-measure set) with respect to the entire axis of real numbers.

The probability that a continuous-type random variable takes values at any point is 0. As a corollary, the probability that a continuous-type random variable takes values on an interval is independent of whether the interval is open or closed. Note that the probability P{x=a} = 0, but {X=a} is not an impossible event.

### How to prove that the area of the probability density function is the limit of 1

dx I guess

So that’s 1/(√x+dx+√x)]/,b^2] right

Then consider the definition of the density function isn’t it a limiting form yah

The probability then that S falls on [x,x+dx] is 1/(√x+dx+√x)

=1/(b-a)*dx/(b-a)*(√x+dx-√x)

=1/(b-a)*(√x+dx-√x)*(√x+dx+√x)/(√x+dx+√x)

And then according to the definition of the density function

The density of S at point x is [1/(b-a)*dx/(b-a)

So the probability that S falls on [x first the area of the circle S falls on [a^2,x+dx] is that R falls on [√x,√x+dx] right

Then the density function of R is f=1/[(b-a)*2√x]

(

That function above is at [a^2

### Why is the area of the normal distribution 1

Because the probability that a continuous-type random variable is in a certain interval is equal to the definite integral over that interval. Because the event that the value of a normally distributed random variable falls on (-00,+00) is a definite event. The probability is 1

, so the integral of the density function of the normal distribution over (-00,+00) is 1, i.e., the area enclosed is 1,

The integral can also be accumulated, which can be deduced by learning to reintegrate. The integral of e^(-x^2) over 0 to +00 is (root pi)/2, and you can use this to draw the conclusion you want.