Given that a random variable is uniformly distributed in an interval, how to find the X probability density function

Given that a random variable is uniformly distributed in an interval, how to find the X probability density function

It is known that x ~ u [a, b], that is, X obeys the uniform distribution on the interval [a, b]
Then the probability density function of X is
p(x)= 1/(b-a) x∈[a,b]
=0 others

Let the random variable x obey the uniform distribution on the interval [a, b], write out its probability density function f (x), and calculate its mathematical expectation ex, variance DX

F(X)=（X-a）/（b-a）f(X)=F'(X)=1/（b-a）E(X)=∫xf(x) dx=∫x/(b-a)dx=x^2/2|(a,b) /(b-a)=(b^2-a^2)/2(b-a)=(a+b)/2D(X)=E(X^2)-E(X)^2=∫x^2f(x) dx-(a+b)^2/4=(b^3-a^3)/3(b-a)-(a+b)^2/4 =(b-a)^2/12

Let the random variable x obey the uniform distribution in the interval [a, b], then its probability density function f (x) =, e (x) =,

f(x)=1/(b-a);a

Let the distribution function of continuous random variable X be f (x) = 1 + 1 / π arctanx. Find the probability of X falling into (0,1)
The range of X is from negative infinity to positive infinity

The probability of X falling into (0,1)
=F(1)-F(0)
=1+1/4-1
=1/4