# cumulative distribution function problem

#### bryce09

##### New Member
I have this question I am trying to get through but i keep coming into trouble. The question is:

Show that the cumulative distribution function from a uniform distribution of the random variable is Fx(y) = (y-a) / (b-a) for some a < y < b

I've started the question but have become stuck going from here

y
∫ x/(b-a) dx
a

when i integrate i come up with y2-a2/2(b-a). When i'm pretty sure it should come out as (y/b-a)-(a/b-a) which would then solve to y-a/b-a

any advice as to what i'm forgetting to do would help me alot, thanks.

#### vinux

##### Dark Knight
There is mistake in the integral formula. To help you, give your answer on the following.
What is the definition of cumulative distribution function?
what is the density function of uniform distribution?

#### bryce09

##### New Member
Is this right? excuse my notation, I hope you understand it

Fx(y)=
= ∫a^y dx/(b-a)
= [x/(1(b-a))]a^y
= [y/(b-a)- a/(b-a)]
= (y-a)/(b-a)

Mod edit: I think this is what you were going for?
$$\int_a^y \frac{1}{b-a}dx$$
$$=\frac{x}{b-a} |_a^y$$
$$=\frac{y}{b-a} - \frac{a}{b-a}$$
$$=\frac{y-a}{b-a}$$

#### vinux

##### Dark Knight
You are right. You could use tex format using $$[/math ]. eg: [math***] \int_{a}^{y}\dfrac{1}{b-a} dx[/math***]. deleting *** gives \( \int_{a}^{y}\dfrac{1}{b-a} dx$$.\)