If the probability density of a random vribale is given by :

f(x)= x for 0

Question

If the probability density of a random vribale is given by :

f(x)= x               for 0<x<1
       2-x            for 1<=x<2
        0              elsewhere

find the probabilities that a random variable having this probability density will take on a value

(a)between .2 and .8

anonymous · Answer

My issue isn't how to integrate between the two, but the process in finding the answer.
Do I have to integrate each function first, plug in the limits from x = 0 to 1 (for function 1),
x=1 to 2 (for function 2) and then integrate that answer for each function and plug in the
two probabilities for both functions?

anonymous · Answer

^^Oh and that word should say "variable" in my first post.

anonymous · Answer

If I'm right then would I have to do the following:
$$f(x)=\int\limits_{0}^{1}x dx= 1/2$$
$$f(x)=\int\limits_{.2}^{.8}1/2dx=.8/2-.2/2=.3$$

zarkon · Answer

.3 is good

zarkon · Answer

though your notation and work is poor

zarkon · Answer

$$\int\limits_{.2}^{.8}xdx=.3$$

zarkon · Answer

Just to see the notation...

$$P(.2<X<.8)=\int\limits_{.2}^{.8}f(x)dx=\int\limits_{.2}^{.8}xdx=.3$$

anonymous · Answer

So I didn't have to find the definite integral of:
$$\int\limits_{0}^{1}xdx$$?

zarkon · Answer

no

zarkon · Answer

if $X$ has density $f(x)$

then 

$$P(a<X<b)=\int\limits_{a}^{b}f(x)dx$$

anonymous · Answer

Oh ok. Thanks, I think I can do the rest from here.