Can somebody give me advice on how to find variance and the expected value; when one considers the function f(Y) = 1/9 is when 0

Question

Can somebody give me advice on how to find variance and the expected value; when one considers the function f(Y) = 1/9 is when  0 <= y <= 9
and the probability is 0 everywhere else when y is not in that interval(0-9). please note that this is a uniform distribution function and Y is continous. Im struggling with the integrals. How can you find an expected value for Y, when all values in the interval [0,9] have the same probability?
thanks for your help!

dumbcow · Answer

$$E(X) = \int\limits_{0}^{9}p(y)*y = \int\limits_{0}^{9}\frac{y}{9} dy$$

dumbcow · Answer

Variance is E(X^2) - E(X)^2

anonymous · Answer

why isn't this 
oh, what dumbcow said

anonymous · Answer

But when I put in 9 for Y: 1/9*9 = 1. But I think the answer is 4.5. Note that I 
am having trouble with integrals, even basic ones. Is the expected value equal to the average value in this case?

dumbcow · Answer

yes it is same as average value since it is uniformly distributed.
E(X) = 4.5

anonymous · Answer

ah I see, thank you!

dumbcow · Answer

no problem :)