Finding Expected Value of fy(Y) = 3(1-y)^2

[semidevil]

ok, so to find the expected value of fy(Y) = 3(1-y)^2 for 0 <= 1 <= 1

I thought the formula is y * fy(Y) which is the intgeral from 0 to 1 of y* 3(1-y)^2. right?

the book says the answer is 1/4...but I get a whole number answer...

 

[assyrian_77]

Hi,

I did the integral of y*f(y) from 0 to 1 and got the answer 1/4...double check your calculation...This is what I did:

Int[3*(1-y)^2] = 3*Int[y^3-2y^2+y] from 0 to 1

 

[assyrian_77]

oops, I forgot the y in the 1st integral expression