Statistics: Let X have the density function f(x) = cx(1-x) for 0

Question

Statistics: Let X have the density function f(x) = cx(1-x) for 0<=x<=1 and f(x)=0 otherwise. Find c. 

Do I integrate the function between the values 0 and 1?

turingtest · Answer

why not?$$\int_{0}^{1}cx(1-x)dx$$

turingtest · Answer

...=1 I guess, right?

anonymous · Answer

How did you get that? I brought the c outside what's being integrated and then multiplied out the function. So now I have $$c \int\limits_{0}^{1}(x^2/2)-(x^3/3)dx$$ Then subbing in 1 and 0 I ended up with c*1/6. Have I done that much correctly?

turingtest · Answer

you have already integrated it, but you left the integral sign o-0 ?

anonymous · Answer

Sorry. That's a typo :D

turingtest · Answer

$$\int_{0}^{1}cx(1-x)dx=1$$$$c\int x-x^2dx=1$$$$c(\frac12x^2-\frac13x^3)|_{0}^{1}=1$$

anonymous · Answer

right you should get c(1/2 - 1/3) =1, then solve for c

turingtest · Answer

$$c(\frac12-\frac13)=c\frac16=1\implies c=6$$

anonymous · Answer

Oh right, sorry Turing, I thought you were getting 1 as the answer for the integration. :D

anonymous · Answer

So that's a rule that applies to all functions in pdf, is it? You just let them equal to 1?

turingtest · Answer

for a probability density functions $$\int_{-\infty}^{\infty}f(x)dx=1$$since the function is zero everywhere except the interval [0,1], we only needed to integrate along those bounds

turingtest · Answer

if you have never seen that formula before, try to think about the logic behind it.

anonymous · Answer

I'm guessing it's equal to one because of the rule in statistics that all probabilities must be between 0 and 1?