1. Normalization constant

Hi,

May anyone help in computing the normalization constant (k) for the function
f(u)=u(1-u)(u-a) where 0<a<1 ?

Thanks

2. Originally Posted by rainy cloud
Hi,

May anyone help in computing the normalization constant (k) for the function
f(u)=u(1-u)(u-a) where 0<a<1 ?

Thanks
For what values of u is the given pdf non-zero?

3. I don't understand the question.
Most of my students laugh when I tell them
I'm allowed to say FU to them, as in $f_U(u)$.

4. Thank you mr fantastic and matheagle for reply.

To be honest I have ever never met what called by narmalization constant.
I googled and found in Wikipedia what called pdf (prabability density function). I do not know here it is the same or not.

What I have for my problem:
f(u)=ku(1-u)(u-a), 0<a<1, u $\in$[0,1] and max f(u)=1 on [0,1].

I tried to compute the $\int_{0}^{1}f(u)du=1$ to find the value of k. Is it true?

When I integrate from $-\infty$ to $\infty$?

Thanks a lot.

5. From what I understand of this question,
you should expand the product by multipling and then integrate wrt u from 0 to 1.
The constant k should be a function of a.

6. Originally Posted by rainy cloud
Thank you mr fantastic and matheagle for reply.

To be honest I have ever never met what called by narmalization constant.
I googled and found in Wikipedia what called pdf (prabability density function). I do not know here it is the same or not.

What I have for my problem:
f(u)=ku(1-u)(u-a), 0<a<1, u $\in$[0,1] and max f(u)=1 on [0,1].

I tried to compute the $\int_{0}^{1}f(u)du=1$ to find the value of k. Is it true?

When I integrate from $-\infty$ to $\infty$?

Thanks a lot.
The pdf is non-zero only for $0 \leq u \leq 1$ so you only integrate f(u) over that interval.

7. Thank you very much for appreciated help and advice.