The question is:
If the probability of a random variable $X$ with space ${R_x} = \left\{ {1,2,...,12} \right\}$ is given by $P\left( x \right) = k\left( {2x - 1} \right)$, what should $k$ be for $P\left( x \right)$ to be a p.d.f. of $X$?
I have the following answer:
For $P\left( x \right)$ to be a p.d.f of the following should hold: $\begin{array}{l}1 = \sum\limits_{x} {P\left( x \right)} \\{\rm{So,}}\\1 = k\left( {2 - 1} \right) + k\left( {4 - 1} \right) + k\left( {6 - 1} \right) + k\left( {8 - 1} \right) + k\left( {10 - 1} \right) + k\left( {12 - 1} \right) + k\left( {14 - 1} \right)\\ + k\left( {16 - 1} \right) + k\left( {18 - 1} \right) + k\left( {20 - 1} \right) + k\left( {22 - 1} \right) + k\left( {24 - 1} \right)\\ = 1k + 3k + 5k + 7k + 9k + 11k + 13k + 15k + 17k + 19k + 21k + 23k\\ = 144k\\\therefore k = \frac{1}{{144}}\end{array}$
Am I correct or should I be using: $1 = \int_1^{12} {k\left( {2x - 1} \right)} \;dx$, to find $k$?
You are correct. For a discrete space, the sum is appropriate.
For the integral to make sense, we would have to be talking about a probability density function (the other "d"), with allowed values in the full range.