I'm having trouble getting the correct answer for the following problem. I need to solve for k such that k makes the following a valid pdf.
f(x) = \begin{cases} k{e}^{-x/\gamma} &\text{ for }x\le 2\\ 2k{e}^{-x/\gamma} &\text{ for }x\gt 2\\ \end{cases}
I don't so much need the answer as I know what the answer is (it has been supplied) but I can't get the methodology down and reach the solution myself. As I understand it, the process should be that I integrate both with the appropriate bounds and the result should equal 1 allowing me to solve for k but I must be doing something wrong as my answer resembles the correct answer but has a bunch of extraneous stuff in it. Sorry for not supplying more info but it took me forever to just write out the question as I'm not familiar with the formatting.
What is $\gamma$? Well, recall that for a $f(x)$ to be a probability density of random variable $X$, we must have $\int_{\mathbb{R}} f(x) = 1 $. In our case, this equates to solve
$$ k \int\limits_{- \infty}^2 e^{-x/ \gamma} + 2k \int\limits_2^{\infty} e^{-x/\gamma} = 1$$