Let's say I have this distribution:
f(x) = k/(x^5) for x > 1, and 0 for x <= 1
How would I solve for the variable k?
Normally I would have an upper bound on x like 1 < x < 5 or something, in this case 5 is the upper bound, and I would solve for k like this:
1 = definite integral from 1 to 5 of k/(x^5) dx
So how to solve in this case since I don't have an upper bound for x?
We have: $\displaystyle \int_{1}^\infty kx^{-5} = 1\implies \dfrac{k}{4} = 1\implies k = 4$ .