posted by Paul on .
I'm doing statistics homework and am stuck on a problem using integration.
The problem gives a distribution where for x>1, f(x) = k x^-6. I am then asked to "Determine the value of k for which f(x) is a legitimate pdf. "
To be a legitimate pdf, the integral of f(x)dx has to be equal to 1 (from -infinity to infinity).
I can do the integral, 1/(-5x^5) or k/(-5x^5), but how do I set this equal to 1 and solve for k without knowing x?
The only thing I have been able to come up with is that x = -.2^(1/5), or -.72477, which would make k = 1, thereby not changing the integral at all..