Calculus
posted by Paul .
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..

Hmmm.
1= INt (1>inf)k/x^6 dx=
= k/5 * 1/x^5 (limits x=1+ to inf)
1= k/inf + k/5
k= 5
check my thinkig 
I have no clue!!!

Thanks bob! That's what I ended up working out and it turned out to be the right answer :)