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September 17, 2014

September 17, 2014

Posted by **Paul** on Thursday, September 27, 2007 at 4:35am.

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..

- Calculus -
**bobpursley**, Thursday, September 27, 2007 at 6:53amHmmm.

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

- Calculus -
**Sidney**, Thursday, September 27, 2007 at 2:26pmI have no clue!!!

- Calculus -
- Calculus -
**Paul**, Thursday, September 27, 2007 at 4:22pmThanks bob! Thats what I ended up working out and it turned out to be the right answer :)

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