anonymous
  • anonymous
Let f(x)=k/(x^3+x) if x>=1, and f(x)=0 otherwise where k is const. (a) For what value of k is f a probability density function? (b) Find the mean of this density function. I'm having a hard time finding my limits of integration, I think it's 1 to infinity. But if that's the case I end up with: k[ln(x) - (1/2)ln(x^2+1)] from 1 to infinity which looks like k(infinity - infinity) - k[0-(1/2)ln(2)] and... i'm lost.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Shayaan_Mustafa
  • Shayaan_Mustafa
are you still there?
Shayaan_Mustafa
  • Shayaan_Mustafa
@Vadatajs
anonymous
  • anonymous
Yes?

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Shayaan_Mustafa
  • Shayaan_Mustafa
so do you know what is the criterion to normalize the probability function?
anonymous
  • anonymous
\[\int\limits_{-\infty}^{\infty} f(x) dx = 1\]
Shayaan_Mustafa
  • Shayaan_Mustafa
absolutely. but in your case. what should be the limit?
anonymous
  • anonymous
1 to infinity is my guess since x >= 1
Shayaan_Mustafa
  • Shayaan_Mustafa
correct. now solve the equation.
anonymous
  • anonymous
\[ k \int\limits_{1}^{\infty} 1/x - x/(x^2+1) dx = k[\ln (x) - (1/2)\ln(x^2+1)] from 1 \to \infty\]i use partial fractions to come out with
Shayaan_Mustafa
  • Shayaan_Mustafa
ok..
anonymous
  • anonymous
This is all in my description...
Shayaan_Mustafa
  • Shayaan_Mustafa
now can you solve it?
anonymous
  • anonymous
No, k(infinity - infinity) - k[0-(1/2)ln(2)]
anonymous
  • anonymous
ln(x) from 1 to infinity -0.5ln(x^2 +1) from 1 to infinity
JamesJ
  • JamesJ
Once you integrate you have \[ \frac{k}{2} \ln\left( \frac{x^2}{x^2+1} \right) \] Evaluating that at x = 1, we clearly have \( -\frac{k}{2}\ln(2) \). In the limit as x --> infinity, it is \[ \frac{k}{2} \ln(1) = 0 \] Therefore the entire integral which must be equal to 1 gives us the equation: \[ \frac{k}{2} \ln 2 = 1 \] Now it's easy to find \( k \).
anonymous
  • anonymous
\[\int\limits_{1}^{\infty}(k/(x^3+x))dx=\int\limits_{1}^{\infty}(k/x)dx-\int\limits_{1}^{\infty}(kx/(x^2+1))dx \] that's by using partial fractions
anonymous
  • anonymous
factor k out and solve the integrals separately as...
anonymous
  • anonymous
\[\lim_{R \rightarrow \infty}(\int\limits_{1}^{R}(1/x)dx-\int\limits_{1}^{R}(x/(x^2+1)dx)=\lim_{R \rightarrow \infty}(\ln R-0.5\ln(R^2+1))\] this gives \[\lim_{R \rightarrow \infty}(lnR-\ln(R^2+1)^(1/2))\] this gives \[\lim_{R \rightarrow \infty}((lnR)/0.5\ln(R^2+1))\] this gives \[\infty/\infty\] hence you think of L'Hopital's Rule
anonymous
  • anonymous
so you differentiate the numerator and the denomenator and simplify...this gives... \[\lim_{R \rightarrow \infty}(R^2/(R^2+1))\]
anonymous
  • anonymous
this still gives \[\infty/\infty\] so you do the differential again... and we have... \[\lim_{R \rightarrow \infty}(2R/2R)\] this is simple to evaluate as the 2R's cancel out to give... \[\lim_{R \rightarrow \infty}(1)=1\]
anonymous
  • anonymous
Note that it's... \[\int\limits_{1}^{\infty}(1/(x^3+x))dx =1\] therefore for an equation of the form... \[\int\limits_{1}^{\infty}(k/(x^3+x))dx=1,\] \[k(1)=1,\] hence k=1...hope this answers your question...
JamesJ
  • JamesJ
What brightantwiboasiako has written is not correct - his solution has dropped the value of the integral end point x = 1 - the way the limit of ln(x^2/(x^2+1) is evaluated is incorrect and gives the wrong answer.

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