Solving Random Variable Work: 0 to Infinity = 0.002?

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SUMMARY

The discussion centers on the integration of a probability density function from 0 to infinity, confirming that the constant \( C \) equals 0.0021. The method used involves integrating the function \( C \, \text{exp}\left(-\frac{2.1x}{1000}\right) \) and setting the result equal to 1. The calculations validate that \( C = \frac{2.1}{1000} \) is indeed correct, establishing the normalization of the distribution.

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Uniman
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View attachment 432

Work done so far...

Integrating from 0 to infinity and equating it to 1, we get

(c/2*10^-3) = 1

c= 2/1000

=0.002

Is it correct?
http://www.chegg.com/homework-help/questions-and-answers/-q3136942#
 

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Uniman said:
https://www.physicsforums.com/attachments/432

Work done so far...

Integrating from 0 to infinity and equating it to 1, we get

(c/2*10^-3) = 1

c= 2/1000

=0.002

Is it correct?
http://www.chegg.com/homework-help/questions-and-answers/-q3136942#


Hi Uniman, :)

Yes the method you have used is correct.

\[\int_{0}^{\infty}C\,\mbox{exp}\left(-\frac{2.1x}{1000}\right)dx=1\]

\[\Rightarrow C\left[-\frac{1000}{2.1}\mbox{exp}\left(-\frac{2.1x}{1000}\right)\right]^{\infty}_{0}=1\]

\[\Rightarrow \frac{1000}{2.1}C=1\]

\[\therefore C=\frac{2.1}{1000}=0.0021\]

Kind Regards,
Sudharaka.
 

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