MHB Solving Random Variable Work: 0 to Infinity = 0.002?

AI Thread Summary
The integration from 0 to infinity is correctly set up to find the constant C, equating it to 1. The calculations show that C equals 0.0021, not 0.002 as initially suggested. The method used for integration and the exponential function is validated by the responses in the discussion. The final value of C is crucial for further applications in probability and statistics. Accurate computation of constants in random variable work is essential for correct results.
Uniman
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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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