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Suppose I have the following probability density function.

$$f(x) = \begin{cases}ce^{-\frac x{200}}&\mathrm{\ if\ } 0<x<\infty\\ 0&\mathrm{\ if\ }x\le0\end{cases}?$$

How do I find $c$? Currently, I have the following.

$$\int_0^\infty ce^{-\frac x{200}}dx = 1$$

$$\lim_{x\to\infty} -200ce^{-\frac x{200}} + 200c = 1$$

$$c = \frac1{200}$$

Is this right?

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2 Answers

up vote 3 down vote accepted

Yes, that is completely correct

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So we want to choose $c$ so that $$\int_0^\infty ce^{-x/200}\, dx=1.$$

An antiderivative of $ce^{-x/200}$ is $-200ce^{-x/200}$. So our definite integral is $200c$. It follows that $c=1/200$.

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Why would you assume when the conditions are listed? –  idealistikz Sep 12 '12 at 5:07
    
No, the conditions were listed. –  idealistikz Sep 12 '12 at 21:22
    
@idealistikz: Yes, you are right, it wasn't in TeX so I missed it. –  André Nicolas Sep 12 '12 at 21:30
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