Question

If F(x)=exp(-kx), 0 12 years ago

Answer 1

Somehow I get the feeling that there's more to this question than you've typed. k is called the "decay constant" or something like that. As you state this problem, k could take on many different values.

Answer 2

if \(F(x)=\exp(-kx)\) is a probability density function on \(\mathbb{R}^+\) then we know:$$\int_\mathbb{R^+} F=1$$i.e.$$\begin{align*}\int_0^\infty F(x)\ dx&=\lim_{a\to\infty}\int _0^a\exp(-kx)\ dx\\&=\lim_{a\to\infty}\left[-\frac1k\exp(-kx)\right]_0^a\\&=-\frac1k\lim_{a\to\infty}\left(\exp(-ka)-1\right)\\&=\frac1k\end{align*}$$

Answer 3

so $$1/k=1$$ therefore \(k=1\)