probability problem

Quote:

Original post by cignox1 How do I integrate f(x) = ke^(-kx)?

Try finding the derivative of e^(ax).

Yes, I believe it's the right way and perhaps integral table would help?

I'm not too familiar with probability but you're integral would be F(x) = -e^(-kx) + C. since the the derivitive of e^(exponent involving x) is just the original function multiplied by the derivitive of the exponent with respect to x

Thank you, I use the integral table (on the one I have on my book there is not this integral, only e^x, and I never understood integrals :-(.

Tell me if I'm right. Now I know that my integral is k/(-k) e^(-kx). This can be simplified and becomes -e^(-kx). I have to calculate in ]4,inf] so:

-e^(-k inf) -(-e^(-k 4)) = lim (R-> inf) of -e^(-kR) - (-e^(-k 4)) = e^(-k 4)

Since k = 1/4, it is e^-1, as in the solution of the prof.

What I wonder is: x should be greater than 4 (strictly) but I calculate as in [4,inf] instead tham ]4,inf]. Does it matter?

Thank you again!

No, it doesn't matter.

edit: Definite integral (Riemann integral)is defined as a limit of a Riemann sum, where intervals of the summation aproach zero. This means that infinitesimal point adds zero to the sum.

Ok, good. Thank you again!