Here's the question you clicked on:
kirbykirby
Find k such that f(x) is a probability density function (pdf):
\[f(x)=ke^{-\frac{x-\mu}{\theta}}\] for \[x>\mu\] Perhaps more legibly as: \[f(x)=k*\exp({-\frac{x-\mu}{\theta}})\]
Usually there's a bound for x like 0<x<2, but here it's x>u, so do I integrate from u to infinity? (And set this integral equal to 1)
I don't need to know how to solve the integral (at least I don't think so yet). I'm just wondering what the bounds on the integral should be.
i think x> mu is just to tell that exponential term has negative exponent always, the bounds will be still -infinity to infinity.
Oh ok I think you are right :) When I think about it now... I wouldn't have found a numerical answer had I used mu as one of the bounds ._.
because if x<mu, the integral might diverge.
Ok well thank you :)!