Yes, a has to be negative. And if so, while t goes to +infinity, the limit of is 0. Which should give you the solution for a...
f(x) = 5e^(ax) for x >= 0.
Find a so that f(x) is a probability density function.
My work:
I can only get up to
(5/a) lim [(e^at)-1] = 1
t-> infinite
Don't really know what to do after, but i think a has to be negative so that e^at can approach 0.