[RESOLVED] Probability density function, help!

A PDF is defined by the function
[Ce^(-t/10), 0 <= t <= infinity]
f(t) ={
[0, elsewhere]
where C is a constant. Find the value of C.
By my reckoning, if f(t) is a PDF, then its area is 1. So the integral from 0 to infinity of Ce^(-t/10) = 1
the antiderivative of Ce^(-t/10) = -10Ce^(-t/10): I'll call this function F(t).
I'm fine up to here, but now to find the area don't I have to let F(infinity)-F(0) = 1, and solve for C? F(0) I can do, but F(infinity)?
My first thought was that although the domain says 0 <= t <= infinity, it's really the values for which the function is positive. So I was going to let the original function f(t) = 0 and find the x-intercept in terms of C, and that would replace the infinity in the domain, but I can't rearrange Ce^(-t/10) = 0 to make t the subject.
I'm really stuck with this, anyone have any ideas?

Hi,

Haven't do maths for years, correct me if I did wrong.

You got F(t) = -10C e^(-t/10)

you said you got F(0), I suppose you did it by -10C e^(-0/10) = -10C

For F(infinity) = -10C e^(-infinity/10)
=> -10C e^(-infinity)
=> (-10C) / (e^infinity)
=> 0 , as (e^infinity) -> infinity

In F(infinity)-F(0) = 1, you could find C.

Hope it is correct and helps you

Yeah, that's correct, but I actually ended up doing it another way. Like you said, F(0) gave me -10C, but F(infinity) I did differently.
F(infinity) = -10Ce^(-infinity/10)
= -10Ce^infinity
= 0, because looking at the graph of e^x, i realised that the assymptote off to the left will never = 0..... that is unless x = -infinity! So yeah, -10C*0 = 0
So putting that back into the integral, --10C = 1, so C = 1/10.
Yay, I got 95% for this test! I think that's the lowest mark I've had all year. Bring on the exam!!!