# Integral of exponential distribution from zero to infinity

1. Jun 13, 2012

### cooper607

1. The problem statement, all variables and given/known data

here's a problem from my assignment

let integral p(x)dx=Ae^-(x/a) dx...........(1)
find value of A, that makes integral p(x)dx=1;
and
find mean x so that integral x*p(x)dx=a.........2)

2. Relevant equations
now
to solve the first one i found out A to be (-1/a*e^(x/a))
but when i am integrating the two i can not find that value 1........plz tell me if my value of A is ok or not........
and for the second problem how can i integrate those three components all together?
what is the formula, plz notify here......

3. The attempt at a solution
A to be (-1/a*e^(x/a))

2. Jun 13, 2012

### tiny-tim

hi cooper607!

you mean $[Ae^{-x/a}]_0^{infty}\ =\ 1$
A is supposed to be a constant, how can it be a function of x ?

3. Jun 13, 2012

### dimension10

No, its not ok. How is it even a value in the first place?

4. Jun 13, 2012

### Ray Vickson

Are A and a supposed to be constants? If so, then why would you equate A to (-1/a*e^(x/a))?

Let me ask a more fundamental question: have you studied integration? Both of the questions you ask are simple integration problems from introductory calculus. However, if you have not yet had that material, it would be understandable that you are having problems.

RGV

5. Jun 13, 2012

### HallsofIvy

Staff Emeritus
The title of this thread says that the integral is from 0 to infinity. Did you not evaluate the anti-derivative at those values?