# Integral involving exponential logarithm

1. Aug 21, 2010

### cielo

1. The problem statement, all variables and given/known data
If X is a random variable with density function: f(x) = $$\lambda$$e$$^{-x \lambda}$$where X>=0.

2. Relevant equations
Why is the expected value of X, or E[X] = $$\frac{1}{\lambda}$$?

3. The attempt at a solution
E[X] = $$\int$$ x*($$\lambda$$e$$^{- \lambda}$$$$^{x}$$) dx, where the integral is from 0 to infinity.

I let u = -$$\lambda$$x
du = -$$\lambda$$ dx

but I can't get the $$\frac{1}{\lambda}$$ as the answer when I performed the integration.

2. Aug 21, 2010

### HallsofIvy

If you let $u= \lambda x$ then $x= u/\lambda$ and the integral becomes
$$\int \frac{u}{\lambda}e^{-u} du= \frac{1}{\lambda}\int ue^{-u}du$$

Did you remember to replace the "x" multiplying the exponential with $u/\lambda$?