Integral involving exponential logarithm

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SUMMARY

The discussion centers on calculating the expected value E[X] of a random variable X with the density function f(x) = λe^{-xλ} for X ≥ 0. The expected value is established as E[X] = 1/λ. The integration process involves substituting u = -λx, leading to the integral E[X] = ∫ (u/λ)e^{-u} du, which simplifies to (1/λ)∫ ue^{-u} du. The key issue highlighted is ensuring proper substitution during integration to achieve the correct result.

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Homework Statement


If X is a random variable with density function: f(x) = \lambdae^{-x \lambda}where X>=0.

Homework Equations


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

The Attempt at a Solution


E[X] = \int x*(\lambdae^{- \lambda}^{x}) dx, where the integral is from 0 to infinity.

I let u = -\lambdax
du = -\lambda dx

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

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