# Integration by parts

1. Aug 14, 2010

### michonamona

Integration by parts - Exponential distribution

1. The problem statement, all variables and given/known data
Solve the following definite integral:

$$\int^{\infty}_{0} \frac{1}{\lambda} x e^{-\frac{x}{\lambda}} dx$$

I'm asked to solve this integral. The solution is $$\lambda$$, although I'm not sure how this was done.

2. Relevant equations

3. The attempt at a solution
$$\int^{\infty}_{0} \frac{1}{\lambda} x e^{-\frac{x}{\lambda}} dx$$

$$= \frac{1}{\lambda} \int^{\infty}_{0} x e^{-\frac{x}{\lambda}} dx$$

$$=\frac{1}{\lambda} \left( \left[ x e^{-\frac{x}{\lambda}} \right] ^{\infty}_{0} - \int^{\infty}_{0} e^{-\frac{x}{\lambda}} dx \right)$$, integration by parts.

The $$\left[ x e^{-\frac{x}{\lambda}} \right] ^{\infty}_{0}$$ term, by fundamental theorem of calculus is 0. Thus,

$$= - \int^{\infty}_{0} e^{-\frac{x}{\lambda}} dx \right)$$,

I don't know what to do at this point, because as far as I know, taking the definite integral of this term will result in $$e^{-\frac{x}{\lambda}}$$ , which, solving for 0 and infinity will yield -1.

Where have I gone wrong?

M

Last edited: Aug 14, 2010
2. Aug 14, 2010

### Whitishcube

Close. You forgot the chain rule. The derivative of e^-u is -e^-u. (or antiderivative.)

3. Aug 14, 2010

### michonamona

Thanks whitish,

I was in the middle of editing the formula after your post. Do you mind looking at what I have posted again?

Thanks

4. Aug 15, 2010

### Whitishcube

nevermind. I see whats happening now, and i'm getting the same answer as you are. with your last integral you can just use U substitution. are you sure just lambda is the right answer?

Last edited: Aug 15, 2010