ATH 101: Integration by Parts - Exponential Distribution

[michonamona]

Integration by parts - Exponential distribution

Homework Statement

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.

Homework Equations

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?

I appreciate your input.

M

 

[Whitishcube]

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

 

[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

 

[Whitishcube]

nevermind. I see what's 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?