# Evaluating an integral when theta>0

1. Mar 23, 2013

### nick.martinez

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

∫ e^(-y/θ) dy
0

2. Relevant equations

The directions say evaluate when theta>0.

3. The attempt at a solution

I have been integrating by using u of substitution and been getting -θ*e^(-y/θ) and evaluating from 0 to ∞ and getting the answer as undefined. I need help with evaluating the integral. Please help. Thanks.

Last edited by a moderator: Mar 24, 2013
2. Mar 23, 2013

### Zondrina

Are you familiar with the anti derivative of $e^{- \frac{x}{a}}$?

3. Mar 23, 2013

### SammyS

Staff Emeritus
Your result should not be "undefined".

You have the correct anti-derivative.

Write $\displaystyle \ \int_0^{\infty}e^{-y/\theta}\,dy\$ as $\displaystyle \ \lim_{a\to\infty} \int_0^{a}e^{-y/\theta}\,dy\ .$

Now use your anti-derivatve with the limits of integration, then take the limit.

4. Mar 23, 2013

### nick.martinez

No, I'm not. How is it used in this situation?

Last edited by a moderator: Mar 24, 2013
5. Mar 23, 2013

### Infrared

You want to evaluate the integral $\int e^{-y/\theta}, dy$. What happens when you make the substituion $u = {\frac{-y}{\theta}}$ ? What is $du$ ?

Can you generalize to $\int f(cx)\, dx$ when you know what $\int f(x)\, dx$ is? This is a useful result to know when evaluating integrals so you don't always have to make these substitutions.

Last edited: Mar 23, 2013
6. Mar 23, 2013

### nick.martinez

How would i take the anti derivative if says to evaluate for all theta> 0 not quite sure how to do this.

Last edited: Mar 23, 2013
7. Mar 23, 2013

### Infrared

What happens to -θ*e^(-b/θ) as b goes to ∞? to 0? Remember that θ is just a constant.

Last edited: Mar 23, 2013
8. Mar 23, 2013

### nick.martinez

is theta a constant even though in the directions it says evaluate from theta>0.

9. Mar 23, 2013

### SammyS

Staff Emeritus
No. It says evaluate when θ > 0 .

If θ ≤ 0, then $\displaystyle \ \int_0^{\infty}e^{-y/\theta}\,dy\$ does not converge.

(Did you not understand my previous post?)

10. Mar 23, 2013

### nick.martinez

For some reason I'm not understanding what you are saying, SammyS. HS scientist just said it was a constant, and I was disagreeing. Is there any way you could give me the answer and answer how you got it?

Thanks

Last edited by a moderator: Mar 24, 2013
11. Mar 23, 2013

### nick.martinez

also since it says evaluate when theta > 0 this means that all values to infinity so doesnt this mean when evaluating ill get infinity/infinity is the exponent.

12. Mar 23, 2013

### SammyS

Staff Emeritus
This just means θ is a constant, and it's positive.

In your Original Post, you said that using u substitution, you got -θ*e^(-y/θ) .

That's the correct result for the indefinite integral, $\displaystyle \ \int e^{-y/\theta}\,dy\ .$ Another way to say that is: The anti-derivative of $\displaystyle \ e^{-y/\theta}\$ is -θ*e^(-y/θ) .

Do you know how evaluate a definite integral (with limits of integration), if you know the indefinite integral (without limits of integration)?

13. Mar 23, 2013

### nick.martinez

yes i know how to evaluate the definite but when the values of theta are : theta>0 how do you evaluate this when the limits of integration are also from 0 to infinity?

14. Mar 23, 2013

### SammyS

Staff Emeritus
Again: θ is simply some constant -- it happens to be positive. The variable of integration is "y" . It's y that goes from 0 to ∞ as specified by the limits of integration.

15. Mar 24, 2013

### Infrared

Let's try to do this step by step:
$\displaystyle\lim_{y\rightarrow \infty} {-\theta e^{-y/\theta}} = -\theta \displaystyle\lim_{y\rightarrow \infty} {e^{-y/\theta}}$ =

$-\theta (\displaystyle\lim_{y\rightarrow \infty} {e^{-y}})^{1/\theta}$.

Since $\theta$ is greater than zero, what can you say about this limit?
Similarly, what about $-\theta (\displaystyle\lim_{y\rightarrow 0} {e^{-y}})^{1/\theta}$?

Last edited: Mar 24, 2013
16. Mar 24, 2013

### Infrared

Yes, thank you. I should be more careful when writing out solutions.

17. Mar 24, 2013

### SammyS

Staff Emeritus
You may still be able to go back and Edit that post.

Now that you've made the change, I'll delete my post regarding the typo.

(I think PF gives you something like a 700 minute window for Editing.)

Last edited: Mar 24, 2013
18. Mar 24, 2013

### Infrared

Wow, I did not realize that there is such a long allowed time to edit posts. Thanks!