Evaluating an integral when theta>0

[nick.martinez]

Homework Statement

∞
∫ e^(-y/θ) dy
0

Homework Equations

The directions say evaluate when theta>0.

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.

 

[STEMucator]

Homework Statement

∞
∫ e^(-y/θ) dy
0

Homework Equations

the directions say evaluate when theta>0

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 intrgal. please help. thanks

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

 

[SammyS]

Homework Statement

∞
∫ e^(-y/θ) dy
0

Homework Equations

the directions say evaluate when theta>0

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

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.

 

[nick.martinez]

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

 

[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.

 

[nick.martinez]

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.

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

 

[Infrared]

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

 

[nick.martinez]

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

 

[SammyS]

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

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?)

 

[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

 

[nick.martinez]

also since it says evaluate when theta > 0 this means that all values to infinity so doesn't this mean when evaluating ill get infinity/infinity is the exponent.

 

[SammyS]

for some reason I am not understand what you are saying sammys. HS scientist just said it was a constant and I was diagreeing. is there any way you could give me the answer and answer how you got it?

Thanks

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)?

 

[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?

 

[SammyS]

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?

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.

 

[Infrared]

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

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

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

 

[Infrared]

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

 

[SammyS]

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

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

Added in Edit:

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.)

 

[Infrared]

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