Improper Integrals & LIMITS help needed (1 Viewer)

[blackops23]

Hey guys, as part of an improper integral question, I ended up with a limit which I don't know how to solve;

Here's the question:

Integrate (x^2)(e^(-x)) INTEGRAL LIMITS: 0(<)x(<)+inf

Now the limit I ended up with was:

as t-->+infinity
[-(t^2)*(e^(-t)) - 2t*e^(-t) - 2e^(-t) + 2]

The answer is 2, which means everything else equated to 0.

Can someone please explain why this happens?

I mean for -(t^2)(e^(-t)), when you put infinity in, would you not have the result: -inf*(0)?

How on earth does one calculate that?
I've heard something like L'Hospital Rule would be used here, can someone verify that, or please outline what other method they would you to calculate this limit?

Thank you very much, immensely appreciate the help

 

[AAEldar]

You should get:



Then when subbing in for the limits, the first part will all cancel as the zero is out the front then the second part all goes when you sub in for 0, except for the 2.

At least I hope that's right.

 

[deterministic]

Hey guys, as part of an improper integral question, I ended up with a limit which I don't know how to solve;

Here's the question:

Integrate (x^2)(e^(-x)) INTEGRAL LIMITS: 0(<)x(<)+inf

Now the limit I ended up with was:

as t-->+infinity
[-(t^2)*(e^(-t)) - 2t*e^(-t) - 2e^(-t) + 2]

The answer is 2, which means everything else equated to 0.

Can someone please explain why this happens?

I mean for -(t^2)(e^(-t)), when you put infinity in, would you not have the result: -inf*(0)?

How on earth does one calculate that?
I've heard something like L'Hospital Rule would be used here, can someone verify that, or please outline what other method they would you to calculate this limit?

Thank you very much, immensely appreciate the help

we have for any n. This is because the rate at which tends towards 0 is much quicker than the rate at which tends towards infinity. Thus dominates and hence the limit is 0.

More formally, using L'Hopital Rule for integer values of n (for non integer values of n, the proof is very similar):

 

[Timothy.Siu]

You should get:



Then when subbing in for the limits, the first part will all cancel as the zero is out the front then the second part all goes when you sub in for 0, except for the 2.

At least I hope that's right.

That logic is a bit flawed, although the answer is right. When you "sub" in infinity, you'll get 0 * (infinity) which is what the original poster was asking about. When you have something like this, you can use L'hopitals rule. Which states that if you have a limit in some indeterminant form, i.e. 0/0, etc, then basically, the limit of f(x)/g(x) equals the limit at that same value of f'(x)/g'(x). So, here were have -(x^2+2x+2)/e^x. Here, you'll need to use L'hopitals rule twice, and you'll have that the original limit equals the limit of -2/e^x as x goes to infinity, which is of course 0.
Alternatively, being a little less rigorous, you could just say because exponential grows faster than polynomials, that -(x^2+2x+2)/e^x = 0 as x goes to infinity.
By the way, this is definitely not Extension 2 material. Another side note is that when you have improper integrals, for this one, you should consider it as the limit as M goes to infinity, with the integral going from 0 to M, because you can't actually sub in infinity, as its not a number or a value, and you would end up with nonsensical values like infinity* 0 which we cannot know the value of.

 

[Shadowdude]

Question: is L'Hopital's rule in MX2 now? (also, L'Hopital's rule is first year maths - I just learned it in the past semester... quite a nice trick, I must say)


Also, the 'formal' way to do this integral - from my experience - is to take the integral from 0 to an arbitrary 'R', which is positive. Evaluating the integral, and then taking the limit as R tends to infinity.

Also remember that e^(-t) = 1/(e^t).

 

[Mature Lamb]

Question: is L'Hopital's rule in MX2 now? (also, L'Hopital's rule is first year maths - I just learned it in the past semester... quite a nice trick, I must say)


Also, the 'formal' way to do this integral - from my experience - is to take the integral from 0 to an arbitrary 'R', which is positive. Evaluating the integral, and then taking the limit as R tends to infinity.

Also remember that e^(-t) = 1/(e^t).

I learnt about L'Hopital's Rule in Year 12, but it was never necessary to use it.

 

[gurmies]

That logic is a bit flawed, although the answer is right. When you "sub" in infinity, you'll get 0 * (infinity) which is what the original poster was asking about. When you have something like this, you can use L'hopitals rule. Which states that if you have a limit in some indeterminant form, i.e. 0/0, etc, then basically, the limit of f(x)/g(x) equals the limit at that same value of f'(x)/g'(x). So, here were have -(x^2+2x+2)/e^x. Here, you'll need to use L'hopitals rule twice, and you'll have that the original limit equals the limit of -2/e^x as x goes to infinity, which is of course 0.
Alternatively, being a little less rigorous, you could just say because exponential grows faster than polynomials, that -(x^2+2x+2)/e^x = 0 as x goes to infinity.
By the way, this is definitely not Extension 2 material. Another side note is that when you have improper integrals, for this one, you should consider it as the limit as M goes to infinity, with the integral going from 0 to M, because you can't actually sub in infinity, as its not a number or a value, and you would end up with nonsensical values like infinity* 0 which we cannot know the value of.

This.

 

[cutemouse]

Well the notation means

Mathematicians tend to be notoriously lazy sometimes