How to evaluate the definite integral?

[Eats Dirt]

How to evaluate the definite integral!?

Homework Statement

Integral test to see if the series converges or diverges.


(infinity)
Sigma ((n^2)/(e^(n/3))
(n=1)

Homework Equations

First I derive to make sure the function is decreasing and positive so it fits the criteria of the integral test.

It decreases when. 7=< x.

The Attempt at a Solution

So I integrate it (using Integration by parts) and get to the point where I have to evaluate the definite integral

[(-3e^(-x/3))*(x^2+6x+18)] from 7 to a lim a-->infinity

This is where I get confused. My textbook says in sigma evaluate from 1 but in the answer it says plug in 7. I was wondering why this was?

My biggest problem is that If I plug these numbers in I have no idea how to simplify it down to a finite number. The answer is apparently 54 but I have no idea how to get there from here.

 

[Simon Bridge]

Urrrgh ... this is where LaTeX is really helpful.
Lets see if I've understood you:

You want to check:
$$ \sum_{n=1}^\infty n^2 e^{-n/3} $$ ... to see if it converges.

You want to use the integral test - so you derived something unspecified to figure out some sort of number that relates to something important about the integral test ...

You find the function is decreasing and positive for $7\leq x$ - would this be correct?

... and you end up trying to evaluate:
$$ \int_7^\infty -3(x^2+6x+18)e^{-x/3}dx$$ ... which you is doable by parts.

You want to know where the 7 comes from?
Clearly from the limits to the integral test... that unspecified thing you derived earlier?

It helps with this sort of confusion to be very formal in your layout ... start with the expression you want to test and what that test is. How does the integral test work?

 

[gabbagabbahey]

So I integrate it (using Integration by parts) and get to the point where I have to evaluate the definite integral

[(-3e^(-x/3))*(x^2+6x+18)] from 7 to a lim a-->infinity

This doesn't follow from integration by parts. You need to show your steps as for how you got this from $\int_7^{\infty} x^2 e^{-\frac{x}{3}} dx$.

This is where I get confused. My textbook says in sigma evaluate from 1 but in the answer it says plug in 7. I was wondering why this was?

You said it yourself, that the integral test requires that the summand is monotone decreasing on the interval, which is only true of $n^2 e^{-\frac{n}{3}}$ for $n \geq 7$. So, the integral test can only tell you if $\sum_7^{\infty} n^2 e^{-\frac{n}{3}}$ converges, but since the rest of your sum consists of 6 finite terms, if $\sum_7^{\infty} n^2 e^{-\frac{n}{3}}$ converges, so does $\sum_1^{\infty} n^2 e^{-\frac{n}{3}} = \sum_1^{6} n^2 e^{-\frac{n}{3}} + \sum_7^{\infty} n^2 e^{-\frac{n}{3}}$.

 

[Dick]

To get the answer 54 they integrated f(x)=x^2*exp(-x/3) from 0 to infinity. The integral test doesn't really require that f(x) is decreasing everywhere. It should just be decreasing in the limit as x->infinity. And it should be integrable on the rest of the interval you choose where it's not decreasing.

 

[Eats Dirt]

Ok here are my steps

$$x^2e^\frac{-x}{3}$$

Then with integration by parts I set$$u=x^2\\ du=2xdx\\ v=-3e^\frac{-x}{3} \\dv=e^\frac{-x}{3}$$

Now I get.

$$-3x^2e^\frac{-x}{3}-\int-3e^\frac{-x}{3}2xdx$$

Use IBP a second time and now I get

$$u=2x\\<br /> du=2dx\\<br /> dv=-3e^\frac{-x}{3}\\<br /> v=9e^\frac{-x}{3}\\$$

and I end up getting the expression

$$-3e^\frac{-x}{3}(x^2+6x+18)$$

Now here's the problem I encounter if If I try to evaluate this from 7 to a I don't know how to simplify it I don't know how to get rid of the other a's because for example $$-3e^\frac{-a}{3}a^2.$$ Since a→∞ I can't see how to simplify it down to a single limit (which is apparently supposed to be 54).

 

[Dick]

Your latex is coming along nicely, keep working on it. It's good practice. But what you have there is an improper integral. To evaluate the value at infinity you need to take the limit of the expression as a->infinity. l'Hopital's rule can tell that it is zero. The 54 part will come from the value at x=0. As I said, they aren't integrating 7->infinity, they are integrating 0->infinity. Do you see why that's ok?

 

[Eats Dirt]

Your latex is coming along nicely, keep working on it. It's good practice. But what you have there is an improper integral. To evaluate the value at infinity you need to take the limit of the expression as a->infinity. l'Hopital's rule can tell that it is zero.

haha thank you for your compliment :D!
http://www.wolframalpha.com/input/?i=integrate+(x^2)*e^(-x/3)

wolframalpha says it should be 54, and I have no idea how it gets from where I am to there lol.

 

[Dick]

http://www.wolframalpha.com/input/?i=integrate+(x^2)*e^(-x/3)

wolframalpha says it should be 54, and I have no idea how it gets from where I am to there lol.

They (and Wolfram) are integrating 0->infinity. Not 7->infinity. Do you see why that's ok? What's the value of your antiderivative at 0?

 

[Eats Dirt]

They (and Wolfram) are integrating 0->infinity. Not 7->infinity. Do you see why that's ok? What's the value of your antiderivative at 0?

the value is 54! I see where the 54 come from now, thanks! but when say I plug in a as a→∞ is there a way to tell that the terms go to zero? Because I can't plug in a number that big.

EDIT wait I'm guessing l'hospital's rule like you said before? I'm going to try it.

 

[Dick]

the value is 54! I see where the 54 come from now, thanks! but when say I plug in a as a→∞ is there a way to tell that the terms go to zero? Because I can't plug in a number that big.

Like I said, you want the limit. Not the value. Remember things like l'Hopital's theorem? What's the limit of x/exp(x) as x->infinity?

 

[Eats Dirt]

Like I said, you want the limit. Not the value. Remember things like l'Hopital's theorem? What's the limit of x/exp(x) as x->infinity?

ok so it is already in an indeterminate form $$\frac{1}{e^\frac{x}{3}}$$ on the bottom which →∞ and the top also →∞.

So I derive twice and find the limit to be 0.

so when we evaluate the definite integral we are plugging in 7 and a as a →∞.

the limit of a is 0 so it will end up being $$\\ 0- -3e^\frac{−x}{3}(x^2+6x+18)$$ With 7 subbed in for x?Or is it good enough to do the integral for the integral test and then do l'hospital's rule to see the limit?

 

[Dick]

ok so it is already in an indeterminate form $$\frac{1}{e^\frac{x}{3}}$$ on the bottom which →∞ and the top also →∞.

So I derive twice and find the limit to be 0.

so when we evaluate the definite integral we are plugging in 7 and a as a →∞.

the limit of a is 0 so it will end up being $$\\ 0- -3e^\frac{−x}{3}(x^2+6x+18)$$ With 7 subbed in for x?Or is it good enough to do the integral for the integral test and then do l'hospital's rule to see the limit?

You've got everything exactly right. But you seem to be missing part of what I've been saying. They didn't put 7 in for the lower limit. You certainly could. They put 0 in for x. Because it's a simpler number. Do you see why that still proves the integral converges?

 

[Eats Dirt]

You've got everything exactly right. But you seem to be missing part of what I've been saying. They didn't put 7 in for the lower limit. You certainly could. They put 0 in for x. Because it's a simpler number. Do you see why that still proves the integral converges?

Oh yes because if you add constants to a convergent function it will still converge eventually. So we can sub in any convenient point while doing the integral test? or is it somehow bounded by when the series starts. Thanks SOO much by the way I am way more clear on this now.

 

[Dick]

Oh yes because if you add constants to a convergent function it will still converge eventually. So we can sub in any convenient point while doing the integral test? or is it somehow bounded by when the series starts. Thanks SOO much by the way I am way more clear on this now.

Yes! You can put in any convenient number as long as the integral is well defined, (for example the integrand is bounded) between the point where it starts to decrease and your point of convenience.