1. ## Integrating r^2 exp(-2r/a)

Integrating in the limits: infinity -> 0

I have tried integration by substitution and get -ar^2/2 exp(-2r/a) + k, which then gives me zero when I apply the limits. Apparently, the final answer is a^3/4.

Please could I have some help? I am new to integration by substitution btw.

Thanks.

2. Originally Posted by Lex
Integrating in the limits: infinity -> 0

I have tried integration by substitution and get -ar^2/2 exp(-2r/a) + k, which then gives me zero when I apply the limits. Apparently, the final answer is a^3/4.

Please could I have some help? I am new to integration by substitution btw.

Thanks.
When you write "exp(-2r/a)" do you mean:

$\displaystyle [A]\;\;\;r^2e^{\frac{-2r}{a}}$

or

$\displaystyle [B]\;\;\;(r^2)^{\frac{-2r}{a}}$

?

3. It's the first one - [A].

4. Try integration by parts-- twice.

Let $\displaystyle u = r^2$, so $\displaystyle du = 2r\,dr$, $\displaystyle dv = \exp(-2r/a)$, and $\displaystyle v = -\frac{a}{2}\exp(-2r/a)$.

$\displaystyle \lim_{b \to \infty}\left[-\frac{ar^2}{2}\exp(-2r/a)\right]_0^b + \lim_{b \to \infty} \int_0^b ar\exp(-2r/a)\,dr$

Now we have to assume a > 0 for the limit in the left term to exist. If we do this, we can compute it to 0.

Use integration by parts on the remaining integral. I computed it to $\displaystyle a^3/4$ as you said the answer should be.

5. Originally Posted by gosualite
Try integration by parts-- twice.

Let $\displaystyle u = r^2$, so $\displaystyle du = 2r\,dr$, $\displaystyle dv = \exp(-2r/a)$, and $\displaystyle v = -\frac{a}{2}\exp(-2r/a)$.

$\displaystyle \lim_{b \to \infty}\left[-\frac{ar^2}{2}\exp(-2r/a)\right]_0^b + \lim_{b \to \infty} \int_0^b ar\exp(-2r/a)\,dr$

Now we have to assume a > 0 for the limit in the left term to exist. If we do this, we can compute it to 0.

Use integration by parts on the remaining integral. I computed it to $\displaystyle a^3/4$ as you said the answer should be.
Hey Lex, just incase you haven't been introduced to "Integration by Parts" yet, just know that this is what it entails (if you have been introduced and understand it, just skip this post):

Given an integral of the form:

$\displaystyle \int f(x)g'(x)dx$

We can go back to the definition of the power rule to minipulate the equation, and add integration symbols to both sides, and do some more minipulation (all of this is "mathematically legal" of course. If you want me to show you how we can arrive at the following formula, just post here and I'll show you why integration by parts works.) We can arrive at the following formula for an integral of the above form:

$\displaystyle \int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx$

The above is more commonly written, with alittle less obvious notation, as the following:

$\displaystyle \int\;u\;dv = uv\;-\;\int\;v\;du$

One thing to notice in both equations is the$\displaystyle g'(x)$ in the first and the $\displaystyle dv$ in the second. Realizing this, it just means that to find $\displaystyle g(x)$ (i.e. $\displaystyle v$ in the second equation) on the right side of the equation, we just need to integrate it (i.e. $\displaystyle g(x) = \int\;g'(x)$. That may seem really obvious but allot of people miss that and get confused. And as for finding the $\displaystyle f'(x)$ (i.e. $\displaystyle du$ in the second equation) it should be obvious from the notation that $\displaystyle f'(x) = \frac{d}{dx}[f(x)]$, but it might be slightly less obvious in the notation of the second equation (du = $\displaystyle \frac{d}{dx}[u]$). Again, this may seem obvious, but I've seen many people get confused with this part of the notation as well. I perfer writing the equation in the first form, because then people do not so easily forget that $\displaystyle u$ and $\displaystyle v$ are both functions of $\displaystyle x$, and they don't so easily mix which "part" of the "parts" to integrate and which to differentiate.

Another note, you may ask "Which part of the equation do I define as $\displaystyle u$ and which as $\displaystyle dv$?" Its simple to decide. Follow the "LIATE" rule (proposed by Herbert Kasube of Bradley University). Out of the two functions of x you have in your integrand, the first function's type to appear on the following list should be $\displaystyle u$. Obviously, the other function is then, by default, $\displaystyle dv$. Heres the LIATE list:

[1] - L - Logarithmic Functions: [i.e. $\displaystyle log(x)\;,\;ln(x)\;,\;etc.$]
[2] - I - Inverse Trigonomic Functions: [i.e. $\displaystyle arctan(x)\;,\;sin^{-1}(x)\;,\;etc.$]
[3] - A - Algebraic Functions: [i.e. $\displaystyle x^2+6x\;,\;4x^16-x^2+12\;,\;etc.$]
[4] - T - Trigonomic Functions: [i.e. $\displaystyle cos(x)\;,\;cot(x)\;,\;etc.$]
[5] - E - Exponential Functions: [i.e. $\displaystyle e^x\;,\;3^{x}\;,\;etc.$]

So, as an example, say you have:

$\displaystyle \int cos(x)e^xdx$

and you want to know if $\displaystyle cos(x)$ or $\displaystyle e^x$ should be $\displaystyle u$. Follow your finger down the list: neither function is logarithmic, so move down. Niether function is inverse trigonomic so move down. Niether function is algebraic (remeber, not just a function involving algebra, but "algebraic" as defined in the list) so we move down. Bingo! We see that $\displaystyle cos(x)$ is a trigonomic function! So then $\displaystyle u=cos(x)$ and by default $\displaystyle e^x = dv$. Also note that $\displaystyle e^x$ is exponential, and last on the list, so it follows that if you went backwards (i.e. up the list) the first function type you ran into would be your $\displaystyle dv$.

I hope this serves as an ample explaination of integration by parts. Now try to finish the integral problem you originally posted. Use the first few steps that gosualite posted and then go from there. Good luck. Any more questions, post here and I'll be happy to help.

6. Thanks to both of you. I managed to get to the solution. mfetch22, please could you explain how to use the LIATE rule in gosualite's method? I got to the answer by choosing u and dv by trial and error, but it seems that dv must be taken as the exponential function in each integration by parts step, right?

7. Originally Posted by Lex
Thanks to both of you. I managed to get to the solution. mfetch22, please could you explain how to use the LIATE rule in gosualite's method? I got to the answer by choosing u and dv by trial and error, but it seems that dv must be taken as the exponential function in each integration by parts step, right?
I'd be glad to. I am assuming you are refering to the first part of his solution, where he makes the definitions of u and dv? This part right

Originally Posted by gosualite
Try integration by parts-- twice.

Let $\displaystyle u = r^2$, so $\displaystyle du = 2r\,dr$, $\displaystyle dv = \exp(-2r/a)$, and $\displaystyle v = -\frac{a}{2}\exp(-2r/a)$.
if you meant some other part of his work, post here and let me know and I'll show the correct part for which you were reffering to. Assuming you meant the above part in the quoation, this is how we apply the LIATE rule. First, make sure to be checking the post where I detailed each part of LIATE as I go through the explaination. So, we have the following integral for which we need to evaluate which "part" to define as $\displaystyle u$ and which as $\displaystyle dv$ :

$\displaystyle \int\;r^2e^{\frac{-2r}{a}}$

Going through the LIATE process, we first look for any logarithmic functions (hence the L), and neither $\displaystyle r^2$ or $\displaystyle e^{\frac{-2r}{a}}$ are logarithimc functions. Moving on. Then we check for inverse trigonomic functions. Same story, niether function fits "I" for inverse trigonomic. Then we check for "A", algebraic functions, and we get a match. $\displaystyle r^2$ falls under the "algebraic" definitions in LIATE. Therefore, as the rules of LIATE dictate, the first type of function that we come across as we move down the LIATE letters is the "algebraic" type, flagged by the expression $\displaystyle r^2$ in the integrand. Thus, $\displaystyle u=r^2$, and as a consequence the other function is $\displaystyle dv$, giving $\displaystyle dv = e^{\frac{-2r}{a}}$. And, as I explained in my other post, finding $\displaystyle v$ and $\displaystyle du$ is simply a matter of integrating $\displaystyle dv$ and differentiating $\displaystyle u$, as gosualite showed a couple posts above.

Lex, if you have any further questions, post back here and I'll be happy to assist

8. Thanks again. Makes sense. For some reason I didn't recognise r^2 as algebraic, but now I do!

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# itegral r/2 x^2 e -ax^2

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