## Help me integrate this!

For the discussion of math. Duh.

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Atre
Posts: 25
Joined: Wed Aug 06, 2008 8:36 pm UTC

### Help me integrate this!

Very quick problem.

integrate
$x^3e^{-bx^2}$

Preview is suggesting the forum imath won't show correctly, this is the sum: http://integrals.wolfram.com/index.jsp?expr=A(x^3e^({bx^2})&random=false

Obviously I go for integration by parts and then rearrange my second term to get something I can do by recognition.

Problem is I get $((x^2/(2b))+(3/(4b^2))Ae^(bx^2)$ disagreeing with Wolfram. I presume that I need $((x^2/(2b))+(2/(4b^2))Ae^{bx^2}$
but I've been staring at my 4lines of working too long and have gone error-blind .

Can anyone pull out my error please?

[my recognition term is [imath]-(3/(4b^2))*(-2bxe^{-bx^2})[/imath] ]
Last edited by Atre on Fri Aug 14, 2009 12:13 am UTC, edited 1 time in total.

skeptical scientist
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### Re: I'm being retarded - Help!

It's a little hard to pinpoint your error when you only show what answer you get, not the steps in your computation. What work did you do to get the answer you got?

Also, you can fix the math/imath tags by surrounding the exponent with braces instead of parentheses:$\int x^3 e^{bx^2} \, dx=?$ Generally speaking, anything that you want the math display software to recognize as a single unit can be surrounded by braces. (For more information, Google "latex help", or else quote people who get math to display correctly to figure out how they did it. If figuring out how to get the math tags to do your bidding is too complicated - latex can be a little daunting to the beginner - feel free to type your math without using them.)
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

Atre
Posts: 25
Joined: Wed Aug 06, 2008 8:36 pm UTC

### Re: I'm being retarded - Help!

integrate
$\int x^3e^{bx^2}$

Do parts and fiddle second term into recognition form
$= -x^2e^{bx^2}/{2b} + \int(2bxe^{bx^2}){3/{4b^2}}$

Final Ans

$= - ({x^2/{2b} +{3/{4b^2}}})e^{bx^2}$

Just some trivial maths from my Quantum course (first step in a Heisenburg proof) but it's just not my day for numeracy...

skeptical scientist
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### Re: I'm being retarded - Help!

Ack, show your work man. When you say, "Do parts and fiddle second term into recognition form," that's like five steps right there. Why must I guess what parts you are using when you integrate by parts? The number one way to avoid stupid mistakes is to write out every step. People will insist on "saving time" by skipping steps, and then repay the time saved, with interest, by redoing the step multiple times to find the error. Writing out every step also makes your work that much more understandable to others.

I'm going to guess you chose your parts as
$\begin{array}{cc} u=x^2/(2b) & v=e^{bx^2} \\ du=x/b \, dx & dv=2bxe^{bx^2} \, dx. \end{array}$

Then $\int x^3 e^{bx^2} \, dx\,=\,\int u dv \,=\, uv-\int v du \,=\, x^2e^{bx^2}/(2b) - \int xe^{bx^2}/b \, dx$
...and I still have no idea where your 3 is coming from, but I do see a few sign errors.

p.s. See how I wrote out everything I did, so you could see exactly how I did the integration by parts and got what I got?
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

Esquilax
Posts: 68
Joined: Wed Dec 24, 2008 3:56 am UTC
Location: Canada

### Re: I'm being retarded - Help!

When integrating by parts, try setting [imath]u = x^2 , dv = xe^{bx^2} dx[/imath]

Edit: Stupidity, and also this is the same as the post above me
Last edited by Esquilax on Thu Aug 13, 2009 5:50 pm UTC, edited 2 times in total.
Spoiler:

skeptical scientist
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Posts: 6142
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### Re: I'm being retarded - Help!

Esquilax wrote:When integrating by parts, try setting [imath]u = x^2 , v = xe^{bx^2}[/imath]

I do believe you mean setting dv=xebx^2dx (at least, if you are using the usual convention [imath]\int udv=uv-\int vdu[/imath]). That's what I did, up to a constant factor.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

Esquilax
Posts: 68
Joined: Wed Dec 24, 2008 3:56 am UTC
Location: Canada

### Re: I'm being retarded - Help!

Yes... yes I did mean that (sorry, I just got up when I wrote that post)
Spoiler:

flangad
Posts: 15
Joined: Thu Aug 13, 2009 7:53 pm UTC

### Re: I'm being retarded - Help!

double substitution?

$u=bx^2, xdx=\frac{du}{2b}$
$\int \frac{ue^{u} du}{2b^2}=\frac{1}{2b^2} \int{ue^u du} = \frac{1}{2b^2}(bx^2e^{bx^2}-e^{bx^2})$

*edit: oops exponent mistake...
Last edited by flangad on Thu Aug 13, 2009 8:36 pm UTC, edited 1 time in total.

skeptical scientist
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### Re: I'm being retarded - Help!

flangad, [imath]e^{bu} \, \neq \, e^be^u[/imath].

You can do something along those lines, but in the end you're still using substitution and integration by parts; you're just reversing the order of the two.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

Atre
Posts: 25
Joined: Wed Aug 06, 2008 8:36 pm UTC

### Re: I'm being retarded - Help!

skeptical scientist wrote:Ack, show your work man. When you say, "Do parts and fiddle second term into recognition form," that's like five steps right there. Why must I guess what parts you are using when you integrate by parts?

Ah jeez, I SO deserved that slap on the wrist.

The error was in my parts selection. "Do parts and fiddle second term into recognition form" was the mistake showing itself, recognition had to be used in the first stage of working.... This the only way to deal with a compound exponent.

I selected

[imath]v = x^3[/imath] giving [imath]dv = 3x^2[/imath]
[imath]du = e^{-bx^2}[/imath] giving [imath]u = -e^{-bx^2}/2bx[/imath]

I think the error was trying to integrate the compound exponent like that... Tis incorrect. Error follows on from the above.

skeptical scientist
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### Re: I'm being retarded - Help!

Atre wrote:[imath]du = e^{-bx^2}[/imath] giving [imath]u = -e^{-bx^2}/2bx[/imath]

Yeah, that was your error. Your choice of du does not have an elementary antiderivative; if [imath]u = -e^{-bx^2}/2bx[/imath], then, by the quotient rule, $\frac{du}{dx} \, \, = \, \, \frac{2bxe^{-bx^2}2bx+2be^{-bx^2}}{(2bx)^2} \, \, = \, \, e^{-bx^2}+\frac{e^{-bx^2}}{2bx^2},$ which is the term you had as du plus another term.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

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