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 06-20-2012, 06:33 PM #1 veteran     Join Date: Oct 2011 Location: Zooom baby. Posts: 3,082 Integration question I got wrong and idk why :( Basically need help with this, probably something really retarrded by me, if someone couuld point out where I went wrong/how to do it right would be cool. sorry for the shadow btw.
 06-20-2012, 07:15 PM #2 veteran     Join Date: Aug 2009 Location: Stanford, CA USA Posts: 3,306 Re: Integration question I got wrong and idk why :( Forget the image or create a better one that is crystal clear because this is not very good (hold camera lens parallel to page or even put in a window that has plenty of light or illuminate it with a lamp and FOCUS! lol ) In any case just type in characters what your integrating function and limits are in a plain post or what the problem is about and try the homework session in the future.
 06-20-2012, 07:53 PM #3 adept   Join Date: Aug 2010 Posts: 887 Re: Integration question I got wrong and idk why :( Ummm -- I took a quick look. It was a pretty obvious mistake, so here goes... I[f(x)] means the indefinite integral of f(x). D[f(x)] means the derivative of f(x). Ok, so you want to find: I[(e^(2x) - 2)^2] Here's basically the way you reasoned through this, I think: You saw this: Let D[blah] = D[(e^(2x) - 2)^3] Then: D[blah] = 3*(e^(2x) - 2)^2 * 2 * e^(2x) D[blah] = [6 * e^(2x)] * [(e^(2x) - 2)^2] You basically realized that the bold terms match, so you just divided [blah] by [6*e^(2x)], hoping this would give you the integral you're looking for. But you didn't account for the fact that when you take the derivative of [blah]/[6*e^(2x)], you have to use the product/quotient rule and differentiate both parts of the product/quotient. In other words, you can't do this kind of thing unless you get a constant coefficient (you got a coefficient in terms of 'x' instead). You could probably do this using integration by parts, but it's easiest just to distribute out the term you're trying to integrate first: I[(e^(2x) - 2)^2] = I[e^(4x) - 4e^(2x) + 4] The right hand side should be pretty easy to deal with...
06-21-2012, 03:30 AM   #4
veteran

Join Date: Oct 2011
Location: Zooom baby.
Posts: 3,082
Re: Integration question I got wrong and idk why :(

Quote:
 Originally Posted by pocketzeroes Ummm -- I took a quick look. It was a pretty obvious mistake, so here goes... I[f(x)] means the indefinite integral of f(x). D[f(x)] means the derivative of f(x). Ok, so you want to find: I[(e^(2x) - 2)^2] Here's basically the way you reasoned through this, I think: You saw this: Let D[blah] = D[(e^(2x) - 2)^3] Then: D[blah] = 3*(e^(2x) - 2)^2 * 2 * e^(2x) D[blah] = [6 * e^(2x)] * [(e^(2x) - 2)^2] You basically realized that the bold terms match, so you just divided [blah] by [6*e^(2x)], hoping this would give you the integral you're looking for. But you didn't account for the fact that when you take the derivative of [blah]/[6*e^(2x)], you have to use the product/quotient rule and differentiate both parts of the product/quotient. In other words, you can't do this kind of thing unless you get a constant coefficient (you got a coefficient in terms of 'x' instead). You could probably do this using integration by parts, but it's easiest just to distribute out the term you're trying to integrate first: I[(e^(2x) - 2)^2] = I[e^(4x) - 4e^(2x) + 4] The right hand side should be pretty easy to deal with...
Cheers man yeah that is exaclty what I did, completely forgot that i would have to differentiate the 6*e^2x.
And yeah does work out fine doing it the easier way, thanks a bunch.

 06-21-2012, 11:17 AM #5 Pooh-Bah   Join Date: Jun 2009 Posts: 5,832 Re: Integration question I got wrong and idk why :( You draw your pi funny :| it's meant to be two separate downstrokes (three strokes total, the horizontal first)
06-23-2012, 09:44 AM   #6
old hand

Join Date: Jul 2010
Posts: 1,835
Re: Integration question I got wrong and idk why :(

Quote:
 Originally Posted by Kittens You draw your pi funny
+1 That's not pi, it's a headless stick figure lol

06-24-2012, 11:40 PM   #7
Carpal \'Tunnel

Join Date: Jun 2006
Location: drinking in a saloon
Posts: 11,900
Re: Integration question I got wrong and idk why :(

Quote:
 Originally Posted by Kittens You draw your pi funny
that's what she said

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