Omnimaga
General Discussion => Other Discussions => Math and Science => Topic started by: Sorunome on November 07, 2012, 06:54:33 pm
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Hey, today in AP Calculus we learned something called "Integration by Ports" for integrating and i don't really get it.
I think it is something like if you have integral(f(x)*g(x)) then the solution is f(x)G(x)+integral(f'(x)G(x))+c
Is that correct?
If not, please help me by explaining :D
Thanks in advice.
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Derp! not in BC, so me not be there yet!
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I don't really know where in calculus i'm currently XD
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AB, BC, or 2/3?
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well, just talked with jacobly of irc and it is integral(f(x)g(x)) = f(x)G(x) - integral(f'(x)G(x))
G(x) is integral(g(x))
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okay... well my AB-ness doesn't know that yet
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*integral(f(x)g(x)) = f(x)G(x) - integral(f'(x)G(x)) + c
And now you know something epic once you get into BC :D
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Your in BC if your doing integration already or an equivalent course.
Integration by parts is the reverse of the product rule you learned for differentiation
If you have an integral of the form integrate(u*dv) it is the same as u*v-integrate(v*du)
an example of how to use this would be like this
take the antiderivative of x*sin(x). You would start by saying that x is u and sin(x) is dv
integrate dv(sin(x)) to get -cos(x)=v. and you differentiate u(x) to get 1(du).
you then have x*-cos(x)-integrate(-cos(x)*1)
Once you work that out you get sin(x)-x*cos(x)
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well, I'm in AB, but I read ahead
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take the antiderivative of x*sin(x). You would start by saying that x is u and sin(x) is dv
integrate dv(sin(x)) to get -cos(x)=v. and you differentiate u(x) to get 1(du).
you then have x*-cos(x)-integrate(-cos(x)*1)
Once you work that out you get sin(x)-x*cos(x)
That was a example we did in class XD
And i like the raw theory more, thank you anyways, i get it now :D
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its a common simple example. If you want to see an interesting example look up sin(x)*e^x its a really interesting use of the rules
pimathbrainiac. How far up are you? If you ever want any help just ask here on omni theres lots of good people here to help you with math problems.
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integral(sin(x)*e^x)
Erm, how is that possible? I mean, you never get rid of e^x or sin/cos multiplication in integral O.o
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no but you end up having the integral be on both sides then its just algebra to work through it
u=sin(x)
dv=e^x
integrate(sin(x)*e^x)=sin(x)*e^x-integrate(cos(x)*e^x)
for this integral you repeat the process
u=cos(x)
dv=e^x
integrate(cos(x)*e^x)= cos(x)*e^x-integrate(-sin(x)*e^x)
move the negative out of the integral
integrate(cos(x)*e^x)= cos(x)*e^x+integrate(sin(x)*e^x)
substitute back in and you get
integrate(sin(x)*e^x)=sin(x)*e^x-(cos(x)*e^x+integrate(sin(x)*e^x))
integrate(sin(x)*e^x)=sin(x)*e^x-cos(x)*e^x-integrate(sin(x)*e^x)
2*integrate(sin(x)*e^x)=sin(x)*e^x-cos(x)*e^x
integrate(sin(x)*e^x)=(sin(x)*e^x-cos(x)*e^x)/2
thats one of my favorite examples my TA showed me on integration.
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no but you end up having the integral be on both sides then its just algebra to work through it
u=sin(x)
dv=e^x
integrate(sin(x)*e^x)=sin(x)*e^x-integrate(cos(x)*e^x)
for this integral you repeat the process
u=cos(x)
dv=e^x
integrate(cos(x)*e^x)= cos(x)*e^x-integrate(-sin(x)*e^x)
move the negative out of the integral
integrate(cos(x)*e^x)= cos(x)*e^x+integrate(sin(x)*e^x)
substitute back in and you get
integrate(sin(x)*e^x)=sin(x)*e^x-(cos(x)*e^x+integrate(sin(x)*e^x))
integrate(sin(x)*e^x)=sin(x)*e^x-cos(x)*e^x-integrate(sin(x)*e^x)
2*integrate(sin(x)*e^x)=sin(x)*e^x-cos(x)*e^x
integrate(sin(x)*e^x)=(sin(x)*e^x-cos(x)*e^x)/2
thats one of my favorite examples my TA showed me on integration.
Cool, thanks, I get it! :D
The funny thing is that jacobly was explaining it to me on the same time via irc, lol
Why is integrating so more complicated than deriving? :D
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Why is integrating so more complicated than deriving? :D
In general, it isn't. With the problems you're working with, it's because integration maps functions to functions of at least the same class, whereas derivatives map functions to functions of at most the same class. (In other words, the same basic reason that division is harder than multiplication).
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same reason taking the square root of something is harder then squaring it. Reversing processes is often harder then the original process. Things get even harder for integrating when you move to multiple variables. I've always liked deriving a lot more then integration though recently I've begun liking integration more. Its much more of an art like logical proof or trig identities
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Yeah I learnt that just this semester too. It was some weird stuff like integrate(u*dv) = u*v+integrate(du*v) or something similar. Pretty useful when you have to integrate something hard to integrate such as ln x. You'll just have to pose u = ln x and dv = whatever else there is in the equation so du would be 1/x and v = integrate(dv), so it'll be easier to integrate stuff.
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Woo, Ultraviolet Voodoo! XD Oh, mnemonics... Also: Ruler, we learned integration in the second half of AB, but then didn't learn things like integration by parts till BC. So maybe you're talking about college courses, but AP-wise, integration definitely does NOT necessarily mean you are in BC. In fact, if you didn't learn to integrate at all, you'd be pretty much screwed on the AP test... =P
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I meant the advanced integration techniques. Most of the advanced techniques are BC only.
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Ah yeah, stuff like that : http://cedprog.pagesperso-orange.fr/prepa/math/Chap_13.pdf :P
(french isn't really important, maths matter here ^^)
Also, I made a program some time ago, that smartly makes integration by parts steps by steps, for the nspire, in basic :
http://tiplanet.org/forum/archives_voir.php?id=4481 (french and english !)
Critor also made one with a GUI and also steps by steps, in Lua, but I can't find it hosted, mayeb it's not ready yet.