Author Topic: Integration by Parts (Read 8299 times)

Sorunome · « **on:** November 07, 2012, 06:54:33 pm »

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

Thanks in advice.

pimathbrainiac · « **Reply #1 on:** November 07, 2012, 06:58:57 pm »

Derp! not in BC, so me not be there yet!

Sorunome · « **Reply #2 on:** November 07, 2012, 07:00:02 pm »

I don't really know where in calculus i'm currently

pimathbrainiac · « **Reply #3 on:** November 07, 2012, 07:00:31 pm »

AB, BC, or 2/3?

Sorunome · « **Reply #4 on:** November 07, 2012, 07:03:24 pm »

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))

pimathbrainiac · « **Reply #5 on:** November 07, 2012, 07:04:09 pm »

okay... well my AB-ness doesn't know that yet

Sorunome · « **Reply #6 on:** November 07, 2012, 07:07:14 pm »

*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

ruler501 · « **Reply #7 on:** November 07, 2012, 07:07:24 pm »

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)

pimathbrainiac · « **Reply #8 on:** November 07, 2012, 07:08:29 pm »

well, I'm in AB, but I read ahead

Sorunome · « **Reply #9 on:** November 07, 2012, 07:09:52 pm »

Quote from: ruler501 on November 07, 2012, 07:07:24 pm

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

And i like the raw theory more, thank you anyways, i get it now

ruler501 · « **Reply #10 on:** November 07, 2012, 07:18:33 pm »

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.

Sorunome · « **Reply #11 on:** November 07, 2012, 07:23:07 pm »

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

ruler501 · « **Reply #12 on:** November 07, 2012, 07:28:25 pm »

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.

Sorunome · « **Reply #13 on:** November 07, 2012, 07:33:48 pm »

Quote from: ruler501 on November 07, 2012, 07:28:25 pm

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!

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?

AngelFish · « **Reply #14 on:** November 07, 2012, 07:36:00 pm »

Quote from: Sorunome on November 07, 2012, 07:33:48 pm

Why is integrating so more complicated than deriving?

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).

Author Topic: Integration by Parts (Read 8299 times)

Sorunome

Integration by Parts

pimathbrainiac

Re: Integration by Ports

Sorunome

Re: Integration by Ports

pimathbrainiac

Re: Integration by Ports

Sorunome

Re: Integration by Ports

pimathbrainiac

Re: Integration by Ports

Sorunome

Re: Integration by Ports

ruler501

Re: Integration by Ports

pimathbrainiac

Re: Integration by Ports

Sorunome

Re: Integration by Parts

ruler501

Re: Integration by Parts

Sorunome

Re: Integration by Parts

ruler501

Re: Integration by Parts

Sorunome

Re: Integration by Parts

AngelFish

Re: Integration by Parts