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Messages - Sorunome
Pages: 1 ... 495 496 [497] 498 499 ... 591
7441
« on: November 08, 2012, 07:16:44 pm »
We don't know, it could be or it could be that it just emulates a z80
7442
« on: November 08, 2012, 06:54:18 pm »
Haha, it looks cool IMO! awesome work  EDIT: How added the calc to wikipedia? 
7443
« on: November 08, 2012, 06:37:29 pm »
Hopefully this will also power up the community activity * Sorunome stares at omnis low activity
7444
« on: November 08, 2012, 05:25:28 pm »
Most Epic News Ever. I'll wait now before I'll buy a new calc 
7445
« on: November 07, 2012, 07:33:48 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?
7446
« 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
7447
« on: November 07, 2012, 07:09:52 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
7448
« 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
7449
« 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))
7450
« on: November 07, 2012, 07:00:02 pm »
I don't really know where in calculus i'm currently
7451
« 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.
7452
« on: November 07, 2012, 06:48:47 pm »
Haha, that's pretty cool! And that are a lot of tracks O.o
7453
« on: November 07, 2012, 06:47:23 pm »
I'm not very good with math
Was Xeda account hacked?? 
Or her brain
7454
« on: November 07, 2012, 06:22:32 pm »
I'm mostly programming, sorry But my record on the 3x3x3 is 36 seconds
7455
« on: November 07, 2012, 04:12:54 pm »
looking epic with the new screenshot! Great work so far
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