HSC 2012-14 MX2 Integration Marathon (archive) (5 Viewers)

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HeroicPandas

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Re: MX2 Integration Marathon

im not saying that im better than you, i know u can beat me in ext and ext2 :D

the question i posted up, u can use that way cos the power of e isnt a linear :p
 

U MAD BRO

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Re: MX2 Integration Marathon

im not saying that im better than you, i know u can beat me in ext and ext2 :D

the question i posted up, u can use that way cos the power of e isnt a linear :p
i hate mx1 !
do we use substitution?
do these then :p

and

:D
 

HeroicPandas

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Re: MX2 Integration Marathon

i hate mx1 !
do we use substitution?
do these then :p

and

:D
LOL all this time i've been doing 4unit integration...i never knew that, feels good!
the first on is easy, i'll do the 2nd one
 

barbernator

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Re: MX2 Integration Marathon

This is a double integral over a keystone shape in polar coordinates. (so limits are finite, and not a complete circle).

I've tried all sorts of things. I am not even sure if there is an analytic solution. Is there away to show there isn't one?
why do you post this, and you can't even do other simple questions?
 

HeroicPandas

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Re: MX2 Integration Marathon

ahh its too hard, the 2nd one and 3rd one, IM MAD! its late now i gotta go to sleep, cya and remember to try to do my question :D
 

barbernator

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Re: MX2 Integration Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\int e^{e^{x}@plus;x}=\int e^{e^{x}}e^{x}\\ let~u~=e^{x}\\ du=e^xdx\\ =e^{e^{x}} @plus; C" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int e^{e^{x}+x}=\int e^{e^{x}}e^{x}\\ let~u~=e^{x}\\ du=e^xdx\\ =e^{e^{x}} + C" title="\int e^{e^{x}+x}=\int e^{e^{x}}e^{x}\\ let~u~=e^{x}\\ du=e^xdx\\ =e^{e^{x}} + C" /></a>
 

barbernator

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Re: MX2 Integration Marathon

is there a fast way to do this question?
the way i do it is with complex numbers and it gets really messy.
u can do partial fracts but it takes bloody ages
 

Sanjeet

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<a href="http://www.codecogs.com/eqnedit.php?latex=\int e^{e^{x}@plus;x}=\int e^{e^{x}}e^{x}\\ let~u~=e^{x}\\ du=e^xdx\\ =e^{e^{x}} @plus; C" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int e^{e^{x}+x}=\int e^{e^{x}}e^{x}\\ let~u~=e^{x}\\ du=e^xdx\\ =e^{e^{x}} + C" title="\int e^{e^{x}+x}=\int e^{e^{x}}e^{x}\\ let~u~=e^{x}\\ du=e^xdx\\ =e^{e^{x}} + C" /></a>
Lol so simple, can't believe no one got it before
 
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Re: MX2 Integration Marathon

Omg I just did the e^x one without anything!
 
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