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HSC 2013-14 MX1 Marathon (archive) (4 Viewers)

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seanieg89

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Re: HSC 2013 3U Marathon Thread

Oh ye of little faith.
 
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Re: HSC 2013 3U Marathon Thread

Ok ok proof coming.



The 'trick' is to recognise that you can split the terms, and you can take out whatever does not depend on the summation. Then it is a matter of recognising an infinite sum. Since the result is true for the 'n' sum, then it is the same for the 'm' sum and the result is (3/2)^2 = 9/4.


You bereive me nao?
 
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Re: HSC 2013 3U Marathon Thread

<a href="http://www.codecogs.com/eqnedit.php?latex=Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" title="Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" /></a>

Edit: This is at a 3U difficulty, right guys?
This is more leaning toward a 4U question.
 

Capt Rifle

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Re: HSC 2013 3U Marathon Thread

This is more leaning toward a 4U question.
Alrighty then what about this:

<a href="http://www.codecogs.com/eqnedit.php?latex=Find \int \frac{x^2dx}{\sqrt{x-1}}\: \: \: using\: the\: substitution\: x=u^2@plus;1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Find \int \frac{x^2dx}{\sqrt{x-1}}\: \: \: using\: the\: substitution\: x=u^2+1" title="Find \int \frac{x^2dx}{\sqrt{x-1}}\: \: \: using\: the\: substitution\: x=u^2+1" /></a>
 

HeroicPandas

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Re: HSC 2013 3U Marathon Thread

<a href="http://www.codecogs.com/eqnedit.php?latex=Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" title="Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" /></a>

Edit: This is at a 3U difficulty, right guys?
A 3U human can do it...
scan0016.jpg
 
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Re: HSC 2013 3U Marathon Thread

Actually sorry - I didn't see that the substitution was given. It is doable by a 3U student.
 

Sy123

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Re: HSC 2013 3U Marathon Thread

 
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seanieg89

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Re: HSC 2013 3U Marathon Thread

By just considering areas you wouldn't need parts.
 

Sy123

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Re: HSC 2013 3U Marathon Thread

By just considering areas you wouldn't need parts.
This.

I'm not silly enough to post a 4U question in a 3U thread.
In fact that integral is 2U.
 

Sy123

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Re: HSC 2013 3U Marathon Thread



 

seanieg89

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Re: HSC 2013 3U Marathon Thread

 
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HeroicPandas

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Re: HSC 2013 3U Marathon Thread

Great question.



Nice work.

=====================



complete the square

int 1/[(x+5)^2 + 25] = 0.2 arctan(x+5/5) +C

woops thought this was 4U marathon...

 
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Sy123

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Re: HSC 2013 3U Marathon Thread





EDIT: Have I already posted this question? I have a feeling I already have.
 
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