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HSC 2012-14 MX2 Integration Marathon (archive) (4 Viewers)

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Makematics

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Re: MX2 Integration Marathon

The fraction is still a unit that remains intact - it doesn't need to be broken up. As long as you think of the fraction as a variable itself, its exactly the same.
Seriously, if anyone sees expanding [1+5/(x+2)]^5 as being more difficult that (u+5)^5, they are going to struggle with a lot in Extension 2.
lol wow calm down, i understand the fraction doesn't have to be broken down, i'm just saying your expansion will look much messier than my expansion. i understand that the algebraic skill required for both expansions is similar...
 

braintic

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Re: MX2 Integration Marathon

lol wow calm down, i understand the fraction doesn't have to be broken down, i'm just saying your expansion will look much messier than my expansion. i understand that the algebraic skill required for both expansions is similar...
Calm down?? You're completely misreading me.
 

Sy123

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Re: MX2 Integration Marathon





No room for the +c :/
 

Kipling

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Re: MX2 Integration Marathon

Integrate: (7x-3)/(6+4x-x^2)^0.5
 

Sy123

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Re: MX2 Integration Marathon

The following question is set in http://en.wikipedia.org/wiki/Three-dimensional_space with the standard coordinates.





















At some y, I think the notation is (x,k,z), then there will be a 2d curve

So treating y like a constant, we find the area of a general cross section.



Then collecting all the cross sections:



Since they are definite integrals of the same thing but different variables.





Is that all? No rigour issues or anything?





Is this right so far?
 
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seanieg89

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Re: MX2 Integration Marathon





















At some y, I think the notation is (x,k,z), then there will be a 2d curve

So treating y like a constant, we find the area of a general cross section.



Then collecting all the cross sections:



Since they are definite integrals of the same thing but different variables.





Is that all? No rigour issues or anything?





Is this right so far?
Pretty much, I would prefer if you did the work to show the epsilon term tends to zero, this is not too hard.

Also it is probably better to describe your slices as being perpendicular to the y-axis (rather than the x-y plane), ie every point in a slice has the same y-coordinate, this makes it unambiguous.

What is nice is that these volume computations are standard mx2 work, even if the notation for the solids is slightly beyond.
 
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Re: MX2 Integration Marathon

If you differentiate twinkie's answer you get the integrand.
 
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Re: MX2 Integration Marathon

Somehow the dx flew up top.
 
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