HSC 2016 MX2 Integration Marathon (archive) (5 Viewers)

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leehuan

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Re: MX2 2016 Integration Marathon

Did they actually give the floor function on the MathSoc Integration Bee?
 

Carrotsticks

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Re: MX2 2016 Integration Marathon

That seems reasonable. Absolute value integrals are only a tiny bit beyond MX2 level.
They can use their curve sketching techniques (the section on absolute values) to quickly sketch the curve and then work with the integral from there.
 

leehuan

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Re: MX2 2016 Integration Marathon

Four consecutive integers can be written in the form (n-1)n(n+1)(n+2)
=(n^2-1)(n^2+2n)
=n^4+2n^3-n^2-2n

One more to this gives

n^4+2n^3-n^2-2n+1
=n^2(n^2+2n-1-2/n+1/n^2)

There is a bit of symmetry. Let u=n-1/n
u^2 = n^2 + 1/n^2 - 2

= n^2(u^2+1+2u)
= n^2(u+1)^2
=n^2(n+1-1/n)^2
=(n^2+n-1)^2
which is a perfect square
They can use their curve sketching techniques (the section on absolute values) to quickly sketch the curve and then work with the integral from there.
Good point. Not HSC standard but that basically says it can be done with HSC techniques.
 

Paradoxica

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Re: MX2 2016 Integration Marathon

This is the volume of a flat based, oblique topped elliptical cylinder with top height 3 and bottom height 1. This is geometrically identical to the volume of a flat topped elliptical cylinder with height 2. The height is 2 and the area of the base is 2pi. Thus the integral is 4pi.
 

InteGrand

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Re: MX2 2016 Integration Marathon

This is the volume of a flat based, oblique topped elliptical cylinder with top height 3 and bottom height 1. This is geometrically identical to the volume of a flat topped elliptical cylinder with height 2. The height is 2 and the area of the base is 2pi. Thus the integral is 4pi.


 
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seanieg89

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Re: MX2 2016 Integration Marathon

Had a quick crack at the integral with the nested square roots in my break.

So the main difficulty in the problem is handling the integrand, it is not at all clear that this formally written object will be a well defined function, and on which domain it will be defined. (These sorts of things can also have different expressions in different parts of their domain).

If we denote this function by g(x), then from the positivity of square roots, we expect that g is a function defined on the interval of integration (at least) that satisfies:

i) (g(x)^2-x)^2=x - g(x)
ii) x < g(x)^2 < x + sqrt(x)

I believe that these conditions uniquely determine the function g on this interval, given by g(x)=(1+sqrt(4x-3))/2.

This function is straightforward to integrate and gives 127/6.


(Upon confirmation of this answer, I will post how I obtained that expression for g(x) later, but until then I won't ruin it for others. I rushed this calculation a bit so an inaccuracy would not surprise me.)
 
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