HSC 2016 MX2 Integration Marathon (archive) (6 Viewers)
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Paradoxica
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Re: MX2 2016 Integration Marathon
} \rfloor} )
Hint: Graph it.
Hint: Graph it.
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Librah
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Re: MX2 2016 Integration Marathon
Tone it down a notch on the integration bee page
Tone it down a notch on the integration bee page
Paradoxica
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Re: MX2 2016 Integration Marathon
}\rfloor} =\int _0 ^1 \frac{\text{d}x}{\lfloor 1- \log_2 {x}\rfloor}$. Note $\log_2 x < 0$ for $0<x<1$, so $1- \log_2 x > 1$ for $0<x<1$. Note that for $i =1,2,3,\ldots $, we have$)
, the integrand is constant, equal to $\frac{1}{i}$. The width of each of these intervals is $2^{1-i}-2^{-i} = 2^{-i}$. So $I = \sum _{i=1} ^{\infty} \frac{2^{-i}}{i}$. But this is a well-known series representation of $\ln 2$ (see link below for more). Hence $I = \ln 2$.$)
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• Some series representations of ln 2: https://en.wikipedia.org/wiki/Natural_logarithm_of_2#Series_representations
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 = - \sum _{k=1} ^{\infty}\frac{x^k}{k}$, valid for $-1\leq x < 1$. Now, let $x=\frac{1}{t}=t^{-1}$, where $x,t\neq 0$, then we have $\ln \left(1-\frac{1}{t}\right) = - \sum _{k=1} ^{\infty}\frac{t^{-k}}{k}$. Since for this, $x$ was in $[-1,1)\backslash \{0\}$, this formula is valid for $t>1$ or $t\leq -1$. So letting $t=2$, we obtain $\underbrace{\ln \left(1-\frac{1}{2}\right)}_{\ln \frac{1}{2}=-\ln 2} = -\sum _{k=1} ^{\infty} \frac{2^{-k}}{k} \Rightarrow \ln 2 = \sum _{k=1} ^{\infty} \frac{2^{-k}}{k}$, as required. (Actually, even easier, just let $x=\frac{1}{2}$ straightaway).$)
Hint: Graph it.
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• Some series representations of ln 2: https://en.wikipedia.org/wiki/Natural_logarithm_of_2#Series_representations
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Paradoxica
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Paradoxica
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Re: MX2 2016 Integration Marathon
 Prove the product of four consecutive$\\ $integers is one less than a perfect square.$ \\ $It is given this property extends to the ring$\\ $of polynomials with integer co-efficients.$ \\ $b) Hence evaluate $\int \frac{2x-3}{x(x-1)(x-2)(x-3)+2})
leehuan
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Re: MX2 2016 Integration Marathon
That seems reasonable. Absolute value integrals are only a tiny bit beyond MX2 level.
Carrotsticks
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Re: MX2 2016 Integration Marathon
They can use their curve sketching techniques (the section on absolute values) to quickly sketch the curve and then work with the integral from there.
leehuan
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Re: MX2 2016 Integration Marathon
=(n^2-1)(n^2+2n)
=n^4+2n^3-n^2-2n
One more to this gives
n^4+2n^3-n^2-2n+1
=n^2(n^2+2n-1-2/n+1/n^2)
There is a bit of symmetry. Let u=n-1/n
u^2 = n^2 + 1/n^2 - 2
= n^2(u^2+1+2u)
= n^2(u+1)^2
=n^2(n+1-1/n)^2
=(n^2+n-1)^2
which is a perfect square
Four consecutive integers can be written in the form (n-1)n(n+1)(n+2) Prove the product of four consecutive$\\ $integers is one less than a perfect square.$ \\ $It is given this property extends to the ring$\\ $of polynomials with integer co-efficients.$)
=(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
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.
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Paradoxica
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Re: MX2 2016 Integration Marathon
ah yes, I forgot the fact that we had a quadratic relation between x and y.
Paradoxica
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Re: MX2 2016 Integration Marathon
^2 + b^2} $ by using the substitution $ x = \frac{a^2+b^2}{y}\\ $It is given that $\{a,b\}\in \mathbb{R}^+ \\ $Hence evaluate $\int_0^\infty \frac{\log{x}\, \text{d}x}{x^2 + 2ax + 2a^2})
Paradoxica
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Re: MX2 2016 Integration Marathon
Drsoccerball
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Paradoxica
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Re: MX2 2016 Integration Marathon
Neither are correct. Also, there is only one solution.One of my answers was 60 and the other was
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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.)
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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