HSC 2016 MX2 Integration Marathon (archive) (2 Viewers)

Paradoxica · Feb 25, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
I think I posted this and got told off because app it wasn't elementary...

This is elementary. It most definitely is.

Paradoxica · Feb 25, 2016

Re: MX2 2016 Integration Marathon

$$\noindent Easy: $\int_0^\infty 2^{-\lfloor x \rfloor} $d$x \\$Hard: $\int_0^1 \frac{\text{d}x}{\lfloor 1-\log_2{(1-x)} \rfloor}$

Hint: Graph it.

Librah · Feb 25, 2016

Re: MX2 2016 Integration Marathon

Tone it down a notch on the integration bee page

leehuan · Feb 25, 2016

Re: MX2 2016 Integration Marathon

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

Paradoxica · Feb 25, 2016

Re: MX2 2016 Integration Marathon

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

No, but they did put in some absolute values.

InteGrand · Feb 25, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent Easy: $\int_0^\infty 2^{-\lfloor x \rfloor} $d$x \\$Hard: $\int_0^1 \frac{\text{d}x}{\lfloor 1-\log_2{(1-x)} \rfloor}$

Hint: Graph it.

$\noindent For the second one, let $I = \int _0 ^1 \frac{\mathrm{d}x}{\lfloor 1-\log_2 {(1-x)}\rfloor} =\int _0 ^1 \frac{\text{d}x}{\lfloor 1- \log_2 {x}\rfloor}$. 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 $, we have$

$\begin{align*}\lfloor 1-\log_2{x\rfloor } & = i \\ \Longleftrightarrow i \leq 1-\log_2{x} < i+1 \\ \Longleftrightarrow i-1 \leq -\log_2{x} < i \\ \Longleftrightarrow -i< \log_2{x} \leq 1-i \\ \Longleftrightarrow 2^{-i}< x \leq 2^{1-i}. \end{align*}$

$\noindent For each such interval of $2^{-i}< x \leq 2^{1-i}$ ($i=1,2,3,\ldots$), 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 $I = \ln 2$.$

_________

• Some series representations of ln 2: https://en.wikipedia.org/wiki/Natural_logarithm_of_2#Series_representations

_________

$\noindent One way to show that $\sum _{k=1} ^{\infty} \frac{2^{-k}}{k}=\ln 2$ is to first start with the Taylor series $\ln (1-x) = - \sum _{k=1} ^{\infty}\frac{x^k}{k}$, 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).$

Paradoxica · Feb 26, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent Easy: $\int_0^\infty 2^{-\lfloor x \rfloor} $d$x$

$$\noindent For the second one: For $0<x<1$, the integrand is equal to $1$. For $1<x<2$, the integrand is equal to $\frac{1}{2}$. Each unit interval is corresponding to a negative power of $2$, and so the integral of that interval is the corresponding negative power of 2. The integral is then equal to the limiting sum of the sequence $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots + \frac{1}{2^n}$. This is a well known geometric series which is equal to $2$. Thus, the integral is equal to 2.$

Paradoxica · Feb 26, 2016

Re: MX2 2016 Integration Marathon

$$\noindent a) 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 · Feb 26, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
No, but they did put in some absolute values.

That seems reasonable. Absolute value integrals are only a tiny bit beyond MX2 level.

Carrotsticks · Feb 26, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
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 · Feb 26, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\noindent a) Prove the product of four consecutive$\\ $integers is one less than a perfect square.$ \\ $It is given this property extends to the ring$\\ $of polynomials with integer co-efficients.$

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

Carrotsticks said:
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.

menodiks · Feb 26, 2016

Re: MX2 2016 Integration Marathon

Paradoxica · Feb 26, 2016

Re: MX2 2016 Integration Marathon

menodiks said:

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 · Feb 27, 2016

Re: MX2 2016 Integration Marathon

menodiks said:

Paradoxica said:
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.

$\noindent This answer ($4\pi$) would be the case if the integrand were $1$, but it is $x^2$ in fact. The triple integral $I$ can be set up as the following iterated integral: $I=\int _{y=-2} ^{y=2} \int _{x=-\frac{1}{2} \sqrt{4-y^2}} ^{x=\frac{1}{2} \sqrt{4-y^2}} \int _{z=0} ^{z=2-x} x^2 \text{ d}z\text{ d}x \text{ d}y$. (By symmetry of the integrand and $S$ in the $y$ direction, we can write $I$ as double this integral with $y$ limits from 0 to 2 instead.)$

$\noindent The reason for this setup is that our region $S$ is this sort of wedge and part of an elliptical cylinder. We fix $y$ first and for a fixed $y\in [-2,2]$, we find what $x$ varies between by solving for $x$ in the cylinder equation. Then for fixed $x,y$, we have $z$ going from 0 to $2-x$ (the plane). Once we have set up this iterated integral, all we need to do is compute it, which is relatively straightforward, just a bit tedious (the main tedious bit being having to integrate something like a $\left(4-y^2\right)^{\frac{3}{2}}$ when doing the $y$ integral at the end). It will evaluate to $\pi$, so the final answer is $I=\pi$.$

Paradoxica · Feb 27, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
.

ah yes, I forgot the fact that we had a quadratic relation between x and y.

Paradoxica · Feb 27, 2016

Re: MX2 2016 Integration Marathon

$$\noindent Evaluate $ \int_{0}^\infty \frac{\log{x}\,\text{d}x}{(x + a)^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 · Feb 28, 2016

Re: MX2 2016 Integration Marathon

$\int_7^{13} \sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\cdots}}}}}}}}}\,$d$x$

Drsoccerball · Feb 28, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int_7^{13} \sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\sqrt{x+\cdots}}}}}}}}}\,$d$x$

One of my answers was 60 and the other was $\int_7^{13}\sqrt{x^2-1}dx$

Paradoxica · Feb 28, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
One of my answers was 60 and the other was $\int_7^{13}\sqrt{x^2-1}dx$

Neither are correct. Also, there is only one solution.

seanieg89 · Feb 29, 2016

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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