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

Paradoxica · Jan 5, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
It can't be that messy... can it? My intuition tells me there is a very simple way to do it, and it is killing me because I don't see anything that can make it a three-liner.

Nvm, I got it. Typing up my solution now

leehuan · Jan 5, 2016

Re: MX2 2016 Integration Marathon

Wow nice question InteGrand

Note that the proof can still be achieved in the counter case where f(x) is concave up over [a,b]

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InteGrand · Jan 5, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Wow nice question InteGrand

Note that the proof can still be achieved in the counter case where f(x) is concave up over [a,b]

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$Nice. Here was my method:$

$Let $I=\int _a ^b \left(f(x) + f^{-1} (x) \right) \text{ d}x$, so $I = \int _a ^b f(x) \text{ d}x + \int _a ^b f^{-1} (x) \text{ d}x$. For the second integral here, let $u=f^{-1}(x)$, so $x = f(u)$, $\mathrm{d}x = f^\prime (u) \text{ d}u$, and the bounds are the same since when $x=a$, $u=f(a)=a$, similarly at $b$. So we have$

$\begin{align*}I&=\int _a ^b f(x) \text{ d}x + \int _a ^b u\cdot f^\prime (u) \text{ d}u \\ &= \int _a ^b f(x) \text{ d}x + \Big{[} f(u)\cdot u \Big{]} _a ^b - \int _a ^b f(u) \text{ d}u \quad (\text{using Integration by Parts on second integral}) \\ &= f(b)\cdot b - f(a) \cdot a \quad (\text{the two integrals cancel}) \\ &= b^2 - a^2 \quad (\text{as }f(a)=a,f(b)=b). \end{align*}$

Paradoxica · Jan 6, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent \textbf{New Question} -- Goodbye 2015$

$$\noindent Evaluate $\int^1_0 \left (\frac{2}{x + 1} - x - 1 \right )^{2014}$d$x.$

Bloody hell that Recurrence formula is a bitch.

$$\noindent Replace $x+1$ with $x$ and shift the border of integration by one unit to obtain:$\\ \int^1_0 \left (\frac{2}{x + 1} - x - 1 \right )^{2014}$d$x \equiv \int_1^2 \left ( \frac{2}{x} -x \right )^{2014}$d$x\\$Claim: $P(n) : \int_1^2 \left ( \frac{2}{x} -x \right )^{2n}$d$x \equiv \frac{1}{2n+1}\; \forall n \in \mathbb{N} \\$Proof: $P(0)$ is trivial. For $n=1$, we have: $\\ \int_1^2 \left (\frac{2}{x}-x \right )^2 $d$x = \int_1^2 \frac{4}{x^2} - 4 + x^2 $d$x =\left [ -\frac{4}{x} -4x + \frac{x^3}{3} \right ]_1^2 \\ = -2-8+\frac{8}{3}+4+4-\frac{1}{3}=\frac{1}{3} \Rightarrow P(1)$ is true.$\\$Assume $P(n)$ is true: $ \rightarrow \int_1^2 \left ( \frac{2}{x} -x \right )^{2n}$d$x = \frac{1}{2n+1} \\I_1 = \int_1^2 \left ( \frac{2}{x} -x \right )^{2n+2}$d$x = \left[x\left(\frac{2}{x}-x\right)^{2n+2}\right]_1^2-(2n+2)\int_1^2 \left(\frac{2}{x}-x\right)^{2n+1}x\left(-\frac{2}{x^2}-1\right)$d$x$

$$\noindent$I_1 = 2\left(\frac{2}{2}-2\right)^{2n+2}-1\left(\frac{2}{1}-1\right)^{2n+2}+(2n+2)\int_1^2\left(\frac{2}{x}-x\right)^{2n+1}\left(\frac{2}{x}-x+2x\right)$d$x\\I_1 = 1+(2n+2)\int_1^2\left(\frac{2}{x}-x\right)^{2n+2}$d$x+(2n+2)\int_1^22x\left(\frac{2}{x}-x\right)^{2n+1}$d$x\\I_1 = 1+(2n+2)I_1 +(2n+2)\int_1^2\left(\frac{2}{x}-x\right)^{2n+1}$d$(x^2)\\I_2 = \int_1^2\left(\frac{2}{x}-x\right)^{2n+1}$d$(x^2) = \left[x^2\left(\frac{2}{x}-x\right)^{2n+1}\right]_1^2-(2n+1)\int_1^2x^2\left(\frac{2}{x}-x\right)^{2n}\left(-\frac{2}{x^2}-1\right)$d$x\\I_2 = 4\left(\frac{2}{2}-2\right)^{2n+1}-1\left(\frac{2}{1}-1\right)^{2n+1}+(2n+1)\int_1^2x^2\left(\frac{2}{x}-x\right)^{2n}\left(\frac{2}{x^2}+1\right)$d$x\\I_2=-5+(2n+1)I_3\\I_3=\int_1^2 x\left(\frac{2}{x}+x\right)\left(\frac{2}{x}-x\right)^{2n}$d$x=\int_1^2 (x^2+2)\left(\frac{2}{x}-x\right)^{2n}$d$x\\I_3=\int_1^2x^2\left(\frac{2}{x}-x\right)^{2n}\left(\frac{2}{x^2}+1\right)$d$x\\$First evaluation of $I_3:\\$

$$\noindent$I_3=\int_1^2 x\left(\frac{2}{x}+x\right)\left(\frac{2}{x}-x\right)^{2n}$d$x=\int_1^2x\left(\frac{2}{x}-x\right)^{2n}\left(\frac{2}{x}-x+2x\right)$d$x\\I_3=\int_1^2 \left(\frac{2}{x}-x\right)^{2n+1}$d$\left(\frac{x^2}{2}\right)+2\int_1^2x^2\left(\frac{2}{x}-x\right)^{2n}$d$x\\I_3=\left[\frac{x^2}{2}\left(\frac{2}{x}-x\right)^{2n+1}\right]_1^2+\frac{2n+1}{2}\int_1^2x^2\left(\frac{2}{x}-x\right)^{2n}\left(\frac{2}{x^2}+1\right)$d$x+2I_4\\I_3=-\frac{5}{2}+\frac{2n+1}{2}I_3+2I_4\\$Second Evaluation of $I_3:\\I_3=\int_1^2 (x^2+2)\left(\frac{2}{x}-x\right)^{2n}$d$x=2\int_1^2\left(\frac{2}{x}-x\right)^{2n}$d$x+\int_1^2x^2\left(\frac{2}{x}-x\right)^{2n}$d$x\\I_3=\frac{2}{2n+1}+I_4\\2I_3-I_3=\frac{4}{2n+1}+2I_4+\frac{5}{2}-\frac{2n+1}{2}I_3-2I_4\\\frac{2n+3}{2}I_3=\frac{10n+13}{2(2n+1)} \Leftrightarrow I_3 = \frac{10n+13}{(2n+1)(2n+3)}\\I_2=-5+(2n+1)I_3=\frac{10n+13}{2n+3}-5\\I_1=1+(2n+2)I_1+(2n+2)I_2=1+(2n+2)I_1+\frac{(2n+2)(10n+13)}{2n+3}-(10n+10)$

$$\noindent$(2n+1)I_1=\frac{(10n+9)(2n+3)-(2n+2)(10n+13)}{2n+3}\\(2n+1)I_1=\frac{20n^2+48n+27-20n^2-46n-26}{2n+3}=\frac{2n+1}{2n+3}\\\therefore \int_1^2 \left(\frac{2}{x}-x\right)^{2n+2}$d$x=\frac{1}{2n+3}\\\therefore P(n) \Rightarrow P(n+1)\\\therefore$ by the Axiom of Induction, $ P(n)$ is true $\forall n \in \mathbb{N}\\\therefore \int_1^2 \left(\frac{2}{x}-x\right)^{2014}$d$x=\frac{1}{2015}\\\therefore \int_0^1 \left(\frac{2}{x+1}-x-1\right)^{2014}$d$x = \frac{1}{2015}\\\textit{QUOD ERAT-FUCKING-DEMONSTRATUM}$

leehuan · Jan 6, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Bloody hell that Recurrence formula is a bitch.

$\\\textit{QUOD ERAT-FUCKING-DEMONSTRATUM}$

Oh WOW. The reduction formula required 2n in the index, that'd explain it.

Where did you get the proposal from before you induced it?

InteGrand · Jan 6, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Oh WOW. The reduction formula required 2n in the index, that'd explain it.

Where did you get the proposal from before you induced it?

Two possibilities:

1) He tried some small values and noticed a pattern

2) His intuition.

InteGrand · Jan 6, 2016

Re: MX2 2016 Integration Marathon

$The reason this pattern doesn't work for odd exponents can be seen from looking at the general term in the binomial expansion. The expansion of $\left(\frac{2}{x} + x\right)^{2n+1}$ ($n \in \mathbb{N}$) has general term $\binom{2n+1}{k} 2^{2n+1-k} \cdot \frac{x^k}{x^{2n+1-k}} = 2^{2n+1-k}\binom{2n+1}{k} \cdot x^{2k - 2n-1}$, for $k=0,1,2,\ldots,2n+1$. As we can see, when $k=n$ (which is a value $k$ takes on since $n<2n +1$ for all natural $n$), the power of $x$ is $-1$, so when integrated, becomes a log, giving something involving $\ln 2$ in the answer. All other terms are powers of $x$ that integrate to give rational numbers. So the final answer has a $\ln 2$ term and is irrational (since irrational plus rational equals irrational, which you can prove by contradiction).$

$Meanwhile, for even exponents, we can similarly show by considering binomial expansion that the power of $x$ in the general term is always even (whereas before it was always odd). This means there is no $x^{-1}$ term, so all terms integrate to give rational values in the end. So the answer is always \emph{rational}, so the pattern is different to the odd exponent case. I haven't investigated whether there's a pattern for the answers in the odd exponents case, but I wouldn't be surprised if there is.$

$So in summary, an odd exponent causes there to be a $\frac{1}{x}$ term in the integrand, which we know integrates differently than other powers of $x$ (causes a log to appear). When the exponent is even, no $\frac{1}{x}$ term exists in the expansion, so all terms just integrate ``normally'', i.e. like $\frac{x^{k+1}}{k+1}$, so no log terms appear.$

Paradoxica · Jan 6, 2016

Re: MX2 2016 Integration Marathon

Omegadot, please provide your "quick" answer, as I am still flummoxed by the length of my own solution.

ze- · Jan 6, 2016

Re: MX2 2016 Integration Marathon

This thread makes me really insecure about my maths

InteGrand · Jan 6, 2016

Re: MX2 2016 Integration Marathon

ze- said:
This thread makes me really insecure about my maths

A lot of these integrals are harder than would appear in the HSC papers (if they did appear, they'd be broken down into sub-parts, making them easier to do), so you shouldn't stress too much.

Paradoxica · Jan 6, 2016

Re: MX2 2016 Integration Marathon

$$By request of DrSoccerball, I will post four indefinite integrals at once.$\\1. \; \int \frac{x\ln{x}}{\sqrt{x^2-1}}$d$x\\2. \; \int \sqrt{\frac{1}{x}-1}$d$x\\3. \; \int \frac{\sqrt{x^3-1}}{x}$d$x\\4. \int \frac{(x-1)\sqrt{x^4+2x^3+4x^2+2x+1}}{x^2(x+1)}$d$x$

Drsoccerball · Jan 6, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
A lot of these integrals are harder than would appear in the HSC papers (if they did appear, they'd be broken down into sub-parts, making them easier to do), so you shouldn't stress too much.

Hsc be like:
$\int{cotx}dx$

Drsoccerball · Jan 6, 2016

Re: MX2 2016 Integration Marathon

1. $\int{\frac{xlnx}{\sqrt{x^2-1}}}dx$

$$ Using IBP with $ u=lnx $ and $ dv=\frac{x}{\sqrt{x^2-1}}$

$I=ln(x)\sqrt{x^2-1} - \int{\frac{\sqrt{x^2-1}}{x}}dx$

$$ Let $ u^2=x^2-1$

$I= ln(x)\sqrt{x^2-1} -\sqrt{x^2-1} +tan^{-1}(\sqrt{x^2-1}) +C$

Drsoccerball · Jan 6, 2016

Re: MX2 2016 Integration Marathon

$2. $ This is just Integrand's question but inversed, do the same process. $$

Drsoccerball · Jan 6, 2016

Re: MX2 2016 Integration Marathon

$3. $ This is just the working out used in (i) to get the final integral...? $$

$$ Let $ u^2=x^3-1$

$\frac{2}{3} \int{\frac{(u^2\pm 1)du}{u^2+1}}$

leehuan · Jan 6, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Hsc be like:
$\int{cotx}dx$

lol sorry I have to.

$\\ \ln {\sin {x}} +C$

leehuan · Jan 6, 2016

Re: MX2 2016 Integration Marathon

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Paradoxica · Jan 6, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
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Slight technicality: u can be zero. apart from that, all good.

Paradoxica · Jan 6, 2016

Re: MX2 2016 Integration Marathon

The method for the last integral is obvious if you know what a symmetric substitution is.

leehuan · Jan 6, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Slight technicality: u can be zero. apart from that, all good.

Good point, whoops.

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