HSC 2012-14 MX2 Integration Marathon (archive) (6 Viewers)

SpiralFlex · Oct 17, 2013

Re: MX2 Integration Marathon

chris_power_96 said:
1) Use the substitution x=sin^2u
2)(1+2(x(1-x))^.5)^.5 then becomes (1+2cosusinu)^.5
3)We the use the fact that sin^2u+cos^2u=1 to make the above square root into sinu+cosu
4) Then the integral becomes 2sinucos^2u +2sin^2ucosu
5) this becomes (2sin^3u)/3 - (2cos^3u)/3
6) this the becomes (2(x)^1.5)/3 - (2(1-x)^1.5)/3

One-two line method.

HeroicPandas · Oct 18, 2013

Re: MX2 Integration Marathon

SpiralFlex said:
One-two line method.

I give up now...please reveal your method

henryyuan · Oct 18, 2013

Re: MX2 Integration Marathon

Ryannnn

HeroicPandas · Oct 19, 2013

Re: MX2 Integration Marathon

SpiralFlex said:
$1) \int \sqrt{1+2\sqrt{x-x^2}}\;dx$

I give up, my brain has been squished! Can you please your one-two line method?

ForOneEon · Oct 19, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
I give up, my brain has been squished! Can you please your one-two line method?

Nooo! If your brain is squished, how will you do well in your exam?! Then again, my brain exploded trying to find the two-line method.
Explosion > Squished

So, Spiral. If you may please guide us to this two-line solution, we will be very grateful!

Sy123 · Oct 19, 2013

Re: MX2 Integration Marathon

$\int_1^e \ln^{n-1}x \left( \ln x + n) \ dx$

$n \geq 1$

rural juror · Oct 19, 2013

Re: MX2 Integration Marathon

chris_power_96 said:
1) Use the substitution x=sin^2u
2)(1+2(x(1-x))^.5)^.5 then becomes (1+2cosusinu)^.5
3)We the use the fact that sin^2u+cos^2u=1 to make the above square root into sinu+cosu
4) Then the integral becomes 2sinucos^2u +2sin^2ucosu
5) this becomes (2sin^3u)/3 - (2cos^3u)/3
6) this the becomes (2(x)^1.5)/3 - (2(1-x)^1.5)/3

i subbed u = x-x^2 in,

solved the quadratic for x (take positvie root, its the same either way i think),
and cancelled difference of two sqaures and solved it throguh. got different answer -1/3(1-2u)^3/2 + (1-2u)^1/2

it might be the same expressiont hough

SpiralFlex · Oct 19, 2013

Re: MX2 Integration Marathon

ForOneEon said:
Nooo! If your brain is squished, how will you do well in your exam?! Then again, my brain exploded trying to find the two-line method.
Explosion > Squished

So, Spiral. If you may please guide us to this two-line solution, we will be very grateful!

I tricked you guys, this is a simple integral that requires the observation: (Level 1 question ;p)

$(\sqrt{x}-\sqrt{1-x})^2= 1-2\sqrt{x-x^2}$

RealiseNothing · Oct 19, 2013

Re: MX2 Integration Marathon

SpiralFlex said:
I tricked you guys, this is a simple integral that requires the observation: (Level 1 question ;p)

$(\sqrt{x}-\sqrt{1-x})^2= 1-2\sqrt{x-x^2}$

hahaha fuck

HeroicPandas · Oct 19, 2013

Re: MX2 Integration Marathon

spiralflex said:
i tricked you guys, this is a simple integral that requires the observation: (level 1 question ;p)

$(\sqrt{x}-\sqrt{1-x})^2= 1-2\sqrt{x-x^2}$

damnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn!

mehulg111 · Oct 20, 2013

Re: MX2 Integration Marathon

What's the answer to Sy123's question? I got e (http://tinyurl.com/lvhxflr)

Sy123 · Oct 20, 2013

Re: MX2 Integration Marathon

mehulg111 said:
What's the answer to Sy123's question? I got e (http://tinyurl.com/lvhxflr)

Yep

bottleofyarn · Oct 20, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_1^e \ln^{n-1}x \left( \ln x + n) \ dx$

$n \geq 1$

IBP to get
$\left [ x\ln^n{x} \right ]_1^e - \int_1^e{\frac{nx\ln^{n-1}x}{x}\mathrm{d}x} + \int_1^e{n\ln^{n-1}x\mathrm{d}x}$
$= e\ln^n{e} - \ln^n{1}$

$= e$

==

also, we all overthought the one from before since
$\sqrt{a\sqrt{b\sqrt{x}}} = a^{\frac{1}{2}}b^{\frac{1}{4}}x^{\frac{1}{8}}$

==

$$Evaluate$ I = \int_0^n{xe^{-ax}}\mathrm{d}x$
$$Hence find I as $ n \to \infty$

RealiseNothing · Oct 20, 2013

Re: MX2 Integration Marathon

bottleofyarn said:
also, we all overthought the one from before since
$\sqrt{a\sqrt{b\sqrt{x}}} = a^{\frac{1}{2}}b^{\frac{1}{4}}x^{\frac{1}{8}}$

There were plus signs in there.

bottleofyarn · Oct 20, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
There were plus signs in there.

Derp. Good thing you don't need to memorise stuff for maths, right.

Another:
$$Evaluate $I=\int_{2+\varepsilon}^4\frac{\mathrm{d}x}{x^2 \sqrt{x^2 -4}}$
$$Hence, find I as $ \varepsilon \to 0^+$

RealiseNothing · Oct 20, 2013

Re: MX2 Integration Marathon

bottleofyarn said:
Derp. Good thing you don't need to memorise stuff for maths, right.

Another:
$$Evaluate $I=\int_{2+\varepsilon}^4\frac{\mathrm{d}x}{x^2 \sqrt{x^2 -4}}$
$$Hence, find I as $ \varepsilon \to 0^+$

Let $x = 2\sec(\theta)$ and you end up with:

$\int \frac{\cos(\theta)}{2} d\theta$

$\frac{1}{2}\sin(\theta)$

$\frac{1}{2}\sin(\cos^{-1}(\frac{2}{x}))$

$\frac{1}{2}\sin(\cos^{-1}(\frac{1}{2})) - \frac{1}{2}\sin(\cos^{-1}(\frac{2}{2+\varepsilon}))$

As $\varepsilon \to 0^{+}$

$\frac{\sqrt{3}}{4} - \frac{1}{2}(\sin(cos^{-1}(1)))$

$\frac{\sqrt{3}}{4}$

twinklegal19 · Oct 20, 2013

Re: MX2 Integration Marathon

bottleofyarn said:
IBP to get
$\left [ x\ln^n{x} \right ]_1^e - \int_1^e{\frac{nx\ln^{n-1}x}{x}\mathrm{d}x} + \int_1^e{n\ln^{n-1}x\mathrm{d}x}$
$= e\ln^n{e} - \ln^n{1}$

$= e$

==

also, we all overthought the one from before since
$\sqrt{a\sqrt{b\sqrt{x}}} = a^{\frac{1}{2}}b^{\frac{1}{4}}x^{\frac{1}{8}}$

==

$$Evaluate$ I = \int_0^n{xe^{-ax}}\mathrm{d}x$
$$Hence find I as $ n \to \infty$

$\\I=\int_{0}^{n}xe^{-ax}dx\\\\=\left [ -\frac{1}{a}xe^{-ax}\right ]_{0}^{n}+\frac{1}{a}\int_{0}^{n}e^{-ax}dx\\\\=-\frac{n}{a}e^{-an}+\frac{1}{a}\left [-\frac{ e^{-ax}}{a} \right ]_{0}^{n}\\\\=-\frac{n}{a}e^{-an}-\frac{1}{a^2}e^{-an}+\frac{1}{a^2}\\\\\therefore \lim_{n\rightarrow \infty }I=\frac{1}{a^2}$

RealiseNothing · Oct 20, 2013

Re: MX2 Integration Marathon

bottleofyarn said:
$$Evaluate$ I = \int_0^n{xe^{-ax}}\mathrm{d}x$
$$Hence find I as $ n \to \infty$

Hence find:

$\lim_{m \to \infty} \lim_{n \to \infty} \sum_{k=1}^{m} \int_{0}^{n} xe^{-kx} dx$

Sy123 · Oct 20, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
Hence find:

$\lim_{m \to \infty} \lim_{n \to \infty} \sum_{k=1}^{m} \int_{0}^{n} xe^{-kx} dx$

Summing the geometric series we get

$\sum_{n=1}^{\infty} \frac{1}{n^2} = \int_0^{\infty} \frac{x}{e^x - 1} \ dx$

Any ideas on how to squeeze this integral to somehow get pi^2/6?

Sy123 · Jan 2, 2014

Re: MX2 Integration Marathon

$\int_0^1 \frac{\ln x}{\sqrt{1-x^2}} \ dx$

HSC 2012-14 MX2 Integration Marathon (archive) (6 Viewers)

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