HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

Drsoccerball · Apr 7, 2015

Re: MX2 2015 Integration Marathon

simpleetal said:
advanced 4u I think. They have some very interesting problems

Dude... youre not even in year 12

integral95 · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Dude... youre not even in year 12

could be accelerated

or it could be a uni student in disguise.

Ekman · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Next Question:

$\int_{0}^{1} \frac{x}{\sqrt{1+x^4}} dx$

Drsoccerball · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Next Question:

$\int_{0}^{1} \frac{x}{\sqrt{1+x^4}} dx$

$$ Let u=x^2$

$\frac{1}{2} \int_0^1{\frac{1}{\sqrt{1+u^2}}}$

$= \frac{1}{2} ln(1+\sqrt{2})$

VBN2470 · Apr 8, 2015

Re: MX2 2015 Integration Marathon

NEW QUESTION:

$Find $\int_{0}^{\frac{\pi}{4}} \frac{\text{ d}x}{\cos{x}\cdot{\sqrt\sec{x}-\tan{x}}}$$

swagswagyoloswag · Apr 8, 2015

Re: MX2 2015 Integration Marathon

$\int_{0}^{ \frac{\pi}{4} } secx \sqrt{secx + tanx} dx $(Multiplying by the conjugate of the square root)$ \\ \int_{0}^{\frac{\pi}{4}} \frac{sec^{2}x + sec^{2}xtan^{2}x }{ \sqrt{secx + tanx} } dx $(Rationalising the numerator)$ \\ 2[\sqrt{secx+tanx} ]_{0}^{ \frac{\pi}{4} } \\ 2(\sqrt{\sqrt{2} + 1} - 1)$

Ekman · Apr 8, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Find $\int_{0}^{\frac{\pi}{4}} \frac{\text{ d}x}{\cos{x}\cdot{\sqrt\sec{x}-\tan{x}}}$$

Well what I did was let u^2=sec (x) - tan (x) and ended up with:

$\left [ \frac{2}{\sqrt{sec(x) - tan(x)}} \right] _{0}^{\frac {\pi}{4}}$

braintic · Apr 8, 2015

Re: MX2 2015 Integration Marathon

I've always wondered with this integration thread ... the questions asked are far above anything asked in an HSC exam, and seem to be based on a love of mindless algebra. If the aim is to maximise the HSC mark, and everyone has clearly mastered HSC level questions, aren't there MUCH more interesting things to focus on, ie. mathematical CONCEPTS? And if pure interest is not the goal, shouldn't you be focusing on your weaknesses instead of your strengths? (ie. ... back to perms and combs!)

Drsoccerball · Apr 8, 2015

Re: MX2 2015 Integration Marathon

braintic said:
I've always wondered with this integration thread ... the questions asked are far above anything asked in an HSC exam, and seem to be based on a love of mindless algebra. If the aim is to maximise the HSC mark, and everyone has clearly mastered HSC level questions, aren't there MUCH more interesting things to focus on, ie. mathematical CONCEPTS? And if pure interest is not the goal, shouldn't you be focusing on your weaknesses instead of your strengths? (ie. ... back to perms and combs!)

I reckon it will help us get better at integration

braintic · Apr 8, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
I reckon it will help us get better at integration

No doubt, for many people it will. But most people here seem to have almost perfected the techniques already. Not that more practice won't help, but it seems that integration gets a disproportionate amount of attention. And as I said ... I really don't understand the enjoyment people get from it ... it is more about following algorithms than thinking mathematically.

Drsoccerball · Apr 8, 2015

Re: MX2 2015 Integration Marathon

braintic said:
No doubt, for many people it will. But most people here seem to have almost perfected the techniques already. Not that more practice won't help, but it seems that integration gets a disproportionate amount of attention. And as I said ... I really don't understand the enjoyment people get from it ... it is more about following algorithms than thinking mathematically.

I struggle with recurence relations for example I_n =I_(n-1) ect... and also other usefull techniques.. but none of them are posted on this forum

VBN2470 · Apr 8, 2015

Re: MX2 2015 Integration Marathon

(NEW) RECURRENCE QUESTION

$Let $I_n$ = $\int_{0}^{2\pi} (1+\cos\theta)^n \text{ d}\theta$, for $n > 0$. Show that $I_{n+1}=\frac{2n+1}{n+1}I_{n}$$

braintic · Apr 8, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
(NEW) RECURRENCE QUESTION

$Let $I_n$ = $\int_{0}^{2n} (1+\cos\theta)^n \text{ d}\theta$, for $n > 0$. Show that $I_{n+1}=\frac{2n+1}{n+1}I_{n}$$

Did you mean for the upper limit to be 2n .... or 2pi ?

VBN2470 · Apr 8, 2015

Re: MX2 2015 Integration Marathon

braintic said:
Did you mean for the upper limit to be 2n .... or 2pi ?

FIXED

Drsoccerball · Apr 9, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
(NEW) RECURRENCE QUESTION

$Let $I_n$ = $\int_{0}^{2\pi} (1+\cos\theta)^n \text{ d}\theta$, for $n > 0$. Show that $I_{n+1}=\frac{2n+1}{n+1}I_{n}$$

wtf is this question :L can someone solve this....

and also $\int_0^1{\sqrt{\frac{1-x}{3+x}}}dx$

seanieg89 · Apr 9, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
wtf is this question :L can someone solve this....

and also $\int_0^1{\sqrt{\frac{1-x}{3+x}}}dx$

$I_n=I_{n-1}+\int_0^{2\pi}(1+\cos(\theta))^{n-1}\frac{d}{d\theta}(\sin(\theta))\, d\,\theta\\ = I_{n-1}+(n-1)\int_0^{2\pi}(1+\cos(\theta))^{n-2}\sin^2(\theta)\, d\theta \\ = I_{n-1}+(n-1)\int_0^{2\pi}(1+\cos(\theta))^{n-1}(2-(1+\cos(\theta)))\, d\theta \\ = I_{n-1}+2(n-1)I_{n-1}-(n-1)I_n\\ \\ \Rightarrow I_{n+1}=\frac{2n+1}{n+1}I_n.$

seanieg89 · Apr 9, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
and also $\int_0^1{\sqrt{\frac{1-x}{3+x}}}dx$

$I=\int_1^2 \sqrt{\frac{2-x}{2+x}}\, dx \\ = \int_1^2 \frac{2}{\sqrt{4-x^2}}-\frac{x}{\sqrt{4-x^2}} \, dx\\ \\ \\= \ldots$

(I am sure you can do it from here.)

Drsoccerball · Apr 9, 2015

Re: MX2 2015 Integration Marathon

seanieg89 said:
$I=\int_1^2 \sqrt{\frac{2-x}{2+x}}\, dx \\ = \int_1^2 \frac{2}{\sqrt{4-x^2}}-\frac{x}{\sqrt{4-x^2}} \, dx\\ \\ \\= \ldots$

(I am sure you can do it from here.)

Howd you get the first line

VBN2470 · Apr 9, 2015

Re: MX2 2015 Integration Marathon

He used the substitution $u = x+1$ .

Drsoccerball · Apr 9, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
He used the substitution $u = x-1$ .

Then his restrictions are wrong if that substitution was used doubt he used it

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