HSC 2012 MX2 Marathon (archive) (1 Viewer)

IamBread · Jan 18, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
=(

=( It's ok, I'm doing as much math as possible

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Guess it's time for a question, don't want this thread to die!

$$If $\int^6_0f(x)dx=42$

$$and $\int^3_0f(x)dx=12,$

$$Find $\int^6_32f(x)-3dx$

nightweaver066 · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

$\int^6_0 f(x)dx - \int^3_0 f(x)dx = 30$

$\int^6_3 f(x)dx = 30$

$\int^6_3 2f(x)dx = 60$

$\int^6_3 2f(x) - 3dx = 60 - [3x]^6_3$

$= 60 - (3^6 - 3^3)$

$= -642$

Unsure lol.

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
$\int^6_0 f(x)dx - \int^3_0 f(x)dx = 30$

$\int^6_3 f(x)dx = 30$

$\int^6_3 2f(x)dx = 60$

$\int^6_3 2f(x) - 3dx = 60 - [3x]^6_3$

$= 60 - (3^6 - 3^3)$

$= -642$

Unsure lol.

It's all right until the last bit

$\int^6_3 2f(x) - 3dx = 2\int^6_3f(x)dx - \int^6_3 3dx$

$= 60 - [3x]^6_3$

$= 60 - (18 - 9)$

$=51$

Have you done integration yet?

nightweaver066 · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
It's all right until the last bit

$\int^6_3 2f(x) - 3dx = 2\int^6_3f(x)dx - \int^6_3 3dx$

$= 60 - [3x]^6_3$

$= 60 - (18 - 9)$

$=51$

Have you done integration yet?

LOL. My bad, i misread it as power.....

I've finished 2U and 3U integration but haven't done 4U integration yet.

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
LOL. My bad, i misread it as power.....

I've finished 2U and 3U integration but haven't done 4U integration yet.

Haha that makes sense. 4U integration is much better

Here's another one:

$$Prove $ \int \frac{1}{1+x^2}dx = tan^{-1}x + c$

Not much harder then 3U, except you have to figure out the substitution

nightweaver066 · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
Haha that makes sense. 4U integration is much better

Here's another one:

$$Prove $ \int \frac{1}{1+x^2}dx = tan^{-1}x + c$

$$Let $ y = tan^{-1}x$

$x = tany$

$Differentiating with respect to x,$

$1 = sec^2y\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{sec^2y}$

$= \frac{1}{1 + tan^2y}$

$= \frac{1}{1 + x^2}$

$\therefore \int \frac{1}{1 + x^2}dx = tan^{-1}x + c$

SpiralFlex · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Or.

Let $x=\tan \theta$

$dx=\tan \theta \: d\theta$

$=\int \frac{1}{1+\tan^2 \theta}*\sec^2 \theta \: d\theta$

$=\int 1 \:d\theta$

$= \theta + C$

$=\tan^{-1}x+C$

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
$$Let $ y = tan^{-1}x$

$x = tany$

$Differentiating with respect to x,$

$1 = sec^2y\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{sec^2y}$

$= \frac{1}{1 + tan^2y}$

$= \frac{1}{1 + x^2}$

$\therefore \int \frac{1}{1 + x^2}dx = tan^{-1}x + c$

That works, can you do it the other way though?

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
Or.

Let $x=\tan \theta$

$dx=\tan \theta d\theta$

$=\int \frac{1}{1+\tan^2 \theta}*\sec^2 \theta d\theta$

$=\int 1 d\theta$

$=\tan^{-1}x+C$

That's it.

Next one.

$$Show that $\int \frac{1}{\sqrt{a^2 - x^2}}dx = sin^{-1}\frac{x}{a} + c$

SpiralFlex · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Let $x=a\sin \theta$

$dx=a\cos \theta \:d\theta$

$=\int \frac{1}{a\cos \theta}*a\cos \theta\: d\theta$

$=\int 1 \:d\theta$

$=\theta+C$

$=\sin^{-1}\frac{x}{a}+C$

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

$$Show that $\int \frac{1}{\sqrt{a^2 + x^2}}dx = ln|x +\sqrt{a^2 + x^2}| + c$

Then

$\int \frac{1}{\sqrt{x^2 - a^2}}dx = ln|x +\sqrt{x^2 - a^2}| + c$

SpiralFlex · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$$Show that $\int \frac{1}{\sqrt{a^2 + x^2}}dx = ln|x +\sqrt{a^2 + x^2}| + c$

Then

$\int \frac{1}{\sqrt{x^2 - a^2}}dx = ln|x +\sqrt{x^2 - a^2}| + c$

Let $x=a\tan \theta$

$dx=a\sec^2 \theta\: d\theta$

$=\int \sec \theta \: d\theta$

$=\int \frac{\sec \theta (\sec \theta+\tan \theta)}{sec \theta+\tan \theta}\: d\theta$

$=\int \ln|\sec \theta +\tan \theta|+C_{1}$

$=\int \ln|\frac{\sqrt{x^2+a^2}}{a}+\frac{x}{a}|+C_{1}$

$=\int \ln|\sqrt{x^2+a^2}+x|+C$

STOP POSTING OR I CAN'T GO TOILET!

Let $x=a\sec \theta$

$dx=a\sec \theta \tan \theta\: d\theta$

$=\int\frac{1}{a\tan \theta}*a\sec \theta \tan \theta\: d\theta$

$=\int \sec \theta \:d\theta$

$=\ln|\sec \theta+ \tan \theta|+C_{1}$

$=\ln|\frac{x}{a}+\frac{\sqrt{x^2-a^2}}{a}|+C_{1}$

$=\ln|x+\sqrt{x^2-a^2}}|+C$

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
STOP POSTING OR I CAN'T GO TOILET!

Haha ok, I'll let you go to the toilet

Plus I'm out of standard integrals to post lol.

SpiralFlex · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
Haha ok, I'll let you go to the toilet

Plus I'm out of standard integrals to post lol.

Well you could always do some Uni integrals.

$\int \frac{1}{a^2-x^2}\:dx=\frac{1}{2a}\log |\frac{a+x}{a-x}|+C$

$\int \frac{1}{x^2-a^2}\:dx=\frac{1}{2a}\log |\frac{x-a}{x+a}|+C$

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

One last question then sleep for me.

$$Prove via mathematical induction $ (cosx + isinx)^n = cosnx + isinnx$

Goodnight.

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
Well you could always do some Uni integrals.

$\int \frac{1}{a^2-x^2}\:dx=\frac{1}{2a}\log |\frac{a+x}{a-x}|+C$

$\int \frac{1}{x^2-a^2}\:dx=\frac{1}{2a}\log |\frac{x-a}{x+a}|+C$

I was thinking of doing things like that, but wasn't too sure if people had done partial fractions.

SpiralFlex · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
One last question then sleep for me.

$$Prove via mathematical induction $ (cosx + isinx)^n = cosnx + isinnx$

Goodnight.

I will do for n>0. Where n is an integer. I can't be bothered.

Let n = 1,

$LHS=(\cos x+i\sin x)^1=\cos x+i\sin x$

$RHS=\cos (1)x+i\sin (1)x$

$\therefore LHS=RHS$

$The statement is true for n=1.$

$Assume true for n=k$

$(\cos x+i\sin x)^k=\cos kx+i\sin kx$

$$R.T.P for n=k+1$

$R.T.P \: (\cos x+i\sin x)^{k+1}=\cos {k+1}x+i\sin {k+1}x$

$LHS=(\cos x+i\sin x)^{k+1}$

$=(\cos x+i\sin x)^k (\cos x+i\sin x)$

$=(\cos kx+i\sin kx)(\cos x+i\sin x)$ (From the assumption.)

$=\cos kx cos x+i\sin x \cos kx +i\sin kx\cos x -\sin kx \sin x$

$=\cos(k+1)x+i\sin (k+1)x$

$=RHS$

$\therefore$ True for n=k+1$$

Insert conclusion here.

Parvee · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Why does this thread make me feel way behind in 4u?

SpiralFlex · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Parvee said:
Why does this thread make me feel way behind in 4u?

Don't feel like that. We just like to post random stuff. Many of the stuff you see here aren't even testable for 4U.

HSC 2012 MX2 Marathon (archive) (1 Viewer)

Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Member

Well-Known Member

(╯°□°）╯︵ ┻━┻

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)