HSC 2015 MX2 Integration Marathon (archive) (3 Viewers)

Ekman · Apr 20, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Next Question:

$\int_{1}^{2} \frac{\sqrt{ln(9-x)}}{\sqrt{ln(9-x)}-\sqrt{ln(x+3)}}dx$

Kaido said:
Is it 1/2, took me a while but i think it looks good

Edit: I searched up some integration rules lol

If you are referring to this question then im sorry to say that 1/2 is incorrect.

Kaido · Apr 20, 2015

Re: MX2 2015 Integration Marathon

Oh fuark, i made an error in my subtraction, brb

Kaido · Apr 20, 2015

Re: MX2 2015 Integration Marathon

fuar now the integral won't work since i made the mistake ._.

hint plox

Drsoccerball · Apr 20, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
If you are referring to this question then im sorry to say that 1/2 is incorrect.

You still havn't told me how to do this question...

Drsoccerball · Apr 20, 2015

Re: MX2 2015 Integration Marathon

i feel as though it's a substitution of something like :
$u=x-3$

swagswagyoloswag · Apr 20, 2015

Re: MX2 2015 Integration Marathon

lol I let u = 3/2 - x to attempt to change it into an integral "property" since it changes borders to -1/2 and 1/2, but still cant do anything

Drsoccerball · Apr 20, 2015

Re: MX2 2015 Integration Marathon

Brain fart ... how do you factorise 1+x^4

InteGrand · Apr 21, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Brain fart ... how do you factorise 1+x^4

Let $P:\mathbb{C}\to \mathbb{C},P(x)=1+x^4$ .

When $P(x)=0$ , $1+x^4=0\Rightarrow x^4 = -1\Rightarrow x^2 = \pm i \Rightarrow x = \sqrt{\pm i}, -\sqrt{\pm i}$

Now that the four roots are found, you know the factors.

seanieg89 · Apr 21, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Brain fart ... how do you factorise 1+x^4

Or if you want real quadratic factors:

$x^4+1=(x^2+1)^2-(\sqrt{2} x)^2=(x^2+\sqrt{2} x+1)(x^2-\sqrt{2} x+1).$

Drsoccerball · Apr 23, 2015

Re: MX2 2015 Integration Marathon

has anyone solved ekmans question

Drsoccerball · Apr 25, 2015

Re: MX2 2015 Integration Marathon

Made this question(probs been integrated before but this is gold here goes: )
$\int_0^1(sin^{-1}x \times cos^{-1}x)dx$

FrankXie · Apr 25, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Made this question(probs been integrated before but this is gold here goes: )
$\int_0^1(sin^{-1}x \times cos^{-1}x)dx$

$2-\frac{\pi}{2}$

Ekman · Apr 26, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Made this question(probs been integrated before but this is gold here goes: )
$\int_0^1(sin^{-1}x \times cos^{-1}x)dx$

Here's a little twist to this question:

$\int_{0}^{1} (cos^{-1}x)^2 dx$

Sy123 · Apr 26, 2015

Re: MX2 2015 Integration Marathon

$\\ \lim_{\alpha \to \infty} \int_{-\alpha}^{\alpha} \frac{dx}{(x^2+a^2)(x^2+b^2)}$

Ekman · Apr 26, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
$\\ \lim_{\alpha \to \infty} \int_{-\alpha}^{\alpha} \frac{dx}{(x^2+a^2)(x^2+b^2)}$

$\lim_{\alpha \to \infty} \int_{-\alpha}^{\alpha} \frac{dx}{(x^2+a^2)(x^2+b^2)} = \frac{1}{b^2-a^2} \lim_{\alpha \to \infty} \int_{-\alpha}^{\alpha} \frac{1}{x^2+a^2} - \frac{1}{x^2+b^2} dx$

$\frac{1}{b^2-a^2} \lim_{\alpha \to \infty} \left[\frac{tan^{-1}\frac{x}{a}}{a} - \frac{tan^{-1}\frac{x}{b}}{b} \right]_{-\alpha}^{\alpha}$

$= \frac{\pi}{ab(a+b)}$

Sy123 · Apr 26, 2015

Re: MX2 2015 Integration Marathon

$\\ $Find$ \ \int_0^1 x^r (\ln x)^n \ dx \\ \\ $Where$ \ r > 0 \ $and$ \ n \ $is a non-negative integer$ \\ \\ $You may assume that$ \ \lim_{x \to 0^+} x^r (\ln x)^n = 0$

Ekman · Apr 26, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
$\\ $Find$ \ \int_0^1 x^r (\ln x)^n \ dx \\ \\ $Where$ \ r > 0 \ $and$ \ n \ $is a non-negative integer$ \\ \\ $You may assume that$ \ \lim_{x \to 0^+} x^r (\ln x)^n = 0$

Is the answer:
n!/(r+1)^n+1

Sy123 · Apr 26, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Is the answer:
n!/(r+1)^n+1

Something like a (-1)^{n-1} or (-1)^n at the front but other than that, that looks correct

Ekman · Apr 26, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
Something like a (-1)^{n-1} or (-1)^n at the front but other than that, that looks correct

Dammit I forgot the (-1)^n .

Chlee1998 · Apr 26, 2015

Re: MX2 2015 Integration Marathon

I doubt there's a solution for Ekman's question, although I might be wrong cause I didn't spend enough time on it

HSC 2015 MX2 Integration Marathon (archive) (3 Viewers)

Well-Known Member

be.

be.

Well-Known Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

This too shall pass

Well-Known Member

This too shall pass

Well-Known Member

This too shall pass

Well-Known Member

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 3)