HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

dunjaaa · Apr 2, 2014

Re: MX2 Integration Marathon

Just sat my half yearly for MX2 today, this was my recurrence question. I reckon it was a good one

Screen shot 2014-04-02 at 7.32.13 PM.png

Sy123 · Apr 2, 2014

Re: MX2 Integration Marathon

Let's generalize that just a little bit

$\\ $Find the re-currence formula in$ \ n \ $for$ \\ \\ I_n = \int_0^1 x^n (1-x^m)^{1/k} \ dx \ \ $for$ \ n > (m+1) \ $for naturals$ \ n$

(its doable I tried it)

Ikki · Apr 3, 2014

Re: MX2 Integration Marathon

Sy123 said:
Let's generalize that just a little bit

$\\ $Find the re-currence formula in$ \ n \ $for$ \\ \\ I_n = \int_0^1 x^n (1-x^m)^{1/k} \ dx \ \ $for$ \ n > (m+1) \ $for naturals$ \ n$

(its doable I tried it)

Ofcourse you did.

dunjaaa · Apr 3, 2014

Re: MX2 Integration Marathon

chucking a carrot I see
I got I(n)={[k(n-m+1)]/[m+k(n+1)]}I(n-m)

Sy123 · Apr 4, 2014

Re: MX2 Integration Marathon

dunjaaa said:
chucking a carrot I see
I got I(n)={[k(n-m+1)]/[m+k(n+1)]}I(n-m)

Yep something like that

integral95 · Apr 4, 2014

Re: MX2 Integration Marathon

Hey guys in dat UNSW ASOC integration bee, there was this weird question (carrot would approve)

$\int \sqrt{x+\sqrt{x+\sqrt{x+...}}} dx$

Carrotsticks · Apr 4, 2014

Re: MX2 Integration Marathon

MethewYan said:
Hey guys in dat UNSW ASOC integration bee, there was this weird question (carrot would approve)

$\int \sqrt{x+\sqrt{x+\sqrt{x+...}}} dx$

I thought it was a rather different one though lol.

The actual integration is very straightforward, it's just finding the closed form that not too many people are familiar with.

Sy123 · Apr 4, 2014

Re: MX2 Integration Marathon

$\int_0^{\pi/2}\frac{x}{\sin x + \cos x} \ dx$

dunjaaa · Apr 4, 2014

Re: MX2 Integration Marathon

Screen shot 2014-04-04 at 11.44.02 PM.png

Sy123 · Apr 5, 2014

Re: MX2 Integration Marathon

MethewYan said:
Hey guys in dat UNSW ASOC integration bee, there was this weird question (carrot would approve)

$\int \sqrt{x+\sqrt{x+\sqrt{x+...}}} dx$

Hint: Let u = (that expression)

dunjaaa · Apr 5, 2014

Re: MX2 Integration Marathon

Screen shot 2014-04-05 at 4.35.33 PM.png

Not sure if correct

Sy123 · Apr 5, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30228 Not sure if correct

Yep well done!

Carrotsticks · Apr 5, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30228 Not sure if correct

You could also do this.

$\\ u=\sqrt{x+\sqrt{x+\sqrt{x+...}}} \\\\ u^2 = x+u \\\\ u^2 - u - x = 0 \\\\ u=\frac{1 \pm \sqrt{1+4x}}{2} $ and then just integrate this instead.$$

Sy123 · Apr 5, 2014

Re: MX2 Integration Marathon

$\int_0^{\pi/4} \frac{x + \sin x}{1 + \cos x} \ dx$

dunjaaa · Apr 5, 2014

Re: MX2 Integration Marathon

change into half angles -> int{[x+2sin(x/2)cos(x/2)]/2cos^2(x/2) dx}
=int{1/2 [xsec^2(x/2)]+tan(x/2) dx}
=[xtan(x/2)] between x=0 and x=pi/4 by reverse product rule
=pi/4tan(pi/8)

Sy123 · Apr 5, 2014

Re: MX2 Integration Marathon

dunjaaa said:
change into half angles -> int{[x+2sin(x/2)cos(x/2)]/2cos^2(x/2) dx}
=int{1/2 [xsec^2(x/2)]+tan(x/2) dx}
=[xtan(x/2)] between x=0 and x=pi/4 by reverse product rule
=pi/4tan(pi/8)

nice work

$\int_0^{\pi/6} \frac{\sqrt{1 + \sin x}}{\cos x} \ dx$

Sy123 · Apr 5, 2014

Re: MX2 Integration Marathon

$\int_0^{\pi} \frac{\sin^3x}{\sin^2x + 8} \ dx$

Davo_01 · Apr 9, 2014

Re: MX2 Integration Marathon

Sy123 said:
nice work

$\int_0^{\pi/6} \frac{\sqrt{1 + \sin x}}{\cos x} \ dx$

Sy123 · Apr 10, 2014

Re: MX2 Integration Marathon

$\\ $For the positive real$ \ x \ $find the minimum value of$ \ \ f(x) = \int_x^{2x} t\ln t - t \ dt$

hit patel · Apr 10, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\\ $For the positive real$ \ x \ $find the minimum value of$ \ \ f(x) = \int_x^{2x} t\ln t - t \ dt$

e^(4/9) ? Looks way off.

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