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

integral95 · Sep 18, 2014

Re: MX2 Integration Marathon

integral95 said:
right......................

$\int_{0}^{\frac{\pi}{2}} \sum\limits_{k=0}^\infty sin^{2k+1}x dx$

dunjaaa said:
in your case, it'll be in terms of 'k' [i.e. (2k+1)/2^(2k+1)], since that's what you chose to represent as the index of summation

You can't use the limiting sum formula here since it diverges at x = pi/2 (or the indefinite integral which is tan x is undefined at pi/2)

dunjaaa · Sep 18, 2014

Re: MX2 Integration Marathon

it's not your simple limiting sum nor is it and indefinite integral, it's a special type of series

dan964 · Sep 19, 2014

Re: MX2 Integration Marathon

challenge q:
integral sin((1+x^2)/(2+x^2)) dx
using the subsitution u=1+x^2

dunjaaa · Sep 26, 2014

Re: MX2 Integration Marathon

i dont think there is an elementary primitive

dunjaaa · Sep 26, 2014

Re: MX2 Integration Marathon

Screen shot 2014-09-26 at 10.54.22 PM.png

-> expecting something of this nature in BOS trial lol

Sy123 · Sep 27, 2014

Re: MX2 Integration Marathon

$\\ \int_{\frac{1}{\alpha}}^{\alpha} \frac{\tan^{-1}x}{x} \ dx$

dunjaaa · Sep 27, 2014

Re: MX2 Integration Marathon

The same question got posted on the 4u marathon I think. Answer should be (πln(a))/2

Fade1233 · Sep 27, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30954 -> expecting something of this nature in BOS trial lol

Too long. Multiply sec^2tan^x, then simplify then substitute and then probably complex partial fractions. Hopefully not in the BOS. And if it is, hopefully it is 5 marks for all that writing.

aDimitri · Sep 27, 2014

Re: MX2 Integration Marathon

Fade1233 said:
Too long. Multiply sec^2tan^x, then simplify then substitute and then probably complex partial fractions. Hopefully not in the BOS. And if it is, hopefully it is 5 marks for all that writing.

this would probably be 3.

Carrotsticks · Sep 28, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30954 -> expecting something of this nature in BOS trial lol

I don't like putting in any tedious integrals in there. All require neat tricks or clever manipulations.

integral95 · Sep 30, 2014

Re: MX2 Integration Marathon

dunjaaa said:
The same question got posted on the 4u marathon I think. Answer should be (πln(a))/2

Sy123 said:
$\\ \int_{\frac{1}{\alpha}}^{\alpha} \frac{\tan^{-1}x}{x} \ dx$

I haven't seen the Q so yeah

I'll assume alpha is positive here otherwise it wouldn't work

$$using integration by parts with$ \ u = \tan^{-1}x \ \ v = 1/x \\ \tan^{-1}(a)ln(a) - \tan^{-1}(1/a)ln(1/a) - \int_{\frac{1}{\alpha}}^{\alpha}\frac{ln(x)}{1+x^2} \ dx= \\ ln(a)(\tan^{-1}(a)+\tan^{-1}(1/a)) - \int_{\frac{1}{\alpha}}^{\alpha}\frac{ln(x)}{1+x^2} \\ \\ = ln(a)(\frac{\pi}{2}) - \int_{\frac{1}{\alpha}}^{\alpha}\frac{ln(x)}{1+x^2} \\ \\ \\ Now \ let \ I = \ \int_{\frac{1}{\alpha}}^{\alpha}\frac{ln(x)}{1+x^2} \ dx \ $Let u = 1/x$ \ \ \\ \\ dx = -1/u^2 du \\ \\ \therefore I = -\int_{\alpha}^{\frac{1}{\alpha}}\frac{ln(\frac{1}{u})}{1+\frac{1}{u^2}} \ \frac{1}{u^2} \ du = -\int_{\frac{1}{\alpha}}^{\alpha}\frac{ln(x)}{1+x^2} \ dx \Rightarrow 2I = 0 \rightarrow I = 0 \\ \therefore \int_{\frac{1}{\alpha}}^{\alpha} \frac{\tan^{-1}x}{x} \ dx = ln(a)(\frac{\pi}{2})$

juantheron · Oct 1, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\\ \int_{\frac{1}{\alpha}}^{\alpha} \frac{\tan^{-1}x}{x} \ dx$

$Assuming\; $\alpha >0$ \;,Now\; let\;\displaystyle I = \int_{\frac{1}{\alpha}}^{\alpha} \frac{\tan^{-1} x}{x}dx$

$Now\; let \; $\displaystyle x = \frac{1}{t}$\;,Then\; $\displaystyle dx = -\frac{1}{t^2}$\; and \; changing \; limits\; we \; get$

$Now \; let \; $\displaystyle I = -\int_{\alpha}^{\frac{1}{\alpha}}\frac{\tan^{-1}\left(\frac{1}{t}\right)}{t^2}\cdot tdt = \int_{\frac{1}{\alpha}}^{\alpha}\frac{\tan^{-1}\left(\frac{1}{t}\right)}{t}dt$

$So \; $\displaystyle I = \int_{\frac{1}{\alpha}}^{\alpha}\frac{\tan^{-1}\left(\frac{1}{x}\right)}{x}dx = \int_{\frac{1}{\alpha}}^{\alpha} \frac{\cot^{-1} x}{x}dx$

$Now \; $\displaystyle 2I = \int_{\frac{1}{\alpha}}^{\alpha}\frac{\tan^{-1}x+\cot^{-1} x}{x}dx$

$$\displaystyle 2I = \frac{\pi}{2}\int_{\frac{1}{\alpha}}^{\alpha}\frac{1}{x}dx = \frac{\pi}{2}\left(\ln \alpha+\ln \alpha \right) = \frac{\pi}{2}\ln \alpha$

$Using \; the\; formula\; $\displaystyle \tan^{-1} x+\cot^{-1} x = \frac{\pi}{2}$\;,where\; $x>0$

$So \; $\displaystyle I = \int_{\frac{1}{\alpha}}^{\alpha} \frac{\tan^{-1} x}{x}dx = \frac{\pi}{2}\ln \alpha$

juantheron · Oct 1, 2014

Re: MX2 Integration Marathon

$Let \; $\displaystyle A = \sum_{k=1}^{n}\frac{n}{n^2+kn+k^2}$\; and $\displaystyle B = \sum_{k=0}^{n-1}\frac{n}{n^2+kn+k^2}.$

$for \;$n=1,2,3,4,..........,$\; Then\; which \; one \; is \; Leargest...$

integral95 · Oct 1, 2014

Re: MX2 Integration Marathon

juantheron said:
$Let \; $\displaystyle A = \sum_{k=1}^{n}\frac{n}{n^2+kn+k^2}$\; and $\displaystyle B = \sum_{k=0}^{n-1}\frac{n}{n^2+kn+k^2}.$

$for \;$n=1,2,3,4,..........,$\; Then\; which \; one \; is \; Leargest...$

The series generates smaller values ... so B?

dunjaaa · Oct 1, 2014

Re: MX2 Integration Marathon

If divide top and bottom by n^2, we obtain a Riemann's sum for approximating the area beneath the curve y=1/(x^2+x+1) in the interval 0≤x≤1. Since f(x) is decreasing in that interval, B represents the sum of 'n' upper rectangles and A represents the sum of 'n' lower rectangles. So B must be larger

dunjaaa · Oct 1, 2014

Re: MX2 Integration Marathon

This is what I did when I answered it in the other marathon: http://community.boredofstudies.org...arathon-screen-shot-2014-08-26-4.48.33-pm.png

integral95 · Oct 1, 2014

Re: MX2 Integration Marathon

dunjaaa said:
If divide top and bottom by n^2, we obtain a Riemann's sum for approximating the area beneath the curve y=1/(x^2+x+1) in the interval 0≤x≤1. A represents the sum of 'n' upper rectangles whereas B represents the sum of 'n' lower rectangles. So A must be larger

But f(x) = 1/(x^2+x+1) is decreasing between 0≤x≤1, so wouldn't B actually be the upper rectangles?

dunjaaa · Oct 1, 2014

Re: MX2 Integration Marathon

yes sorry my bad LOL

integral95 · Oct 1, 2014

Re: MX2 Integration Marathon

dunjaaa said:
yes sorry my bad LOL

Haha that's alright, that's a pretty neat approach though.

juantheron · Oct 11, 2014

Re: MX2 Integration Marathon

$$Calculate\; the \; Integrals$

$(a)\;\; \displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{n}(x)\cdot \sin (nx)dx$\;\;\;\;\;\;(b)\;\; \int_{0}^{\frac{\pi}{2}}\cos^{n}(x)\cdot \cos (nx)dx$

$Where \; $n\in \mathbb{N}$

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