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Thread: HSC 2017 MX2 Integration Marathon (archive)

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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by InteGrand View Post
    www.imgur.com/nGjEjX8

    Quote Originally Posted by Paradoxica View Post
    Let a+b=c where a, b, c are real numbers. Evaluate

    www.imgur.com/O3sZNpH

    EDIT: That final c should be swapped with a k, but I can't be bothered to edit it and upload another photo.
    Last edited by calamebe; 13 Dec 2016 at 5:21 PM.
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    Re: HSC 4U Integration Marathon 2017



    Last edited by juantheron; 23 Dec 2016 at 6:19 PM.

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    Re: HSC 4U Integration Marathon 2017

    I hope this one isn't too bad.


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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by stupid_girl View Post
    I hope this one isn't too bad.

    A bit tedious though, so after using log laws, u = cos2x, u = tan theta, f(x) = f(a-x), I got -1/64. is that right?
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    -insert title here- Paradoxica's Avatar
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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by integral95 View Post
    A bit tedious though, so after using log laws, u = cos2x, u = tan theta, f(x) = f(a-x), I got -1/64. is that right?
    By inspection, you are correct.
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    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by juantheron View Post


    Wubba-lubba-dub-dub!
    HSC 2015: 2U Math (98)
    HSC 2016: 4U Math (98) 3U Math (97) Eco (95) Adv. Eng (95) Phys (92)
    ATAR: 99.85

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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by integral95 View Post
    A bit tedious though, so after using log laws, u = cos2x, u = tan theta, f(x) = f(a-x), I got -1/64. is that right?
    It's correct, but I wonder how Paradoxica did it by inspection.

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    Re: HSC 4U Integration Marathon 2017

    This definite integral must be positive because the integrand is a complete square.

    (Sorry for the typo in numerator this afternoon.)
    Last edited by stupid_girl; 28 Dec 2016 at 10:54 PM.

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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by stupid_girl View Post
    This definite integral must be positive because the integrand is a complete square.
    61root(2)/840 right? What I did was break up the sin part, get three different integrals, one of which is easy to work out, then for the other two I created a recursion formula for the integral of sec(x)^n from 0 to pi/4, and added/subtracted the three integrals, and got that. I might post working, but it is unbelievably messy.
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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by calamebe View Post
    61root(2)/840 right? What I did was break up the sin part, get three different integrals, one of which is easy to work out, then for the other two I created a recursion formula for the integral of sec(x)^n from 0 to pi/4, and added/subtracted the three integrals, and got that. I might post working, but it is unbelievably messy.
    Sorry it's my fault. I accidentally missed out the exponent on the numerator when I typed it in latex.

    Nevertheless, the mistyped integral is not messy. I just attempted it and the integrand can be broken into 0.5 sin x/(cos x)^7 and -0.5(sec x)^6. Substitue u=cos x and u=tan x respectively.
    Last edited by stupid_girl; 29 Dec 2016 at 12:38 AM.

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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by stupid_girl View Post
    Sorry it's my fault. I accidentally missed out the exponent on the numerator when I typed it in latex.

    Nevertheless, the mistyped integral is not messy. I just attempted it and the integrand can be broken into 0.5 sin x/(cos x)^7 and -0.5(sec x)^6. Substitue u=cos x and u=tan x respectively.
    Oh yeah I meant my working was messy as in illegible haha.
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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by stupid_girl View Post
    This definite integral must be positive because the integrand is a complete square.

    (Sorry for the typo in numerator this afternoon.)
    Well this one is 22/7 - pi, I just did some algebra to break down the expression to tan(x)^4(1-tan(x))^4, made a substitution u=tanx and just some relatively easy integration from there. I did have to use the binomial theorem in the last integral though, I feel like there would be a more elegant way to do it.
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    Re: HSC 4U Integration Marathon 2017

    This is of course a proof that 22/7 > pi.
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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by InteGrand View Post
    This is of course a proof that 22/7 > pi.
    This is obviously false because we all know that pi = 22/7.
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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by Carrotsticks View Post
    This is obviously false because we all know that pi = 22/7.
    What???? I though pi = 3.14!!! Mind = blown.
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    Senior Member integral95's Avatar
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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by Carrotsticks View Post
    This is obviously false because we all know that pi = 22/7.
    My life has been a lie.
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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by calamebe View Post
    Well this one is 22/7 - pi, I just did some algebra to break down the expression to tan(x)^4(1-tan(x))^4, made a substitution u=tanx and just some relatively easy integration from there. I did have to use the binomial theorem in the last integral though, I feel like there would be a more elegant way to do it.
    That's correct!!!
    I am not aware of any more elegant way to do it.
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    Re: HSC 4U Integration Marathon 2017

    This question relies on Fundamental Theorem of Calculus and substitution.

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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by stupid_girl View Post
    This question relies on Fundamental Theorem of Calculus and substitution.
    http://imgur.com/a/YGobh

    I hope that is clear enough. If not I'm trying to learn LaTeX so if it's too blurry/messy I can transcribe it to be readable haha.
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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by stupid_girl View Post
    This question relies on Fundamental Theorem of Calculus and substitution.
    http://imgur.com/a/YGobh

    I hope that is clear enough. If not I'm trying to learn LaTeX so if it's too blurry/messy I can transcribe it to be readable haha.
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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by calamebe View Post
    http://imgur.com/a/YGobh

    I hope that is clear enough. If not I'm trying to learn LaTeX so if it's too blurry/messy I can transcribe it to be readable haha.
    The approach is correct. Well done!

    Just a few minor things to point out.
    In (a)(i),
    f(G(x))g(x)>=f(x)g(x) because g(x)>=0

    In (a)(ii),
    phi(x)>=0 for all 0
    In (b)(ii),
    the result of (a)(ii) can be used because 0<=1-g(t)<=1
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    Re: HSC 4U Integration Marathon 2017

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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by stupid_girl View Post
    http://imgur.com/a/QAcfP

    I feel as though my method is very convoluted, is there a much easier way?
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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by calamebe View Post
    http://imgur.com/a/QAcfP

    I feel as though my method is very convoluted, is there a much easier way?
    What's your final answer? Can't seem to spot it.

    Edit: Oh found it.
    Last edited by InteGrand; 4 Jan 2017 at 9:18 PM.

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    Re: HSC 4U Integration Marathon 2017

    Quote Originally Posted by calamebe View Post
    http://imgur.com/a/QAcfP

    I feel as though my method is very convoluted, is there a much easier way?










    Last edited by InteGrand; 5 Jan 2017 at 9:05 AM.
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