HSC 2017 MX2 Integration Marathon (archive) (2 Viewers)

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stupid_girl

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Re: HSC 4U Integration Marathon 2017

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.:headbang:
 

stupid_girl

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Re: HSC 4U Integration Marathon 2017

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

(Sorry for the typo in numerator this afternoon.)
 
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calamebe

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Re: HSC 4U Integration Marathon 2017

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

stupid_girl

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Re: HSC 4U Integration Marathon 2017

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.
 
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calamebe

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Re: HSC 4U Integration Marathon 2017

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.
 

calamebe

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Re: HSC 4U Integration Marathon 2017

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

(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.
 

stupid_girl

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Re: HSC 4U Integration Marathon 2017

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!!! :lol:
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.:tongue:
 

stupid_girl

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Re: HSC 4U Integration Marathon 2017

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<x<=1

In (b)(ii),
the result of (a)(ii) can be used because 0<=1-g(t)<=1
 
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