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HSC 2015 MX2 Integration Marathon (archive) (8 Viewers)

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Drsoccerball

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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
There is just try my question and then do his its genius
HINT:
 
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InteGrand

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Re: MX2 2015 Integration Marathon

Here's a little twist to this question:

The integral can be written as .

Replace with :

.

Expand and break up the integral, using integration by parts on the first one and the answer to Drsoccerball's original integral for the second one.

Or, you don't need integration by parts for the first one, you can do it based on comparing the corresponding area to the inverse function.
 
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Ekman

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Re: MX2 2015 Integration Marathon

how did you get a (r+1)^(n+1) at the bottom? I got a (r+1)^n
I_n = -n/(r+1)I_n-2 so I_n-1 = -(n-1)/(r+1)I_n-3 and so on. didn't get a n+1
I counted I_0 which is just 1/r+1 , so that's the extra +1. Maybe I might be wrong because I have trouble visualising how many times it gets multiplied.
 

Ekman

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Re: MX2 2015 Integration Marathon

I only had a (r+1)^n-1 and then multiplied with I_0
What was your I_n? Mine was:

The way I see it for the denominator is that I_0=1/r+1. Hence I_1 = 1/(r+1)^2, I_2=2/(r+1)^3 ... I_n=n!/(r+1)^n+1 (Note, I ignored the minus one for the sake of demonstrating how i got the (r+1)^n+1)
 
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Re: MX2 2015 Integration Marathon

What was your I_n? Mine was:

The way I see it for the denominator is that I_0=1/r+1. Hence I_1 = 1/(r+1)^2, I_2=2/(r+1)^3 ... I_n=n!/(r+1)^n+1 (Note, I ignored the minus one for the sake of demonstrating how i got the (r+1)^n+1)
Mine was the same. So





.
.
.



I multiplied 2nd eqn. by then subtract from eqn. above and multiplied 3rd eqn by and added and so forth. So for it would have a fraction of while had so the pattern for the denominator is where c = subscript of I. So will have denominator of n-1 while has a denominator of n? Idk if there was something wrong with it
 

Ekman

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Re: MX2 2015 Integration Marathon

Mine was the same. So





.
.
.



I multiplied 2nd eqn. by then subtract from eqn. above and multiplied 3rd eqn by and added and so forth. So for it would have a fraction of while had so the pattern for the denominator is where c = subscript of I. So will have denominator of n-1 while has a denominator of n? Idk if there was something wrong with it
Im not sure how you got C to be the representation of the subscript of I when n is the that representation. And if the denominator is: then I_n would have a denominator as 1, basically cancelling out the r+1
 

Ekman

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Re: MX2 2015 Integration Marathon

Next Question:

 

Ekman

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Re: MX2 2015 Integration Marathon

Next Question:

 

Ekman

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Re: MX2 2015 Integration Marathon

Next Question:

 

VBN2470

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Re: MX2 2015 Integration Marathon

^ Your method is correct, but you shouldn't be mixing u's with x's, keep your variable the same throughout as your manipulating the integrand as this is not mathematically sound which could result in a loss of marks in the HSC exam.
 

WhoStanLeee

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Re: MX2 2015 Integration Marathon

Yeah, it does look a bit strange...how would I go about undergoing the substitution in one line though? To prevent having u's and x's at the same time?
 

Ekman

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Re: MX2 2015 Integration Marathon

I used the substitution

And I was left with:
 

InteGrand

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Re: MX2 2015 Integration Marathon

^ Your method is correct, but you shouldn't be mixing u's with x's, keep your variable the same throughout as your manipulating the integrand as this is not mathematically sound which could result in a loss of marks in the HSC exam.
There's surely nothing wrong with that, e.g. in the volume integration formula, we often write .

His u is a function of x, he's just doing an algebraic manipulation
Yeah, it does look a bit strange...how would I go about undergoing the substitution in one line though? To prevent having u's and x's at the same time?
To reduce chances of markers taking off marks (which they shouldn't, but very well may), you could explicitly write instead of to remind them it's a function of x.
 
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