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HSC 2012 MX2 Marathon (archive) (1 Viewer)
- Thread starter nightweaver066
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largarithmic
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Re: 2012 HSC MX2 Marathon
uhhhhh... how can this be true? the right hand series (the harmonic series) diverges
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Re: 2012 HSC MX2 Marathon
3000th post!
Woops, almost thought I stumbled on something neat there which was why it was modified in the first place. Ah wells, I'll stick to the original question
3000th post!
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someth1ng
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Re: 2012 HSC MX2 Marathon
I'm lost. I know that it works but how did you do it like that?Ewwww
Carrotsticks
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Re: 2012 HSC MX2 Marathon
After skipping a few steps too. Once you've done a fair few substitution method questions, you can jump from the question directly to the answer.
Substitution method u = 1-x^2.
After skipping a few steps too. Once you've done a fair few substitution method questions, you can jump from the question directly to the answer.
someth1ng
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Re: 2012 HSC MX2 Marathon
I see, how do you know which substitution to use? I probably don't have a chance at it since my school is doing some harder 3U topic first before integration.
Carrotsticks
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Re: 2012 HSC MX2 Marathon
Usually, it is let u = *inside of the square root* for the basic questions.
Alternatively, you could let u^2 = inside of square root.
However you cannot blindly do this. You have to see that it can lead to the correct answer, and this can only be done by carefully observing the integral and doing lots of questions to familiarise yourself with such questions.
Usually, it is let u = *inside of the square root* for the basic questions.
Alternatively, you could let u^2 = inside of square root.
However you cannot blindly do this. You have to see that it can lead to the correct answer, and this can only be done by carefully observing the integral and doing lots of questions to familiarise yourself with such questions.
Re: 2012 HSC MX2 Marathon
You should recognise it's almost f'(x)/sqrt[f(x)], and with a little manipulation you can get it there, once it's like that you can just quickly integrate. But as Carrotsticks said, you need to do a lot of questions to get yourself familiar to this.
You should recognise it's almost f'(x)/sqrt[f(x)], and with a little manipulation you can get it there, once it's like that you can just quickly integrate. But as Carrotsticks said, you need to do a lot of questions to get yourself familiar to this.
Carrotsticks
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Re: 2012 HSC MX2 Marathon
It just so happens that the one Trebla provided was convergent that it was not true.
Trebla, your question looks very interesting. Will try.
Some definite integrals can be divergent, one of the most trivial cases being:
It just so happens that the one Trebla provided was convergent that it was not true.
Trebla, your question looks very interesting. Will try.
Carrotsticks
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Re: 2012 HSC MX2 Marathon
For example, consider this integral:
The above integral in fact converges for all p E (0,1)
These integrals (sometimes called the p-integrals) are perfect examples of this.
Even though the function may not be defined, the area may well be.
An interesting figure sorta related to this is Gabriel's Horn: http://en.wikipedia.org/wiki/Gabriel's_Horn
Oh and I forgot to put the dx in the first p integral I provided.
Trebla's original question with limits 0 --> pi/2 was actually convergent. Just because a point is undefined, does not necessarily mean that the area is automatically infinite.
For example, consider this integral:
The above integral in fact converges for all p E (0,1)
These integrals (sometimes called the p-integrals) are perfect examples of this.
Even though the function may not be defined, the area may well be.
An interesting figure sorta related to this is Gabriel's Horn: http://en.wikipedia.org/wiki/Gabriel's_Horn
Oh and I forgot to put the dx in the first p integral I provided.
SpiralFlex
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Re: 2012 HSC MX2 Marathon
Rule, the best tool in 4u integration,
Substitution is hardly ever needed with
It
It was done using the reverse power
Rule, the best tool in 4u integration,
Substitution is hardly ever needed with
It