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You guys won't need to bring your own paper. I've sorted out the printing shenanigan now, that is what has been taking up most of my time for the past few hours. As for student numbers, I'll have to release them tonight when I'm done sorting this.

You may take the question booklet home =)

Re: MX2 2015 Integration Marathon Haha you guys. Similar idea to what I did, though I'm not too sure why you needed t formula substitutions at all. Use double angle formulas to cancel as much as you can to get 1/cos(x)cos(2x). Multiply top/bottom by cos, turn the denominator into sines, and...

Re: HSC 2015 4U Marathon - Advanced Level I've never seen this question in the context of dragons. I usually see it in the context of monks and either leaving the island or committing suicide.

Re: MX2 2015 Integration Marathon Could you also stop misusing capitals whilst you're at it? =)

My apologies all. The post here reminded me that today is indeed Tuesday, and not Wednesday. So the numbers will be distributed tomorrow, as they were scheduled to be given out on the Wednesday.

Yes your friend is able to come, as it is for 2U. Of course 100 is achievable, by many I believe. The question now is, can you handle dealing with harder questions as well as a fairly pressuring time limit, or are you only able to deal with harder questions? I cannot say which paper it is...

One per subject.

I will be sending out the numbers tonight! Very excited for Friday. To compensate for my delayed distribution of student numbers, I'll answer one question each about 2U/3U/4U. But of course I won't be answering questions like "What's the last question proving?" etc.

Re: MX2 2015 Integration Marathon Okay, add an X to my example from earlier. What is the significance of the line y=x now?

Re: MX2 2015 Integration Marathon If we define asymptotes to be limiting tangents, then the curve I have mentioned above will have no asymptote. What would be the significance of the line y=0 then? Clearly it has some significance, but it cannot be called an asymptote.

That's not really where students have issues. As separate things, as you have presented above, both integrals are clearly the same. However, give a relationship between the two variables, and now you've got a whole lot of students saying "Can we just do that...? But there's a relationship...

Re: MX2 2015 Integration Marathon Plenty. A classic example is the below, where the limit of the derivative does not exist, but it most certainly has a horizontal asymptote. f(x) = \frac{\sin(x^2)}{x}

Re: MX2 2015 Integration Marathon What do you mean 'just integrate both' and 'show it'? Show what? Integrate what? Please provide more specific arguments, not just wish washy vague claims.

Re: MX2 2015 Integration Marathon It doesn't seem obvious at all to me. For the sake of simplicity let's presume f is continuous and positive within the domain provided. Now your LHS multiplies each function value by a factor of C. By doing so, we acquire a new curve and the area under the...

Re: MX2 2015 Integration Marathon I remember that proof now. A bit ashamed that it didn't stick with me first time I read it. It's neat proof.

Re: MX2 2015 Integration Marathon Actually I think I know what you are referring to. I remember seeing something like that a while ago, and iirc braintic made a pdf about it too.

Discipline. You should try it some time.

Re: MX2 2015 Integration Marathon Continuing on from what you have, that integral becomes -2 \int_{0}^{\frac{\pi}{2}} \ln \left ( \sin \theta \right ) d\theta And this integral (ignoring the -2 at the front) evaluates to -\frac{\pi}{2} \ln 2, which gives the answer. But the only way I know...

If you mean using IBP more than twice on a single integral, I do not think so. The reason why we use integration by parts twice for some problems is often to take advantage of the fact that when the sinusoids get differentiated twice, they become the same thing (ignoring constants). For example...