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Does anyone know wtf happened to eco? Who got a band 6 here?

Heyyyy Congrats Soulful! :)

Why don't you just download the app versions? Arent those fine to use since they aren't hooked up to the net?

He is semi-retired, so obviously he wouldn't post as much.

Hahaha gotta love the hate band6man is receiving. I hope everyone realises that he is a troll and people should just ignore him...

Cant decide between integration and harder 3u, can I vote twice?

Please don't, I thought we were over this...

I didn't say that you did take it in any way either, you just auto-corrected him and I suggested that you could of taken it as a compliment rather than auto-correcting him.

You could have taken it as a compliment... Edit: Thanks Integrand :tongue:

Unfortunately curiosity got the better of me and so I looked...

I only looked through the economics one, I dare not look at the chem one...

Re: MX2 2016 Integration Marathon $ Sub in: $ u=3-x \therefore \int_{2}^{4} \frac{\sqrt{\ln{(9-x)}}}{\sqrt{\ln{(9-x)}} + \sqrt{\ln{(3+x)}}} dx \Rightarrow \int_{-1}^{1} \frac{\sqrt{\ln{(6+u)}}}{\sqrt{\ln{(6+u)}} + \sqrt{\ln{(6-u)}}} du \int_{-1}^{1}...

99.25

Alternate Alternate Method: \frac{z}{z^2+r^2} = \frac{z(\bar{z}^2 + r^2)}{z^2 \bar{z}^2 + r^2(z^2+\bar{z}^2) + r^4} $ (multiplying top and bottom by the conjugate of the denominator) $ \frac{z(\bar{z}^2 + r^2)}{z^2 \bar{z}^2 + r^2(z^2+\bar{z}^2) + r^4} = \frac{r^2(z+\bar{z})}{2r^4 +...

Re: MX2 2016 Integration Marathon The actual answer with the boundaries are on my compilation. I cant type it out atm, so you can check it out there

Hey everyone, I decided to take the liberty of collecting all the hard MX2 questions that I solved during my HSC year and develop a more organised compilation for students that study MX2...

Re: MX2 2016 Integration Marathon \int_{0}^{\pi} \cos^{2}(\tan^{-1}(\sin(\cot^{-1}(x)))) dx

Re: HSC 2016 3U Marathon Can confirm, we sometimes argue the mathematical validity of his jokes

Re: MX2 2016 Integration Marathon ALT METHOD: $ Let: $ I =\int^\pi_0 \frac{(x + 3) \sin{(x)}}{1 + \cos^{2}{(x)}} \, dx = \int^\pi_0 \frac{(\pi-x + 3) \sin{(\pi-x)}}{1 + \cos^{2}{(\pi-x)}} \, dx \therefore 2I = \int_{0}^{\pi} \frac{(\pi+6)\sin{(x)}}{1+\cos^{2}{(x)}} dx 2I = -(\pi+6)...