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Re: HSC 2014 4U Marathon Let f_1(x)=ln(x) f_2(x)=ln(ln(x)) f_3(x)=ln(ln(ln(x))) f_n(x)=ln(ln(...ln(x))) Prove by Mathematical Induction for n>1 : \dfrac{d(f_{n})}{dx}=\dfrac{1}{x\cdot f_1\cdot f_2 \cdot ...\cdot f_{n-1}}

Re: HSC 2014 4U Marathon \int{\dfrac{x^2}{x^6-1} dx}

Re: HSC 2014 4U Marathon Have a look at some of the questions I posted earlier in the advance thread

Re: HSC 2014 4U Marathon No wait i see how the symetry method works, both partial volumes are meant to be equal so you should add them and divide by 2 to get v, you didn't divide by 2 so your answer would have been double but you made ur limits 0 to 2 instead of -2 to 2 so the volume halved and...

Re: HSC 2014 4U Marathon Hmm ive never used this symmetry method before but if it works it works, anyways man i gotta go to sleep, work tomoro so il catch ya later. take care bro :)

Re: HSC 2014 4U Marathon Ooh why did you take two seperate partial volumes

Re: HSC 2014 4U Marathon Again remember to multiply by shell radius, your answer was still right because the extra part of the integral happened to be zero but in general ur answer will probably change. Other than that good work :)

Re: HSC 2014 4U Marathon ahahha no worries, there is a button 3rd from the right

Re: HSC 2014 4U Marathon Thing is i think you forgot to multiply by the shell radius. Remember V=2\pi xy , It looks like ur missing the x, if you shift the axis of rotation by 2 then instead of V=2\pi xy , V=2\pi(x--2)y Its actually similar to using substitution u=x-2

Re: HSC 2014 4U Marathon V= 2\pi\int_0^4 x\sqrt{4-(x-2)^2}dx Let u=x-2 dx=du Limits change to -2 and 2 V=2\pi\int_{-2}^{2}(u+2)\sqrt{4-u^2}du V=2\pi\int_{-2}^{2}u\sqrt{4-u^2}du +4\pi\int_{-2}^{2}\sqrt{4-u^2}du =0+4\pi(\frac{4\pi}{2}) (Odd function, Area of semicircle) =8\pi^2 units^3

Re: HSC 2014 4U Marathon Sure, while you make one il type up my solution, I know two different pathways that work but one works very nicely in this example.

Re: HSC 2014 4U Marathon Im not sure if the working out is correct, as the preferred substitutions were either x-2=2\sin{\theta} or u=x-2

Re: HSC 2014 4U Marathon Its right but the way i would have preferred is dividing by \vert x \vert (=\sqrt{x^2}) And using \lim_{x \to -\infty}\frac{x}{\vert x \vert }=-1

Re: HSC 2014 4U Marathon you forgot the x 2\pi\int_0^4 x\sqrt{4-(x-2)^2}dx edit: my bad my answer is 8\pi^2

Re: HSC 2014 4U Marathon answer is -2 btw

Re: HSC 2014 4U Marathon hmm my answer was 8\pi^2 what was ur integral

Re: HSC 2014 4U Marathon This is why i chose the question :). What you need to think about is when you bring the x inside the square root bracket, are you still dividing by x? Like does the x still remain as x?

Re: HSC 2014 4U Marathon Maybe, I have never tried it, but I have two different methods, one of them being rationalizing the numerator

Re: HSC 2014 4U Marathon Find the volume produced by rotating the area bounded by the ellipse: \frac{(x-2)^2}{4}+y^2=1 about the y-axis, using cylindrical shells.

Re: HSC 2014 4U Marathon OKay let me have a quick look around