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It's my favourite topic in 2U! http://www.boredofstudies.org/courses/maths/2u/1119956014_2004_Mathematics_Notes_Emily_Meers.pdf

Re: MX2 Integration Marathon AHHH yes, thank u for spotting that

Re: MX2 Integration Marathon e^x = \cos2u \\ $By implicit differentiation with respect to x, \\ \\ e^x .dx = -2\sin2u.du \\ \\dx = $\frac{-2sin2u}{e^x}du \\ $But e^x = cos2u \\ \\ \therefore dx = $\frac{-2sin2u}{cos2u}du So this is forbidden and incorrect?

Hint:Quadratic Polynomials (topic in 2U)

It is meant to say: Show that 1 < k < \frac{25}{24}

lol the last question (2(c)) requires a useful tool in complex numbers and he must see it himself or he wont do well

yeh u are correcto! i misread it haha its a geometric progression! :D S_n = \frac{a(1-r^n)}{1-r} = \frac{a(r^n - 1)}{r-1} OR z^{n+1} - 1 = (z-1)(1+ z + \dots + z^{n}) \\ 1 + z + \dots + z^n = \frac{z^{n+1}-1}{z-1} \\ \\ \therefore z+ z^2 + \dots + z^n = \frac{z^{n+1}-1}{z-1} -1 \\ \\...

Re: HSC 2013 3U Marathon Thread Very nice! Excellently done, same method as mine

No problemo! goodluck in 2U

U wont have to solve these types of equations algebraically :) Usually the only way to check these types of questions is to do them again (by accurately sketching the 2 graphs again - not too accurate)

Re: MX2 Integration Marathon \int \frac{1+sinx}{cosx}dx = \int (secx+tanx)dx = ln(secx + tanx) - ln(cosx) + C

Do 4U if u love maths!

$sinx = $\frac{x}{2} \\ \\ \therefore x = 2\sin{x} Draw out y = sinx from x = -pi to x = 2pi Draw out y = x from x=-pi to x = 2pi (a)u just look at the graphs from x = 0 to x = 2pi and count how many intersections (b)u just look at the graphs from x = -pi to x = pi and count how many...

Re: HSC 2013 4U Marathon no... i'm afraid not

sorry u have made a little mistake! 3y = e^x (1-y) e^x = \frac{3y}{1-y}

Re: HSC 2013 4U Marathon hm...

ignore me, i was being a dickhead (was a bad day)

Who is 'they'? and yeh i think u just find the volume of the entire sphere... using... V = \pi \int_a^b y^2 dx Unless the ________ is rotated around the y-axis, then u find the area of a hemisphere

Re: MX2 Integration Marathon $Let I = $\int_0^{\pi/2} \frac{dx}{1+ \tan^{\pi}(x)} \\ I= \int_0^{\pi/2} \frac{cos^{\pi}x }{cos^{\pi}x + sin^{\pi}x} dx \\ $ Use the greatest rule in integration: $\int_0^a f(x)dx = \int_0^a f(a-x)dx \\ \\ \therefore I = \int_0^{\pi/2} \frac{sin^{\pi}x}{sin^{\pi}x...

Re: MX2 Integration Marathon lol nice! but why Sy's face? xD