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yes im always wrong on the internet

\int \frac{x^3}{2x+1}dx \\ $Let u = 2x +1, du = xdx \\ \\ = $\int \frac{x^2}{u}du \\=\int \frac{(\frac{u-1}{2})^2}{u}du \\ = \frac{1}{4}\int \frac{u^2 - 2u+1 }{u}du \\ = \frac{1}{4} \int (u - 2 + \frac{1}{u})du \\ = \frac{1}{4}\left ( \frac{u^2}{2} - 2u + ln(u)\right )+C \\ = \frac{1}{4}\left (...

Get a pen/pencil and a nice piece of paper and draw out the DEMAND CURVE (has a negative gradient)

this is 3U

Re: HSC 2013 2U Marathon very nice! u got it, yeh the hint didnt look that helpful yes i encourage all 2U to participate in this marathon - it'll help u greatly

Re: HSC 2013 2U Marathon Another hint...for these 2 integration questions \frac{a+b+c+ \dots + y}{z} = \frac{a}{z}+ \frac{b}{z}+ \frac{c}{z} + \dots + \frac{y}{z}

Its: \int \frac{s-2}{s\sqrt{s}}ds

Split it and it'll be easily integrated \frac{a+b}{c} = \frac{a}{c} + \frac{b}{c} ^ this is what i mean by splitting

\int x^{\frac{-3}{2}}dx = \frac{x^{\frac{-3}{2}+1}}{\frac{-3}{2}+ 1} +C = \frac{x^{\frac{-1}{2}}}{\frac{-1}{2}}+C = -2x^{\frac{-1}{2}} +C i think u made a mistake when adding fractions OR u subtracted the power by 1

Re: MX2 Integration Marathon did u get part of this from a sydney boys trial paper?

Re: HSC 2013 2U Marathon \int \frac{x^3}{x^2 + 1}dx = \int \frac{x^3 +x - x}{x^2+1}dx \int \frac{x}{\sqrt{x-1}}dx = \int \frac{x-1+1}{\sqrt{x-1}}dx Here are some bad hints so we can increase the activity in this 2u marathon

damnnnn nice thanks

http://community.boredofstudies.org/showthread.php?t=63912

yep

Re: HSC 2013 2U Marathon i agree

Re: HSC 2013 2U Marathon yeh yeh i understand, i misread ur post haha Finding the derivative of x^{x^{x}} would be hard for 2U students

Re: HSC 2013 2U Marathon Nice! EDIT: misread ur post xD

Re: HSC 2013 2U Marathon haha

Re: HSC 2013 2U Marathon (a)$ If $y = x^x,$ find $\frac{dy}{dx} \\ \\ (b)$If z = x^{x^{x}}, $ $ hence using part a or otherwise find $\frac{dz}{dx}

Very nice! "Now we can sacrifice the k inequalities..." i like this