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Re: MX2 Integration Marathon I wasn't sure how to latex the limits when subbing into the primitive so I sort of left that part out (I'm used to solving just indefinite integrals on BOS in the past lol) \lim_{t \to \infty}\int_0^{t} \frac{x}{x^4+1} \ dx \\=\lim_{t \to...

congratulations! Haven't heard back from them yet, fingers crossed.

Re: MX2 Integration Marathon just use the t-formula? Should end up with an inverse tan

I got a simpler result? \ln\left | x-\frac{a+b}{2} + \sqrt{(x-a)(x-b)} \right |

Re: HSC 2013 3U Marathon Thread this lol

Re: MX2 Integration Marathon \int \frac{x^{-\frac{1}{2}}}{1+x^{\frac{1}{3}}}=\int \frac{dx}{\sqrt{x}\left ( 1+x^{\frac{1}{6}} \right )}\\$Let $u^6=x\\6u^5du=dx\\\\\int \frac{6u^5du}{u^3(1+u^2)}\\=6\int\frac{u^2}{1+u^2}\\=6\int\left ( 1-\frac{1}{1+u^2} \right...

EY opens today :)

I think most students would dw

haha

Don't they close on the 23rd of may or something for pwc? Edit: 23rd June*

^ thank you so much for that helpful advice :)

^

Just checking, how far back should awards go? Should we only include recent ones or the ones throughout our entire high school life? (Not sure how much the HR needs/want to know)

It's the same format at the 2U one

Some are already open I think

Re: HSC 2013 4U Marathon Solve: \frac{4x^2}{(1-\sqrt{1+2x})^2} \leq 2x+9

Re: HSC 2013 4U Marathon I think you are missing a \frac{1}{2}

When you sub u=x^2 you divide by 2x which cancels one of the x"s and you are left with x^2, which becomes u

Re: HSC 2013 4U Marathon \begin{align*}I&=\int \frac{dx}{(x+2)\sqrt{x^2+4x-5}}\\u^2&=x^2+4x-5\\&=(x+2)^2-9\\u^2+9&=(x+2)^2\\dx&=u\frac{du}{x+2}\\I&=\int \frac{u\cdot du}{(x+2)^2u}\\&=\int \frac{du}{u^2+9}\\&=\frac{1}{3}\tan^{-1}\left ( \frac{\sqrt{x^2+4x-5}}{3} \right )+C\end{align*}

Re: HSC 2013 4U Marathon \begin{align*}\int \frac{\sin2x}{(1+\sin^2x)^2}dx&=\int \frac{ 2\sin{x}\cos{x}}{(1+\sin^2x)^2}dx\\&=\int \frac{ d\left ( \sin^2{x} \right )}{(1+\sin^2x)^2}\\&=-\frac{1}{\left ( 1+\sin^2x\right )}+C\end{align*} edit: lol sy beat me to it