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Re: MX2 2016 Integration Marathon Why?

Re: HSC 2016 3U Marathon $\textbf{NEW QUESTION}$ $\noindent (i) Let $n$ and $k$ be positive integers with $k\leq n$ and $k\geq 1$. Show that $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$ (this identity is known as \textit{Pascal's Rule}).$ $\noindent (ii) For a non-negative integer $k$, the...

Re: MX2 2016 Integration Marathon $\noindent \textbf{NEW EASY QUESTION}$ $\noindent Find $\int \frac{\sqrt{1+x}}{x}\text{ d}x$.$

Re: MX2 2016 Integration Marathon Wasn't the –1 just a typo? (By the way, this wasn't the fundamental theorem of calculus. The symbol ∫ on its own (without any limits of integration) just means an antiderivative, so ∫d/dx (f(x)) dx = f(x) (+C) simply because ∫ and d/dx are essentially inverse...

Re: MX2 2016 Integration Marathon $Correct!$

Re: MX2 2016 Integration Marathon \noindent \textbf{EASY QUESTION}$^\dagger$$ $\noindent Find $\int \frac{1}{\sqrt{1+\mathrm{e}^{x}}}\text{ d}x$.$ $\noindent $^\dagger$ \small{Relative to most questions in this Marathon.}$

Re: HSC 2016 2U Marathon That has to be one of the most tedious 2U questions I've ever seen (especially the second derivative one). :)

Re: HSC 2016 2U Marathon $\noindent We wouldn't multiple by $\frac{c}{c}$, but rather we'd multiply both sides of the equation by $c$ (I guess that's what you meant). Another way to structure the proof is to say that the average of the new list is:$ $$\begin{align*}\mu _{\text{new}} &=...

Re: HSC 2016 2U Marathon $\noindent Don't need to prove it, it's assumed.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent To make it even clearer, we can let $u=\mathrm{e}^{x}\Rightarrow \mathrm{d}u = \mathrm{e}^{x}\text{ d}x$, so the integral is $\int \frac{\mathrm{d}u}{1+u^2}=\tan^{-1} u = \tan^{-1} \left(\mathrm{e}^{x}\right)$.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent The numerator just needed to be the derivative of $\mathrm{e}^x$ for us. I.e. it was (leaving out the limits) $\int \frac{\mathrm{e}^{x}}{1+\left(\mathrm{e}^{x}\right)^2}\text{ d}x =\int \frac{f^\prime...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent For (b), first multiply top and bottom of the integrand by $\mathrm{e}^{x}$. So we obtain $ $$\begin{align*}\int_0 ^1 \frac{1}{\mathrm{e}^{-x}+\mathrm{e}^{x}}\text{ d}x &=\int_0 ^1...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $The other method is to \textsl{first} argue that $\alpha$ lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, and then use the inverse trigonometric identity. We can use the identity $\tan^{-1} A + \tan^{-1} B = \tan^{-1}...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $So in other words, let $\alpha = \tan^{-1} \frac{3}{5}-\tan^{-1} \left( -\frac{1}{4}\right)\Rightarrow \alpha=\tan^{-1} \frac{3}{5} + \tan^{-1} \left( \frac{1}{4}\right) $. Then $ $$\begin{align*}\tan \alpha &= \frac{\tan...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $So basically we want to show that $\tan^{-1} \frac{3}{5}-\tan^{-1} \left(-\frac{1}{4} \right)=\frac{\pi}{4}$. So what we do is, we show that $\tan LHS = \tan RHS$ (using the compound angle formula for tan on the L.H.S., which...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Use this inverse trigonometric identity: $\tan^{-1} A - \tan^{-1}B = \tan^{-1} \left(\frac{A-B}{1+AB} \right)$, which is true provided that $\tan^{-1} A - \tan^{-1}B$ lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Did you find the exact area from $-a$ to $a$? Use this expression and let $a\to \infty$. Also use the facts that $\lim_{\psi \to \pm \infty} \tan^{-1} \psi = \pm \frac{\pi}{2}$.$

Re: MX2 2016 Integration Marathon Did I? Must have forgotten I did it!

Yes it's possible (if it weren't, you wouldn't have got an interview I think haha). Good luck!

Re: MX2 2016 Integration Marathon $\noindent I think so, just a bit tedious to get the expression to look nice.$ $\noindent Let $f(x)=\frac{1}{x(x+1)(x+2)\ldots (x+n)}$. We let $\frac{1}{x(x+1)(x+2)\ldots (x+n)}\equiv \sum_{k=0}^{n}\frac{A_k}{x+k}$ (partial fractions). Our goal is essentially...