Search results

14(b)(ii) The graphs suggest that as c increases from c=0.8 to c=1, there should be some point where the graph has exactly one x-intercept. Furthermore, that x-intercept will be a double root, as indicated by there is also a stationary point. Therefore, let f(x) = x^4 - 2cx^3 + 1, and f'(x) =...

Lol my bad. Idk why I decided to switch the cos out for a sin when posting

\int_0^\infty \frac{\sin x}{x^2+1}dx

As correctly stated, A and B are good because they both equal to 0, soo they are ruled out. C fails because by taking the odd function f(x) = -x^3, we have \begin{align*} &\quad\int_{-a}^0 f(x)\,dx + \left| \int_0^a f(x)\,dx \right|\\ &= \int_{-a}^0 -x^3\,dx + \left| \int_0^a -x^3\,dx...

Re: HSC 2018 MX2 Integration Marathon $\noindent Take $f(x)$ to be any arbitrary odd function well defined for all $ -\frac\pi2 \leq x \leq \frac\pi2\\ $and let $ I = \int_{-\pi/2}^{\pi/2} \frac{\cos x}{e^{f(x)}+1}\,dx. $\noindent Considering the substitution $u=-x,\\ \begin{align*}I&=...

Re: HSC 2018 MX2 Integration Marathon \text{Let }I = \int_0^{\pi/6} \ln (\tan x + \sqrt{3} ) \,dx \begin{align*} I &= \int_0^{\pi/6} \ln \left( \frac{ \frac{1} {\sqrt3} - \tan x }{1 + \frac{\tan x}{\sqrt3} } + \sqrt{3} \right)dx\\ &= \int_0^{\pi/6 } \ln \left( \frac{4}{\sqrt3 + \tan x}...

\int_0^{10} x^{\lim_{n\to \infty} f^n(x)}\,dx \text{ for }f(x) = \frac12 \left( x + \frac2x\right) \text{where }f^n(x) = \underbrace{f \cdots f}_{n\text{-times}}(x) Shouldn't be hard assuming I didn't mess up typing the question.

@Paradoxica I think I finally figured what it should've been. \int_1^e x \sqrt[6]{x^{-1} \sqrt[20]{x \sqrt[42]{x^{-1} \dots}}}\,dx Well, hopefully.

I'd recommend InteGrand's rearrangement. But it's still fairly easy just using what they give you. \text{Closure condition: Let }x,y\in C_g\text{, so we have} \\ \begin{align*} gx &= xg \\ gy &= yg \end{align*} \noindent\text{Then, with the aid of the associative rule,} \\ \begin{align*} g(xy)...

This was in the final exam and I never figured it out. \text{Let }f:\mathbb{R}\to \mathbb{R}\text{ be a convex function. Prove that for any }x,y,z\in \mathbb{R}, \frac{ f(x)+f(y)+f(z) }3 + f \left( \frac{x+y+z}3 \right) \ge \frac{2}{3} \left[ f \left( \frac{x+y}{2} \right) + f \left(...

Re: HSC 2018 MX2 Integration Marathon Which isn't even 4U. Savage,

Re: HSC 2018 MX2 Integration Marathon That being said is there a way of doing it using only 4U methods

\int \frac{\sin^2 x - 4\sin x \cos x + 3\cos^2 x}{\sin x +\cos x}dx

\text{The substitution }x=\exp (-u)\text{ turns the first integral into }\int_0^\infty u^n \log u \exp (-u)\,du \text{Call it }I_n.\text{ From IBP,}\\ \begin{align*}I_n&= -e^{-u}u^n \log u \Big |_0^\infty + \int_0^\infty e^{-u}u^{n-1}\left(1 + n \log u\right)\,du\\ &= J_{n-1}+ n...

\int_0^\pi \ln(1+2a\cos x + a^2)\,dx\text{ splitting into appropriate cases} (Not sure if already asked)

At least quick partial fractions is possible this time round lol u^2 = \tanh x \implies 2u\,du = \text{sech}^2 x\,dx = (1-\tanh^2 x)\,dx\quad (u\ge 0) \begin{align*} I & = \int \sqrt{ \tanh x} \, dx\\ &= \int \frac{2u^2}{1-u^4} \, du\\ &= \int \frac{2u^2}{(1-u^2)(1+u^2)}\,du\\ &=...

Re: HSC 2018 MX2 Integration Marathon \text{For }a>0,\text{ find }\int_0^{\infty} \frac{dx}{(1+x^a)(1+x^2)} Also spoiler for the one above:

Yes.

It's a bit lame but it's not unjustified, because a lot of people rote-learn the (n-1)! formula without understanding why it works. Thus they can't adapt it to other circular arrangements (e.g. those, but two must sit next to each other).

Easy difficulty \int \frac{\tanh x}{\exp x}\,dx