HSC 2016 MX2 Integration Marathon (archive) (1 Viewer)

Drsoccerball · Nov 29, 2015

Re: MX2 2016 Integration Marathon

New Question:
$\int_2^4{\frac{\sqrt{ln(9-x)}}{\sqrt{ln(9-x)}+\sqrt{ln(x+3)}}}dx$

omegadot · Nov 30, 2015

Re: MX2 2016 Integration Marathon

Ekman said:
$\int_{0}^{\pi} \cos^{2}(\tan^{-1}(\sin(\cot^{-1}(x)))) dx$

InteGrand said:
$Note that the integrand given in Ekman's question can be simplified to $\frac{1+x^2}{\left(2+x^2\right)^2}$, which you can show using right-angled triangles.$

$$\noindent InteGrand - I think the integrand simplies to $\frac{1 + x^2}{2 + x^2}.$

$$\noindent Note that $\sin (\cot^{-1} x) = \frac{1}{\sqrt{1 + x^2}} = \alpha$ and $\cos (\tan^{-1} \alpha ) = \frac{1}{\sqrt{1 + \alpha^2}}.$

$$\noindent So$\\\begin{align*}\cos \left (\tan^{-1} \left [\sin (\cot^{-1} x)\right ] \right ) = \frac{1}{\sqrt{1 + \left (\frac{1}{\sqrt{1 + x^2}} \right )^2}} = \sqrt{\frac{1 + x^2}{2 + x^2}}.\end{align*}.$

$$\noindent Thus$\\\begin{align*}\int^\pi_0 \cos^2 \left (\tan^{-1} \left [\sin (\cot^{-1} x) \right ]\right ) \, dx &= \int^\pi_0 \frac{1 + x^2}{2 + x^2} \, dx\\&= \int^\pi_0 \frac{(2 + x^2) - 1}{2 + x^2} \, dx\\&= \int^\pi_0 dx - \int^\pi_0 \frac{dx}{2 + x^2}\\&= \left [x - \frac{1}{\sqrt{2}} \tan^{-1} \left (\frac{x}{\sqrt{2}} \right ) \right ]^\pi_0\\&= \pi - \frac{1}{\sqrt{2}} \tan^{-1} \left (\frac{\pi}{\sqrt{2}} \right ).\end{align*}$

InteGrand · Nov 30, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent InteGrand - I think the integrand simplies to $\frac{1 + x^2}{2 + x^2}.$

$$\noindent Note that $\sin (\cot^{-1} x) = \frac{1}{\sqrt{1 + x^2}} = \alpha$ and $\cos (\tan^{-1} \alpha ) = \frac{1}{\sqrt{1 + \alpha^2}}.$

$$\noindent So$\\\begin{align*}\cos \left (\tan^{-1} \left [\sin (\cot^{-1} x)\right ] \right ) = \frac{1}{\sqrt{1 + \left (\frac{1}{\sqrt{1 + x^2}} \right )^2}} = \sqrt{\frac{1 + x^2}{2 + x^2}}.\end{align*}.$

$$\noindent Thus$\\\begin{align*}\int^\pi_0 \cos^2 \left (\tan^{-1} \left [\sin (\cot^{-1} x) \right ]\right ) \, dx &= \int^\pi_0 \frac{1 + x^2}{2 + x^2} \, dx\\&= \int^\pi_0 \frac{(2 + x^2) - 1}{2 + x^2} \, dx\\&= \int^\pi_0 dx - \int^\pi_0 \frac{dx}{2 + x^2}\\&= \left [x - \frac{1}{\sqrt{2}} \tan^{-1} \left (\frac{x}{\sqrt{2}} \right ) \right ]^\pi_0\\&= \pi - \frac{1}{\sqrt{2}} \tan^{-1} \left (\frac{\pi}{\sqrt{2}} \right ).\end{align*}$

Oh yes, I realise I forgot to square-root a hypotenuse for one of them.

Ekman · Nov 30, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
New Question:
$\int_2^4{\frac{\sqrt{ln(9-x)}}{\sqrt{ln(9-x)}+\sqrt{ln(x+3)}}}dx$

$$ Sub in: $ u=3-x$

$\therefore \int_{2}^{4} \frac{\sqrt{\ln{(9-x)}}}{\sqrt{\ln{(9-x)}} + \sqrt{\ln{(3+x)}}} dx \Rightarrow \int_{-1}^{1} \frac{\sqrt{\ln{(6+u)}}}{\sqrt{\ln{(6+u)}} + \sqrt{\ln{(6-u)}}} du$

$\int_{-1}^{1} \frac{\sqrt{\ln{(6+u)}}}{\sqrt{\ln{(6+u)}} + \sqrt{\ln{(6-u)}}} du = \int_{0}^{1} \frac{\sqrt{\ln{(6+u)}}}{\sqrt{\ln{(6+u)}} + \sqrt{\ln{(6-u)}}} + \frac{\sqrt{\ln{(6-u)}}}{\sqrt{\ln{(6-u)}} + \sqrt{\ln{(6+u)}}} du$

$= \int_{0}^{1} 1 du = 1$

omegadot · Nov 30, 2015

Re: MX2 2016 Integration Marathon

$$\noindent \textbf{Next Question}$

$$\noindent Evaluate $\int^1_{-2} \sqrt{x^4 + 2x^3 + x^2} \, dx.$

Drsoccerball · Nov 30, 2015

Re: MX2 2016 Integration Marathon

$$ Alternatively, Let $ u=2-x $ and use the fact that $ \int_0^a{f(x)}dx=\int_0^a{f(a-x)}dx$

Drsoccerball · Nov 30, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent \textbf{Next Question}$

$$\noindent Evaluate $\int^1_{-2} \sqrt{x^4 + 2x^3 + x^2} \, dx.$

$Take out $ x^2 $ from the square root and notice that the inside is a perfect square$

$\therefore I=\int_{-2}^1{x(x+1)}dx $ Which is easily solveable.$

kawaiipotato · Nov 30, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$Take out $ x^2 $ from the square root and notice that the inside is a perfect square$

$\therefore I=\int_{-2}^1{x(x+1)}dx $ Which is easily solveable.$

There should be absolute values so we need to integrate from -2 to -1 of x(x+1) adding it with integral from -1 to 0 of -x (x+1) and finally adding with the integral of x (x+1) from 0 to 1

omegadot · Nov 30, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$Take out $ x^2 $ from the square root and notice that the inside is a perfect square$

$\therefore I=\int_{-2}^1{x(x+1)}dx $ Which is easily solveable.$

kawaiipotato said:
There should be absolute values so we need to integrate from -2 to -1 of x(x+1) adding it with integral from -1 to 0 of -x (x+1) and finally adding with the integral of x (x+1) from 0 to 1

Correct and is exactly why I chose the limits of integration I did. The final answer is: 11/6.

InteGrand · Nov 30, 2015

Re: MX2 2016 Integration Marathon

$$\textbf{NEW QUESTION}$

$Find $\int _{-1} ^{1} \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{1 + 2^{\frac{1}{x}}}\right)\text{ d}x$.$

leehuan · Nov 30, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
$$\textbf{NEW QUESTION}$

$Find $\int _{-1} ^{1} \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{1 + 2^{\frac{1}{x}}}\right)\text{ d}x$.$

Improper integrals?

Have fun with limits if you're a 2016er

InteGrand · Nov 30, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
Improper integrals?

Have fun with limits if you're a 2016er

Yes. Although of course, such a question would be unlikely to be asked directly in the HSC, and if it were asked, students would probably be guided through the steps.

dan964 · Nov 30, 2015

Re: MX2 2016 Integration Marathon

old solution had an error (noted by Integrand below)
new solution View attachment ddx11 take 2.pdf

InteGrand · Nov 30, 2015

Re: MX2 2016 Integration Marathon

dan964 said:
View attachment 32697
is this wrong?

$The answer should be $\frac{2}{3}$; also, you don't need to actually differentiate it, because the integrand is continuous in the split integrals when making the bounds replaced by $\epsilon$ and then taking the limit, so we can just use the fundamental theorem of calculus and then take the limit.$

$Basically, the mistake you made was that you should have split it up into two limits with one limit going to 0 from $\textsl{below}$ and the other from \textsl{above}. Note that this gives different results, since $\lim _{\xi \to 0^\pm}\frac{1}{\xi}=\pm \infty$, so $2^{\frac{1}{\xi}}$ approaches different things depending on which way $\xi$ approaches 0 \Big{(}as $\lim _{\Xi \to +\infty}2^\Xi = +\infty$ whilst $\lim _{\Xi \to -\infty}2^{\Xi} = 0$\Big{)}.$

dan964 · Nov 30, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
$The answer should be $\frac{2}{3}$; also, you don't need to actually differentiate it, because the integrand is continuous in the split integrals when making the bounds replaced by $\epsilon$ and then taking the limit, so we can just use the fundamental theorem of calculus and then take the limit.$

$Basically, the mistake you made was that you should have split it up into two limits with one limit going to 0 from $\textsl{below}$ and the other from \textsl{above}. Note that this gives different results, since $\lim _{\xi \to 0^\pm}\frac{1}{\xi}=\pm \infty$, so $2^{\frac{1}{\xi}}$ approaches different things depending on which way $\xi$ approaches 0 \Big{(}as $\lim _{\Xi \to +\infty}2^\Xi = +\infty$ whilst $\lim _{\Xi \to -\infty}2^{\Xi} = 0$\BIg{)}.$

The only reason I differentiated it was to get the graph.
But I see what you are saying.

Doesn't the limit approach 1? As (1+2^-(1/x)) is 1 when x approaches 0.

I'll post a fixed solution

InteGrand · Nov 30, 2015

Re: MX2 2016 Integration Marathon

dan964 said:
The only reason I differentiated it was to get the graph.
But I see what you are saying.

$Note that we don't need the graph of the integrand to see that it is not continuous over the interval in question, since we can tell that the function $f(x)=\frac{1}{1+2^{\frac{1}{x}}}$ is not continuous here as it is undefined at $x=0$. So the fact that $f(x)$ is undefined there tells us that $f^\prime (x)$ is undefined there too, so is not continuous on the interval in question.$

dan964 · Nov 30, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
$Note that we don't need the graph of the integrand to see that it is not continuous over the interval in question, since we can tell that the function $f(x)=\frac{1}{1+2^{\frac{1}{x}}}$ is not continuous here as it is undefined at $x=0$. So the fact that $f(x)$ is undefined there tells us that $f^\prime (x)$ is undefined there too, so is not continuous on the interval in question.$

Just wanted to be sure. But it wasn't for continuity; I was really wanting see the shape of the graph to be certain of the limits I needed which is yes, unnecessary.
Fixed solution: http://community.boredofstudies.org...x2-2016-integration-marathon-ddx11-take-2.pdf

I also noted in my original answer I forgot to invert the other values. (So I would've ended up with 1/2 instead of 2/3)

Drsoccerball · Dec 1, 2015

Re: MX2 2016 Integration Marathon

Can someone do this integral I remember paradoxica solved it before but forgot how to solve:

$\int_0^{\frac{\pi}{2}}{xcotx}dx$

leehuan · Dec 1, 2015

Re: MX2 2016 Integration Marathon

I think after the IBP and the ln(sin(x)) appears, that's when you use the f(x)dx=f(a-x)dx integral property. Then something could be done about the sin(2x) due to the altered boundaries and the reflection of ln(sin(u)) about u=pi/2 I think (let u=2x)

omegadot · Dec 1, 2015

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Can someone do this integral I remember paradoxica solved it before but forgot how to solve:

$\int_0^{\frac{\pi}{2}}{xcotx}dx$

$$\noindent Note the integral is improper since the integrand is undefined at $x = 0$. Since it converges we can use a clever little boundary switch to avoid having to evaluate the integral directly at this point (in which case one would need to take a limit as $x \rightarrow 0^+$ and requires the use of l'H\^{o}pital's rule in order to calculate the limit -- something that is beyond the scope of the Extension 2 course).$

$$\noindent Let $I = \int^{\frac{\pi}{2}}_0 x \cot x \, dx$. Now$\\\begin{align*}I &= \int^{\frac{\pi}{4}}_0 x \cot x \, dx + \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} x \cot x \, dx = \int^{\frac{\pi}{4}}_0 x \cot x \, dx + I_1\end{align*}$

$$\noindent For the integral $I_1$, set $x = \frac{\pi}{2} - u, dx = - du$ while for the limits of integration at $x = \frac{\pi}{4}, u = \frac{\pi}{4}$ while at $x = \frac{\pi}{2}, u = 0$. Thus$

$\begin{align*}I_1 &= - \int^0_{\frac{\pi}{4}} \left (\frac{\pi}{2} - u \right ) \cot \left (\frac{\pi}{2} - u \right ) \, du\\&= \int^{\frac{\pi}{4}}_0 \left (\frac{\pi}{2} - u \right ) \tan u \, du\\&= \frac{\pi}{2} \int^{\frac{\pi}{4}}_0 \tan u \, du - \int^{\frac{\pi}{4}}_0 u \tan u \, du\\&= \frac{\pi}{2} \Big{[}-\ln |\cos u| \Big{]}^{\frac{\pi}{4}}_0 - \int^{\frac{\pi}{4}}_0 u \tan u \, du\\&= \frac{\pi}{4} \ln 2 - \int^{\frac{\pi}{4}}_0 u \tan u \, du\end{align*}$

$$\noindent So for integral $I$ we have\\\begin{align*}I &= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 x \cot x \, dx - \int^{\frac{\pi}{4}}_0 x \tan x \, dx\\&= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 x \left (\cot x - \tan x \right ) \, dx\\&= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 x \left (\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} \right ) \, dx\\&= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 x \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} \, dx\\&= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 2x \frac{\cos 2x}{\sin 2x} \, dx\\&= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 2x \cot 2x \, dx\end{align*}$

$$\noindent For the integral to the right, set $u = 2x, du = 2 dx$ while for the limits of integration, at $x = 0, u = 0$, while at $x = \frac{\pi}{4}, u = \frac{\pi}{2}$. Thus$

$\begin{align*}I &= \frac{\pi}{4} \ln 2 + \frac{1}{2} \int^{\frac{\pi}{2}}_0 u \cot u \, du\\&= \frac{\pi}{4} \ln 2 + \frac{1}{2} I\\\Rightarrow \frac{1}{2} I &= \frac{\pi}{4} \ln 2\\\Rightarrow I &= \frac{\pi}{2} \ln 2.\end{align*}$

$$\noindent Thus $\int^{\frac{\pi}{2}}_0 x \cot x \, dx = \frac{\pi}{2} \ln 2.$

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