Higher Level Integration Marathon & Questions (1 Viewer)

Carrotsticks · Feb 21, 2016

Re: Extracurricular Integration Marathon

Drsoccerball said:
Is this apart of Matrices ? Doesn't seem to hard So can we treat one of the changing functions as a constant while we integrate the 'flat' part ?

Typically it's taught in an first course in vector calculus, not linear algebra where matrices tend to live.

The initial treatments tend to be highly elementary in nature and are very much as you have described above. A Year 11 student could compute some double integrals, simply treating them as a "2 questions in 1" style problem.

However, the difficulty usually comes in the construction of the integral and then spotting clever substitutions/manipulations to invoke Fubini or something that will help simplify the computation.

Drsoccerball · Feb 21, 2016

Re: Extracurricular Integration Marathon

Carrotsticks said:
Typically it's taught in an first course in vector calculus, not linear algebra where matrices tend to live.

The initial treatments tend to be highly elementary in nature and are very much as you have described above. A Year 11 student could compute some double integrals, simply treating them as a "2 questions in 1" style problem.

However, the difficulty usually comes in the construction of the integral and then spotting clever substitutions/manipulations to invoke Fubini or something that will help simplify the computation.

Like the first question on this marathon ? $\int_{-\infty}^{\infty}{e^{-x^2}}dx$ Thanks Carrot!

Carrotsticks · Feb 21, 2016

Re: Extracurricular Integration Marathon

Drsoccerball said:
Like the first question on this marathon ? $\int_{-\infty}^{\infty}{e^{-x^2}}dx$ Thanks Carrot!

Pretty much. Though for that question we'd use a combination of the polar and cartesian systems to evaluate it.

In the Volumes problem Q14 of 2014 BOS Trials Extension 2, I provided a VERY brief glimpse into evaluating that integral via double integration (Cartesian).

leehuan · Feb 21, 2016

Re: Extracurricular Integration Marathon

I think I ignored all the volumes questions in the BoS trials...they looked scarier than inequalities and mechanics...

Drsoccerball · Feb 21, 2016

Re: Extracurricular Integration Marathon

Carrotsticks said:
Pretty much. Though for that question we'd use a combination of the polar and cartesian systems to evaluate it.

In the Volumes problem Q14 of 2014 BOS Trials Extension 2, I provided a VERY brief glimpse into evaluating that integral via double integration (Cartesian).

Yes, I didn't even pick that up when Sy123 did it for us

Paradoxica · Feb 26, 2016

Re: Extracurricular Integration Marathon

$\int_0^1 \log{[\Gamma{(x+\alpha)}]} \quad \for \alpha \in \mathbb{R}^+$

Paradoxica · Mar 12, 2016

Re: Extracurricular Integration Marathon

$$\noindent$I(\alpha) = \int_0^1 \log{(\Gamma{(x+\alpha)})}$d$x \\ I'(\alpha) = \int_0^1 \frac{\Gamma'{(x+\alpha)}}{\Gamma{(x+\alpha)}}$d$x = \int_0^1 $d$(\log{\Gamma{(x+\alpha)}}) = \log{\Gamma{(x+\alpha)}}\bigg|_0^1\\I'(\alpha) = \log{\Gamma{(\alpha+1)}}-\log{\Gamma{(\alpha)}} \\ $Using the properties of the Gamma function, we have: $\Gamma{(\alpha+1)}=\alpha\Gamma{(\alpha)} \\ \therefore I'(\alpha) = \log{\alpha} \\ \Rightarrow I(\alpha) = \alpha\log{\alpha} - \alpha + \mathcal{C} \\ $It now remains to find a single value of $\alpha$ for which we can evaluate independently to determine the constant of integration.$ \\ I(0) = \int_0^1 \log{\Gamma{(x)}}$d$x = \int_0^1 \log{\Gamma{(1-x)}}$d$x = \frac{1}{2}\int_0^1 \log{\Gamma{(x)}\Gamma{(1-x)}}$d$x \\ $Using the well-known Euler's Reflection Formula for the Gamma Function: $\\ \Gamma{(x)}\Gamma{(1-x)} = \frac{\pi}{\sin{\pi x}}$

$$\noindent$ 2I(0) = \int_0^1 \log{\pi} - \log{(\sin{\pi x})}\,$d$x = \log{\pi} - \frac{1}{\pi} \int_0^\pi \log{\sin{x}}$d$x = \log{\pi} - \frac{\text{J}}{\pi} \\$The value of $J$ is well known here on BOS, and it's value is $-\pi\log{2} \\ 2I(0) = \log{\pi} + \log{2} \Leftrightarrow I(0) = \log{\sqrt{2\pi}} $ , or, to satisfy KoA, $I(0) = \log{\sqrt{\tau}} \\ I(\alpha) = \alpha \log{\alpha} - \alpha + \log{\sqrt{\tau}}$

Paradoxica · Mar 23, 2016

Re: Extracurricular Integration Marathon

Prove the following result:

$\int_0^\infty \frac{(x^2-1)\tan^{-1}{(x^2)}}{x^4 + 4x^2 +1}$d$x= \frac{\pi^2}{12\sqrt{2}}$

leehuan · Mar 23, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
Prove the following result:

$\int_0^\infty \frac{(x^2-1)\tan^{-1}{(x^2)}}{x^4 + 4x^2 +1}$d$x= \frac{\pi^2}{12\sqrt{2}}$

Was this the last question?

leehuan · Mar 23, 2016

Re: Extracurricular Integration Marathon

Also, this was a perplexing question for a Q2...

$\int _{ 0 }^{ 1 }{ \int _{ x }^{ { x }^{ \frac { 1 }{ 3 } } }{ \sqrt { 1-{ y }^{ 4 } } \, dy\, dx } }$

Paradoxica · Mar 23, 2016

Re: Extracurricular Integration Marathon

leehuan said:
Was this the last question?

No, go to like first or second page for q20

seanieg89 · Mar 24, 2016

Re: Extracurricular Integration Marathon

leehuan said:
Also, this was a perplexing question for a Q2...

$\int _{ 0 }^{ 1 }{ \int _{ x }^{ { x }^{ \frac { 1 }{ 3 } } }{ \sqrt { 1-{ y }^{ 4 } } \, dy\, dx } }$

Swap the order of integration to arrive at $\int_0^1 (y-y^3)\sqrt{1-y^4}\, dy.$

This is killed by the substitution $y^2=\sin(\theta)$ , and the result is $\frac{\pi}{8}-\frac{1}{6}.$

Paradoxica · Mar 24, 2016

Re: Extracurricular Integration Marathon

seanieg89 said:
Swap the order of integration to arrive at $\int_0^1 (y-y^3)\sqrt{1-y^4}\, dy.$

This is killed by the substitution $y^2=\sin(\theta)$ , and the result is $\frac{\pi}{8}-\frac{1}{6}.$

I finally get how swapping order works... it only took me four months...

Paradoxica · Mar 24, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
Prove the following result:

$\int_0^\infty \frac{(x^2-1)\tan^{-1}{(x^2)}}{x^4 + 4x^2 +1}$d$x= \frac{\pi^2}{12\sqrt{2}}$

Also, looking for real methods of proving the above. I have seen the complex method, but that's not helpful since I don't know anything about complex analysis yet.

integral95 · Mar 24, 2016

Re: Extracurricular Integration Marathon

$\int \frac{e^{2x}+e^{3x}}{coshx} dx$

Paradoxica · Mar 24, 2016

Re: Extracurricular Integration Marathon

integral95 said:
$\int \frac{e^{2x}+e^{3x}}{\cosh{x}} dx$

$$\noindent$ \int \frac{2e^{2x}(1+e^x)}{e^x+e^{-x}}$d$x = \int \frac{2v^2(1+v)}{1+v^2}$d$v\\ = \int 2(1+v)-\frac{2(1+v)}{1+v^2}\,$d$v = 2v + v^2 - \log{(1+v^2)}-2\tan^{-1}{v} +\mathcal{C} \\= 2e^x + e^{2x}-\log{(1+e^{2x})}-2\tan^{-1}{e^x} + \mathcal{C}$

Nothing Extracurricular about this. Just tedious.

leehuan · Mar 24, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
$$\noindent$ \int \frac{2e^{2x}(1+e^x)}{e^x+e^{-x}}$d$x = \int \frac{2v^2(1+v)}{1+v^2}$d$v\\ = \int 2(1+v)-\frac{2(1+v)}{1+v^2}\,$d$v = 2v + v^2 - \log{(1+v^2)}-2\tan^{-1}{v} +\mathcal{C} \\= 2e^x + e^{2x}-\log{(1+e^{2x})}-2\tan^{-1}{e^x} + \mathcal{C}$

Nothing Extracurricular about this. Just tedious.

You forget that when they give it as cosh the HSC students don't know what it is.

integral95 · Mar 24, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
$$\noindent$ \int \frac{2e^{2x}(1+e^x)}{e^x+e^{-x}}$d$x = \int \frac{2v^2(1+v)}{1+v^2}$d$v\\ = \int 2(1+v)-\frac{2(1+v)}{1+v^2}\,$d$v = 2v + v^2 - \log{(1+v^2)}-2\tan^{-1}{v} +\mathcal{C} \\= 2e^x + e^{2x}-\log{(1+e^{2x})}-2\tan^{-1}{e^x} + \mathcal{C}$

Nothing Extracurricular about this. Just tedious.

wow I can't believe I didn't take out a v when I did it,

Paradoxica · Mar 24, 2016

Re: Extracurricular Integration Marathon

leehuan said:
You forget that when they give it as cosh the HSC students don't know what it is.

That's just information. The difficulty of the integral is completely within the realms of the HSC.

leehuan · Mar 24, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
That's just information. The difficulty of the integral is completely within the realms of the HSC.

You'd have to define cosh(x) if this were to be put in the MX2 integration marathon.

Of course, if this was done in advance then I'd agree with what you mean by difficulty.

Higher Level Integration Marathon & Questions (1 Viewer)

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