# Thread: Higher Level Integration Marathon & Questions

1. ## Re: Extracurricular Integration Marathon

How do double integrals even work ?

2. ## Re: Extracurricular Integration Marathon

Originally Posted by Drsoccerball
How do double integrals even work ?
You just evaluate them variable by variable.

How you generate them is a more interesting story.

3. ## Re: Extracurricular Integration Marathon

Originally Posted by Drsoccerball
How do double integrals even work ?
Originally Posted by leehuan
You just evaluate them variable by variable.

How you generate them is a more interesting story.
Yeah, knowing when to revert a function into the definite integral of a simpler function is not at all obvious for the most part.

4. ## Re: Extracurricular Integration Marathon

Originally Posted by Drsoccerball
How do double integrals even work ?
You're probably clever enough to get the general gist of it (double integration via Cartesian coordinate system) from this image. I'm guessing you just want a rough idea, not a more rigorous construction.

Just imagine throwing in an extra dimension to everything. Instead of 1 dimensional partitions (the subintervals) of a domain, we have 2 dimensional partitions of a region. Instead of taking 2 dimensional 'strips', we have 3 dimensional 'strips'. Instead of swooping over the domain once, we swoop over it twice (once in the x direction, again in the y direction). Instead of finding an 'area', we find a 'volume'.

5. ## Re: Extracurricular Integration Marathon

Originally Posted by Carrotsticks
You're probably clever enough to get the general gist of it (double integration via Cartesian coordinate system) from this image. I'm guessing you just want a rough idea, not a more rigorous construction.

Just imagine throwing in an extra dimension to everything. Instead of 1 dimensional partitions (the subintervals) of a domain, we have 2 dimensional partitions of a region. Instead of taking 2 dimensional 'strips', we have 3 dimensional 'strips'. Instead of swooping over the domain once, we swoop over it twice (once in the x direction, again in the y direction). Instead of finding an 'area', we find a 'volume'.

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

6. ## Re: Extracurricular Integration Marathon

Originally Posted by Drsoccerball
Is this apart of Matrices ? Doesn't seem to hard :P 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.

7. ## Re: Extracurricular Integration Marathon

Originally Posted by Carrotsticks
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!

8. ## Re: Extracurricular Integration Marathon

Originally Posted by Drsoccerball
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).

9. ## Re: Extracurricular Integration Marathon

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

10. ## Re: Extracurricular Integration Marathon

Originally Posted by Carrotsticks
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 :P

11. ## Re: Extracurricular Integration Marathon

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

12. ## Re: Extracurricular Integration Marathon

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

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

13. ## Re: Extracurricular Integration Marathon

Prove the following result:

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

14. ## Re: Extracurricular Integration Marathon

Prove the following result:

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

15. ## Re: Extracurricular Integration Marathon

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

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

16. ## Re: Extracurricular Integration Marathon

Originally Posted by leehuan
No, go to like first or second page for q20

17. ## Re: Extracurricular Integration Marathon

Originally Posted by leehuan
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 $\frac{\pi}{8}-\frac{1}{6}.$

18. ## Re: Extracurricular Integration Marathon

Originally Posted by seanieg89
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...

19. ## Re: Extracurricular Integration Marathon

Prove the following result:

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

20. ## Re: Extracurricular Integration Marathon

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

21. ## Re: Extracurricular Integration Marathon

Originally Posted by integral95
$\int \frac{e^{2x}+e^{3x}}{\cosh{x}} dx$
$\noindent \int \frac{2e^{2x}(1+e^x)}{e^x+e^{-x}}dx = \int \frac{2v^2(1+v)}{1+v^2}dv\\ = \int 2(1+v)-\frac{2(1+v)}{1+v^2}\,dv = 2v + v^2 - \log{(1+v^2)}-2\tan^{-1}{v} +\mathcal{C} \\= 2e^x + e^{2x}-\log{(1+e^{2x})}-2\tan^{-1}{e^x} + \mathcal{C}$

22. ## Re: Extracurricular Integration Marathon

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

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

23. ## Re: Extracurricular Integration Marathon

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

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

24. ## Re: Extracurricular Integration Marathon

Originally Posted by leehuan
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.

25. ## Re: Extracurricular Integration Marathon

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.

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