HSC 2012-14 MX2 Integration Marathon (archive) (5 Viewers)

Carrotsticks · Aug 10, 2012

Re: MX2 Integration Marathon

U MAD BRO said:
I'm really interested to know how to do these integrals, I'll post them in extracurricular =)

Half of them have already been done in the BOS Seminar booklet I gave you.

U MAD BRO · Aug 10, 2012

Re: MX2 Integration Marathon

Carrotsticks said:
Half of them have already been done in the BOS Seminar booklet I gave you.

So they are MX2 level

Carrotsticks · Aug 10, 2012

Re: MX2 Integration Marathon

U MAD BRO said:
So they are MX2 level

When broken up into parts as I have done, THEN it is MX2 level.

However, just having it out there as a "Find the value of this integral" kinda question, it is no longer MX2 level.

Anyway, lets try to keep this thread on topic shall we? =)

U MAD BRO · Aug 10, 2012

Re: MX2 Integration Marathon

Carrotsticks said:
When broken up into parts as I have done, THEN it is MX2 level.

However, just having it out there as a "Find the value of this integral" kinda question, it is no longer MX2 level.

Anyway, lets try to keep this thread on topic shall we? =)

ok =)

U MAD BRO · Aug 10, 2012

Re: MX2 Integration Marathon

It's not that complicated actually, and I think these techniques are pretty useful in MX2:
Let $I = \int_{0}^{\infty }e^{-x^{2}}dx$ then we have by the square of an integral:
$I^{2}=\left ( \int_{0}^{\infty }e^{-x^{2}}dx \right )\left ( \int_{0}^{\infty }e^{-y^{2}}dy \right ) = \int_{0}^{\infty }\int_{0}^{\infty }e^{-x^{2}-y^{2}}dydx$
Shifting to polar coordinates:
$\int_{0}^{\frac{\pi }{2}}\int_{0}^{\infty }e^{-r^{2}}rdrd\theta$
For the interior integral, we use the transformation:
$u=r^{2}, \frac{du}{2}=rdr$ to obtain:
$\frac{1}{2}\int_{0}^{\infty }e^{-u}du = -\frac{1}{2}\left [ e^{-u} \right ]\infty to 0 = \frac{1}{2}$
Plugging this into our integral, we're left with:
$I^{2}=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}d\theta = \frac{\pi }{4}$
By taking the square root of both sides, we have the answer:
$I=\int_{0}^{\infty }e^{-x^{2}}dx=\frac{\sqrt{\pi }}{2}$

Also, this thread is painfully slow, because everyone is posting in the nonsense thread

Carrotsticks · Aug 10, 2012

Re: MX2 Integration Marathon

Evaluating the Gaussian Integral can be done much more easily using Double Integrals (compared to the elementary version which requires using the Fundamental Theorem of Calculus), but steps that you skimmed over are not at all obvious.

For example, how can you be sure that multiplying two integrals means that they can be iterated within one another?

Also, the change to polar coordinates is not exactly simple.

C'mon, try to keep this thread MX2-friendly.

Here's a nice integral for some people do do:

$\int \sqrt{\frac{x}{1-x}} dx$

U MAD BRO · Aug 10, 2012

Re: MX2 Integration Marathon

I read that when you evaluate an integral and get a numerical value, you can do a rough sketch the integrand to see whether your solution is reasonable for the area under the curve. Is that a legit way of checking whether ur answer is correct or not?

Carrotsticks · Aug 10, 2012

Re: MX2 Integration Marathon

U MAD BRO said:
I read that when you evaluate an integral and get a numerical value, you can do a rough sketch the integrand to see whether your solution is reasonable for the area under the curve. Is that a legit way of checking whether ur answer is correct or not?

Of course not. Everything must be proven, that's what Mathematics is all about. Drawing a sketch can tell you if you're on the right track or not, but most certainly doesn't constitute a proof.

U MAD BRO · Aug 10, 2012

Re: MX2 Integration Marathon

Carrotsticks said:
Here's a nice integral for some people do do:

$\int \sqrt{\frac{x}{1-x}} dx$

Do you make $x=sin^{2}\theta$ ?

Carrotsticks · Aug 10, 2012

Re: MX2 Integration Marathon

Maybe, maybe not =)

U MAD BRO · Aug 10, 2012

Re: MX2 Integration Marathon

$\theta -\frac{sin2\theta }{2}+C$

Is that the answer?
Wanna make sure it is before I type the working on LaTeX

do i have to make it in terms of x? =(

Carrotsticks · Aug 10, 2012

Re: MX2 Integration Marathon

Yep, in terms of x please.

U MAD BRO · Aug 10, 2012

Re: MX2 Integration Marathon

$sin^{-1}\sqrt{x}-\frac{sin2(sin^{-1}\sqrt{x})}{2}$

=(

deswa1 · Aug 10, 2012

Re: MX2 Integration Marathon

U MAD BRO said:
$sin^{-1}\sqrt{x}-\frac{sin2(sin^{-1}\sqrt{x})}{2}$

=(

You should be able to simplify the second bit using double angle formulae

U MAD BRO · Aug 10, 2012

Re: MX2 Integration Marathon

Do I lose marks if I don't simplify it?

$sin^{-1}\sqrt{x}-\frac{2sin(sin^{-1}\sqrt{x})cos(sin^{-1}\sqrt{x})}{2}=sin^{-1}\sqrt{x}-\frac{2(\sqrt{x})cos(sin^{-1}\sqrt{x})}{2}=sin^{-1}\sqrt{x}-\frac{2(\sqrt{x})cos(\sqrt{1-x})}{2}$

Carrotsticks · Aug 10, 2012

Re: MX2 Integration Marathon

You can further simplify the cos (arcsin rootx)

U MAD BRO · Aug 10, 2012

Re: MX2 Integration Marathon

Carrotsticks said:
You can further simplify the cos (arcsin rootx)

NO!

I edited the solution ^^

asianese · Aug 10, 2012

Re: MX2 Integration Marathon

In fact, the trig sub is a little messy, so you could times top and bottom by sqrt x, and then completing the square in the denominator. It is all algebraic then and no need to mega simplify in terms of theta.

Umad, as we've said keep this on topic, MX2 integration. Also unless Carrot is your tutor outside of BOS, don't hammer him with so many questions! I know he is willing to help, but he is just doing it out of the mathematical goodness of his heart - so respect that.

Carrotsticks · Aug 10, 2012

Re: MX2 Integration Marathon

U MAD BRO said:
Carrot, can you teach me how to do counter integration, looks like a pretty cool technique.
For example, I'm doing this bad ass nasty little integral and I couldn't continue because I have a counter integral:

$\int_{0}^{\infty }\frac{cos(x)}{1+(x)^{2}}dx$

$\int_0^{\infty}\frac{\cos(x)}{(x+i)(x-i)}dx=1/2 \int_{-\infty}^{\infty} \frac{\cos(x)}{(x+i)(x-i)}dx$

then consider the contour integral:

$\oint \frac{e^{iz}}{(z+i)(z-i)}dz$

where the contour is the half-disc in the upper half plane, then just use the Residue Theorem.

I think this is really interesting, I never though integration gets any more advanced =)

If you make a thread in Extracurricular topics, I will answer your question.

asianese said:
In fact, the trig sub is a little messy, so you could times top and bottom by sqrt x, and then completing the square in the denominator. It is all algebraic then and no need to mega simplify in terms of theta.

Yep, this was what I was looking for, except it turns out more nicely if you multiply top and bottom by sqrt(1-x) so you complete the square in the numerator.

This is because when you eventually use the trig sub after completing the square, the dx will cancel with the bottom.

U MAD BRO · Aug 10, 2012

Re: MX2 Integration Marathon

I got the answer anyway

OK, I'll make a thread there =)

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