HSC 2015 MX2 Integration Marathon (archive) (2 Viewers)

InteGrand · Aug 31, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Doesnt it give you 0 ? And also wot u mean by converging and diverging etc..

$It was $\int_{-\infty}^{\infty}\frac{1}{x^2}\text{ d}x$, not $\int_{-\infty}^{\infty}\frac{1}{x}\text{ d}x$ if that's what you may have thought by mistake. And no, it doesn't give 0 (it can't, since $\frac{1}{x^2}$ is positive for all real $x\neq 0$), it diverges to $\infty$. Converging basically means the integral's limit exists and is finite. Diverging means it does not converge.$

Drsoccerball · Aug 31, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$It was $\int_{-\infty}^{\infty}\frac{1}{x^2}\text{ d}x$, not $\int_{-\infty}^{\infty}\frac{1}{x}\text{ d}x$ if that's what you may have thought by mistake. And no, it doesn't give 0 (it can't, since $\frac{1}{x^2}$ is positive for all real $x\neq 0$), it diverges to $\infty$. Converging basically means the integral's limit exists and is finite. Diverging means it does not converge.$

but if you integrate 1/x^2 is becomes 1/x which is odd therefore =0?

InteGrand · Aug 31, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
but if you integrate 1/x^2 is becomes 1/x which is odd therefore =0?

$No, e.g. integrate $x^2$ from $-1$ to $1$. The primitive is odd, but the integral is not 0. If the \emph{integrand} were odd, then the integral would be 0.$

Paradoxica · Aug 31, 2015

Re: MX2 2015 Integration Marathon

$$And\;besides,\;you\;can't\;integrate$\;\int_{-\infty}^{\infty}{\frac{dx}{x^2}}$
$The\;curve\;has\;a\;singularity\;at\;$x=0$.$

glittergal96 · Sep 1, 2015

Re: MX2 2015 Integration Marathon

To get people started on Paradoxica's integral:

$\int_\mathbb{R} \frac{\sin^2(ax)}{x^2}\, dx\\ \\ = \int_\mathbb{R} \sin^2(ax)\frac{d}{dx}(-x^{-1})\, dx\\ \\ = \int_\mathbb{R} \frac{2a\sin(ax)\cos(ax)}{x}\, dx\\ \\ =\int_\mathbb{R} \frac{a\sin(2ax)}{x}\, dx\\ \\ =2a^2\int_\mathbb{R} \frac{\sin(2ax)}{2ax}\, dx\\ \\=a\int_\mathbb{R}\frac{\sin(x)}{x}\, dx.$

Which is a more well known integral. If people don't solve it from here today, I will post the simplest solution for this new integral that I know of.

glittergal96 · Sep 1, 2015

Re: MX2 2015 Integration Marathon

Okay, I will continue. Note in the below that R ranges over non-negative integers.

$\int_\mathbb{R} \frac{\sin(x)}{x}\, dx \\ \\ = \lim_{R \rightarrow \infty}\int_{-(2R+1)\pi/2}^{(2R+1)\pi/2}\frac{\sin(x)}{x}\, dx\\ \\ =\lim_{R\rightarrow\infty}\int_{-\pi/2}^{\pi/2} \frac{\sin((2R+1)x)}{x}\, dx\\ \\=\lim_{R\rightarrow\infty}\int_{-\pi/2}^{\pi/2} \sin((2R+1)x)(\frac{1}{x}-\frac{1}{\sin(x)})\, dx+\lim_{R\rightarrow\infty}\int_{-\pi/2}^{\pi/2} \frac{\sin((2R+1)x)}{\sin(x)}\, dx\\ \\=-\lim_{R\rightarrow \infty}\left[\frac{\cos((2R+1)x}{2R+1}( \frac{1}{x}-\frac{1}{\sin(x)}))\right]^{\pi/2}_{-\pi/2}\\ +\lim_{R\rightarrow \infty}\frac{1}{2R+1}\int_{-\pi/2}^{\pi/2} \cos((2R+1)x)\frac{d}{dx}(\frac{1}{x}-\frac{1}{\sin(x)})\, dx\\+\lim_{R\rightarrow\infty}\int_{-\pi/2}^{\pi/2} \frac{\sin((2R+1)x)}{\sin(x)}\, dx$

Lets look at each of the terms on the final line.

The first is just zero, as the function inside vanishes at the boundary.

The second tends to zero. (We can use basic integral inequalities to show that the derivative term exists and is bounded on the interval of integration. this means that the integral itself is bounded, and the fraction out the front gives us the claimed decay.) I am being a bit lazy here, so ask if there is anything you would like me to clarify or show in more detail.

So it suffices to compute the final term, which we can do by reduction.

$J_R-J_{R-1}=\int_{-\pi/2}^{\pi/2}\frac{\sin((2R+1)x)-\sin((2R-1)x)}{\sin(x)}\, dx\\ \\ =\int_{-\pi/2}^{\pi/2}\cos(2Rx)\, dx\\ \\ =0.$

So

$I=J_0=\pi.$

This gives us an answer of $\pi a$ for the original integral.

glittergal96 · Sep 1, 2015

Re: MX2 2015 Integration Marathon

To be precise, the things I was lazy about can be summarised in the following exercise:

Let $f(x)=\frac{1}{\sin(x)}-\frac{1}{x}$ for $x\in [-\pi/2,\pi/2]\setminus \{0\}$ and $f(0)=0.$

Prove that f is continuous and differentiable on the interval $[-\pi/2,\pi/2]$ . Moreover, show that its derivative is bounded on this interval.

I invite current students to try this! It is fairly easy.

Paradoxica · Sep 2, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
To be precise, the things I was lazy about can be summarised in the following exercise:

Let $f(x)=\frac{1}{\sin(x)}-\frac{1}{x}$ for $x\in [-\pi/2,\pi/2]\setminus \{0\}$ and $f(0)=0.$

Prove that f is continuous and differentiable on the interval $[-\pi/2,\pi/2]$ . Moreover, show that its derivative is bounded on this interval.

I invite current students to try this! It is fairly easy.

Is this the same as showing that $x-\sin{x}$ tends to zero faster than $x\sin{x}$ ?

glittergal96 · Sep 2, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Is this the same as showing that $x-\sin{x}$ tends to zero faster than $x\sin{x}$ ?

Your statement is equivalent to the statement of continuity. Ie f(x) -> 0 as x -> 0.

What is important though is that we can also differentiate f everywhere on this interval (this is easy away from the point 0, but we can also find f'(0) by first principles).

Using this value for f'(0), we can then show that f'(x) is also continuous. (This isn't strictly necessary for our purposes, but as this is MX2 it is probably best to avoid talking about integrating things that are potentially discontinuous).

This means that the integral containing f'(x) in my solution makes sense, the only thing left to do is to show that |f'(x)| < M for some positive real number M for all x in the interval of integration. This tells us that our integral is bounded, as required.

glittergal96 · Sep 8, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
A kind of open-ended question slightly related to my second method above:

Explain why Simpson's rule with equal spacing is "better" as a tool to numerically approximate definite integrals of smooth functions (functions with continuous derivatives of all orders) than the trapezoidal rule with equal spacing.

Try to use calculus and rigorous statements to justify your explanation, rather than just saying vague qualitative things like "quadratics can approximate curves better than lines".

bump.

juantheron · Sep 8, 2015

Re: MX2 2015 Integration Marathon

$Evaluation of $\displaystyle \int\frac{2x^3-1}{x^6+2x^3+9x^2+1}dx\;\; $ and $\;\; \displaystyle \int\frac{x^2\cos^2 x-x\sin x-\cos x-1}{(1+x\sin x)^2}dx$$

AztecWarrior · Sep 8, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
bump.

Because simpson's rule uses the addition seperate curves (across 3 function values) to approximate whereas trapazoidal rule uses the area of trapizoids
Look in the deriving of the formula in the back of ur textbook to see how

braintic · Sep 8, 2015

Re: MX2 2015 Integration Marathon

AztecWarrior said:
Because simpson's rule uses the addition seperate curves (across 3 function values) to approximate whereas trapazoidal rule uses the area of trapizoids
Look in the deriving of the formula in the back of ur textbook to see how

She doesn't need to look it up - I guarantee she knows a lot more maths than you (or me).
This is the integration marathon - she is setting a question for people to attempt, not asking how to do it.
And your answer was the exact type of response she asked respondents to avoid.

glittergal96 · Sep 8, 2015

Re: MX2 2015 Integration Marathon

AztecWarrior said:
Because simpson's rule uses the addition seperate curves (across 3 function values) to approximate whereas trapazoidal rule uses the area of trapizoids
Look in the deriving of the formula in the back of ur textbook to see how

I know how they work and what kind of answer I am looking for.

I am asking students to justify the accuracy of Simpsons rule being greater than that of the trapezoidal rule, and quantify to what extent it is better. Eg, something like a statement about how many function values are required for a prescribed degree of accuracy.

Nothing like this is in the deriving of the formula in high school textbooks that I have seen.

AztecWarrior · Sep 8, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
I know how they work and what kind of answer I am looking for.

I am asking students to justify the accuracy of Simpsons rule being greater than that of the trapezoidal rule, and quantify to what extent it is better. Eg, something like a statement about how many function values are required for a prescribed degree of accuracy.

Nothing like this is in the deriving of the formula in high school textbooks that I have seen.

Soz

leehuan · Sep 9, 2015

Re: MX2 2015 Integration Marathon

I have no idea how people can so easily see the 'reverse product rule' within a reverse quotient rule...

$\\ \int { \frac { { x }^{ 2 }\cos ^{ 2 }{ x } -x\sin { x } -\cos { x } -1 }{ { \left( 1+x\sin { x } \right) }^{ 2 } } dx } \\ =\int { \frac { { x }^{ 2 }-{ x }^{ 2 }\sin ^{ 2 }{ x } -2x\sin { x } -1+x\sin { x } -\cos { x } }{ { \left( 1+x\sin { x } \right) }^{ 2 } } dx } \\ =-\int { \frac { { \left( 1+x\sin { x } \right) }^{ 2 } }{ { \left( 1+x\sin { x } \right) }^{ 2 } } dx } +\int { \frac { { x }^{ 2 }+x\sin { x } -\cos { x } }{ { \left( 1+x\sin { x } \right) }^{ 2 } } dx } \\ =-x-\int { \frac { \cos { x } -x\sin { x } -{ x }^{ 2 }\sin ^{ 2 }{ x } -{ x }^{ 2 }\cos ^{ 2 }{ x } }{ { \left( 1+x\sin { x } \right) }^{ 2 } } dx } \\ =-x-\int { \frac { \cos { x } +x\sin { x } \cos { x } -x\sin { x } -{ x }^{ 2 }\sin ^{ 2 }{ x } -x\sin { x } \cos { x } -{ x }^{ 2 }\cos ^{ 2 }{ x } }{ { \left( 1+x\sin { x } \right) }^{ 2 } } dx }$

$\\ =-x-\int { \frac { \left( \cos { x } -x\sin { x } \right) \left( 1+x\sin { x } \right) -x\cos { x } \left( \sin { x } +x\cos { x } \right) }{ { \left( 1+x\sin { x } \right) }^{ 2 } } dx } \\ =-x-\int { \frac { \left( 1+x\sin { x } \right) \cdot \frac { d }{ dx } \left( x\cos { x } \right) -\left( x\cos { x } \right) \cdot \frac { d }{ dx } \left( 1+x\sin { x } \right) }{ { \left( 1+x\sin { x } \right) }^{ 2 } } dx } \\ =-x-\frac { x\cos { x } }{ 1+x\sin { x } } +C$

Librah · Sep 9, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
Try to use calculus and rigorous statements to justify your explanation, rather than just saying vague qualitative things like "quadratics can approximate curves better than lines".

AztecWarrior said:
Because simpson's rule uses the addition seperate curves (across 3 function values) to approximate whereas trapazoidal rule uses the area of trapizoids
Look in the deriving of the formula in the back of ur textbook to see how

lol

Paradoxica · Sep 15, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
I have no idea how people can so easily see the 'reverse product rule' within a reverse quotient rule...

$\\ \int { \frac { { x }^{ 2 }\cos ^{ 2 }{ x } -x\sin { x } -\cos { x } -1 }{ { \left( 1+x\sin { x } \right) }^{ 2 } } dx }$

What would be the "normal" way of doing this?

Drsoccerball · Sep 15, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
What would be the "normal" way of doing this?

Yeah thats the normal way

Paradoxica · Sep 15, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Yeah thats the normal way

So you literally have no choice but to stare at the question until you see reverse quotient rule?
...
Wow
...

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