HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

[Drsoccerball]

Re: MX2 2015 Integration Marathon

But why is the area under sin(x) between 0 and pi/2 the same as that between pi/2 and pi?

This transformation is exactly how you would prove this rigorously (*).


(*) At least as rigorously as possible in high school. A more subtle point here is that area is not really rigorously defined in high school. Before getting to more advanced notions of volume in measure theory, we typically define the area only for certain regions. (Such as those bounded by the x axis, two vertical lines, and the graph of a positive continuous function f(x). For such regions we would literally DEFINE area as being the definite integral of this function between the two vertical lines.)

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[seanieg89]

Re: MX2 2015 Integration Marathon

Bruh its a sin graph we know that its the same area...

This is a VERY bad attitude to have in maths, there are so many counterintuitive facts and so many plausible statements that are decidedly false. You should always ask yourself the question "why?" about statements you make, and you should be able to answer this question. That is precisely what a proof is.

 

[seanieg89]

Re: MX2 2015 Integration Marathon

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Sure, but we are not integrating sin, we are integrating the log of sin. It is logical and indeed true that this does not change anything, but as we cannot explicitly compute these two integrals by antidifferentiation, a proof has to exploit properties of the integral to show WHY taking the log does not change things.

 

[Librah]

Re: MX2 2015 Integration Marathon

Why couldn't math just be made simpler, i mean like, i'm trying to show this

Dunno what mathematicians see, but that looks pretty obvious enough to me

 

[kawaiipotato]

Re: MX2 2015 Integration Marathon

Why couldn't math just be made simpler, i mean like, i'm trying to show this

Dunno what mathematicians see, but that looks pretty obvious enough to me

Is there a formal proof for that??
If 'c' is a constant independent of x, then you can just bring it out of the integral

 

[Drsoccerball]

Re: MX2 2015 Integration Marathon

Why couldn't math just be made simpler, i mean like, i'm trying to show this

Dunno what mathematicians see, but that looks pretty obvious enough to me

Just integrate both and show it

 

[kawaiipotato]

Re: MX2 2015 Integration Marathon

Why couldn't math just be made simpler, i mean like, i'm trying to show this

Dunno what mathematicians see, but that looks pretty obvious enough to me

Is there a formal proof for that??
If 'c' is a constant independent of x, then you can just bring it out of the integral

edit: https://en.wikipedia.org/wiki/Constant_factor_rule_in_integration

 

[Librah]

Re: MX2 2015 Integration Marathon

It's a question from my MATH1903 assignment, had to use epsilon delta continuity arguments, limits and the Riemann integrability definition.

 

[Librah]

Re: MX2 2015 Integration Marathon

Is there a formal proof for that??
If 'c' is a constant independent of x, then you can just bring it out of the integral

edit: https://en.wikipedia.org/wiki/Constant_factor_rule_in_integration

Yes it's a constant.

 

[Carrotsticks]

Re: MX2 2015 Integration Marathon

Why couldn't math just be made simpler, i mean like, i'm trying to show this

Dunno what mathematicians see, but that looks pretty obvious enough to me

It doesn't seem obvious at all to me.

For the sake of simplicity let's presume f is continuous and positive within the domain provided.

Now your LHS multiplies each function value by a factor of C. By doing so, we acquire a new curve and the area under the new curve may increase/decrease by some factor too, say D.

Is it really that obvious that C=D? In other words, that the two factors are the same? Because that is not obvious at all to me.

There are a lot of 'obvious' things that are not always true, and your intuition often can betray you.

For example, Gabriel's horn is a solid with finite volume, so obviously it also has finite surface area.

There are obviously more rational numbers than integers.

There are obviously fewer numbers in the unit interval, than the entire number line.

If a function has a horizontal asymptote, then obviously the limit of the derivative, as x approaches infinity, is 0.

Before we continue further, if you wish to do so, can you first tell me if you see where I am getting at?

 

[Carrotsticks]

Re: MX2 2015 Integration Marathon

Just integrate both and show it

What do you mean 'just integrate both' and 'show it'?

Show what?

Integrate what?

Please provide more specific arguments, not just wish washy vague claims.

 

[kawaiipotato]

Re: MX2 2015 Integration Marathon

If a function has a horizontal asymptote, then obviously the limit of the derivative, as x approaches infinity, is 0.

Is there a case when it's not?

 

[Carrotsticks]

Re: MX2 2015 Integration Marathon

Is there a case when it's not?

Plenty. A classic example is the below, where the limit of the derivative does not exist, but it most certainly has a horizontal asymptote.

 

[Librah]

Re: MX2 2015 Integration Marathon

It doesn't seem obvious at all to me.

For the sake of simplicity let's presume f is continuous and positive within the domain provided.

Now your LHS multiplies each function value by a factor of C. By doing so, we acquire a new curve and the area under the new curve may increase/decrease by some factor too, say D.

Is it really that obvious that C=D? In other words, that the two factors are the same? Because that is not obvious at all to me.

There are a lot of 'obvious' things that are not always true, and your intuition often can betray you.

For example, Gabriel's horn is a solid with finite volume, so obviously it also has finite surface area.

There are obviously more rational numbers than integers.

There are obviously fewer numbers in the unit interval, than the entire number line.

If a function has a horizontal asymptote, then obviously the limit of the derivative, as x approaches infinity, is 0.

Before we continue further, if you wish to do so, can you first tell me if you see where I am getting at?

Yeah i can see the problem with the integrals now. It just seemed intuitive that you can just multiply into the integral since you've done that without any thought in HS math.

 

[leehuan]

Re: MX2 2015 Integration Marathon

It doesn't seem obvious at all to me.

For the sake of simplicity let's presume f is continuous and positive within the domain provided.

Now your LHS multiplies each function value by a factor of C. By doing so, we acquire a new curve and the area under the new curve may increase/decrease by some factor too, say D.

Is it really that obvious that C=D? In other words, that the two factors are the same? Because that is not obvious at all to me.

There are a lot of 'obvious' things that are not always true, and your intuition often can betray you.

For example, Gabriel's horn is a solid with finite volume, so obviously it also has finite surface area.

There are obviously more rational numbers than integers.

There are obviously fewer numbers in the unit interval, than the entire number line.

If a function has a horizontal asymptote, then obviously the limit of the derivative, as x approaches infinity, is 0.

Before we continue further, if you wish to do so, can you first tell me if you see where I am getting at?

Suddenly, I feel scared to do more maths at uni.

 

[Librah]

Re: MX2 2015 Integration Marathon

Suddenly, I feel scared to do more maths at uni.

Here's gabriels horn from my tute if your interested.

 

[leehuan]

Re: MX2 2015 Integration Marathon

Here's gabriels horn from my tute if your interested.

Well, I'm always interested in harder maths. But I get scared at the same time looking at parts (b) and (c)

 

[Librah]

Re: MX2 2015 Integration Marathon

Well, I'm always interested in harder maths. But I get scared at the same time looking at parts (b) and (c)

You need a surface area formula for integrals for part b and comparison test, that's easily googleable though.

 

[Drsoccerball]

Re: MX2 2015 Integration Marathon

Im thinking that the integral to find the surface area becomes something like 1/x-1 and therefore infinite?

 

[seanieg89]

Re: MX2 2015 Integration Marathon

Yeah i can see the problem with the integrals now. It just seemed intuitive that you can just multiply into the integral since you've done that without any thought in HS math.

That's because when you "learn" integrals in HS you do not do so rigorously.

You are just shown a symbol for it, given a loose handwavy definition in terms of approximation by a finite number of rectangles (for what kinds of functions does this thing converge to a limit? what does it even mean to take a limit? what even is a real number? *), and taught how to calculate some of them using a few simple rules that are given to you but not proven.

There are several different notions of integration, all with slightly different scope / properties and all based on slightly different foundations. Things like being able to bring the constant out are properties we would hope any notion of integration would satisfy, but you would still need to prove this rigorously for each particular notion of integration, and the proof would be different depending on how that integral is even defined.

(*) This may seem like a silly question to ask, but a rigorous definition of real numbers is actually remarkably subtle and pretty much every fact in calculus hinges on this definition. How would you even explain what a real number is to someone who had never heard of them before and had only ever worked with rational numbers? What do you need to add to the rationals to get the reals? (And answering "the irrationals" is very circular lol.)