Partitions and integrability

[Wernilou]

Hello, I'm back with a little bit of homework i really dont understand... I don't know how to even start or where to grab this

1) Be f(x)=sin(x) in [a,b] and Pₙ a partition in [a,b] that divides this interval in equal parts. Demonstrate that:



Where S(f,Pₙ) is the upper sum of f in Pₙ and I(f,Pₙ) is the lower sum of f in Pₙ

2) Use the integrability criterion to demonstrate that f(x) is integrable in [a,b]

I am quite confused and the teacher is really busy doing preparation courses, plus my classmates are equally confused with this.
Any help is accepted...Thanks a lot.

 

[Steven G]

You are told nothing about a bound for a and b?

Lets look at one of those (b-a)/n bases. Now draw the rectangle over this base. Depending upon where we are at the curve using the left endpoint as the height might give us an over estimate of the area and using the right endpoint will give us an underestimate OR I have that reversed. Since we want S(f,P) - I(f,P) which is always positive we can calculate this value for any base as [(b-a)/n] |f(left endpoint) - f(right endpoint)|

I haven't done analysis in over 30 years and I never had the pleasure of teaching it. Having said that I feel what I said above sounds good. Please work with it over each of the n intervals and see what you get. Please post back.

 

[Wernilou]

-This is quite confusing honestly...

 

[Steven G]

You need to know that [math]\left| \sin b - \sin a\right|\leq\left| b-a\right|[/math]
[math] Let \ \triangle x \dfrac{b-a}{n}[/math]
Now the n-subdivisions of a to b will be:[math] (a, a+ 1\triangle x), (a+1\triangle x, a+2\triangle x), (a+2\triangle x, a+3\triangle x), ... (a + (n-1)\triangle x, a+n\triangle x=b)[/math]
Now for each subdivision we want the max and min value of the inscribed rectangles. Since the sin function is either increasing or decreasing (what happens at the max and min points????) the max and mins will occur at the endpoints in each subdivision-one endpoint will be the max and the other endpoint will be the min. The problem is that we do not know in any given subdivision which endpoint will be the max and which endpoint will be the min. Now we luck out here since we want the difference between Sup(f, p) and Inf( f, p). We know that Sup (f, p) - Inf (f, p) = | Inf(f, p) - Sup (f, p)| so it does not matter if in any subdivision we get the subtraction backwards!

Now we get using the left endpoints: [math]\dfrac{b-a}{n}[ f(a) + f(a+1\triangle x) + f(a+2\triangle x) + ... + f(a+(n-1)\triangle x)][/math]
For the right endpoints we get [math]\dfrac{b-a}{n}[ f(a+1\triangle x) + f(a+2\triangle x) + ... + f(a+n\triangle x=b)][/math]
Try to finish up from here.