/sci/ - Science & Math

[Anonymous No.7308031]

Can anyone post a proof that taking the definite integral of a range is the same as doing a Riemann sum over that range as n goes to infinity?

I've been looking but can't find one. Thanks.

[Anonymous No.7308032]

>>7308031
here ya go

[Anonymous No.7308036]

>>7308031
Here ya go, buddy
https://www.youtube.com/watch?v=gFpHHTxsDkI

[Anonymous No.7308045]

>>7308036
that isn't general, that is just one specific function.

[Anonymous No.7308051]

>>7308031
I don't understand what you're asking.
What's your definition of an integral if it isn't the limit of a Riemann sum?

[Anonymous No.7308060]

>>7308051
The anti derivative of the function.

[Anonymous No.7308067]

is integration (Riemann integration at least) not defined that way?

[Anonymous No.7308070]

>>7308060
http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus

[Anonymous No.7308074]

Let <span class="math">\epsilon > 0[/tex] and [tex]f: [0,1] \to \mathbb R[/tex] be a continuous function.
Use uniform continuity of <span class="math">f[/spoiler] to find a <span class="math">\delta > 0[/spoiler] such that <span class="math">|x-y| < \delta \Rightarrow [f(x) - f(y)| < \epsilon[/tex].
Let N = 1+\lfloor 1/\delta \rfloor. For all n \ge N, we can write <div class="math">\left|\int_0^1 f(x)dx - \sum_{k=0}^{n-1} \frac{1}{n}f\left(\frac{k}{n}\right)\right| \le \sum_{k=0}^{n-1} \int_{k/n}^{(k+1)/n}|f(x) - f\left(\frac{k}{n}\right)| \le \epsilon</div>[/spoiler][/spoiler]

[Anonymous No.7308075]

Are you asking for a proof of the FTC?

If you don't know one why don't you just look it up? There are so many different versions with different levels of generality fitting varying levels of sophistication out there.

[Anonymous No.7308078]

Let <span class="math">\epsilon > 0[/spoiler] and <span class="math">f: [0,1] \to \mathbb R[/spoiler] be a continuous function.
Use uniform continuity of <span class="math">f[/spoiler] to find a <span class="math">\delta > 0[/spoiler] such that <span class="math">|x-y| < \delta \Rightarrow [f(x) - f(y)| < \epsilon[/tex].
Let <span class="math">N = 1+\lfloor 1/\delta \rfloor[/spoiler]. For all <span class="math"> n \ge N[/spoiler], we can write <div class="math">\left|\int_0^1 f(x)dx - \sum_{k=0}^{n-1} \frac{1}{n}f\left(\frac{k}{n}\right)\right| \le \sum_{k=0}^{n-1} \int_{k/n}^{(k+1)/n}|f(x) - f\left(\frac{k}{n}\right)| \le \epsilon</div>[/spoiler]

[Anonymous No.7308079]

>>7308060
kek

[Anonymous No.7308080]

>>7308074
>>7308078
:^(

[Anonymous No.7308081]

>>7308070
Thank you.

This was the section I was satisfied with

http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Proof_of_the_second_part

[Anonymous No.7308083]

>>7308081
There are better proofs out there. Literally every calculus book and analysis book would have a version, not counting the hundreds google will find for you.

[Anonymous No.7308086]

>>7308080
Yeah sorry, here, I hope this one works:
Let <span class="math">\epsilon > 0[/spoiler] and <span class="math">f: [0,1] \to \mathbb R[/spoiler] be a continuous function.
Use uniform continuity of <span class="math">f[/spoiler] to find a <span class="math">\delta > 0[/spoiler] such that <span class="math">|x-y| < \delta \Rightarrow [f(x) - f(y)| < \epsilon[/spoiler].
Let <span class="math">N = 1+\lfloor 1/\delta \rfloor[/spoiler]. For all <span class="math"> n \ge N[/spoiler], we can write <div class="math">\left|\int_0^1 f(x)dx - \sum_{k=0}^{n-1} \frac{1}{n}f\left(\frac{k}{n}\right)\right| \le \sum_{k=0}^{n-1} \int_{k/n}^{(k+1)/n}|f(x) - f\left(\frac{k}{n}\right)| \le \epsilon</div>

[Anonymous No.7308088]

>>7308083
link a better one? I'm not a good judge of these things

[Anonymous No.7308092]

>>7308086
>>7308086
Can we go ahead and talk about how bad /sci/ needs a TeX preview button?

[Anonymous No.7308104]

>>7308036
fail

[Anonymous No.7308115]

>>7308086
oh, okay, yeah, now I totally understand

*rolls eyes*

[Anonymous No.7308121]

>>7308115
If you don't understand that, then you have no business studying FTC. What he laid out is the most basic proof based pretty much only on the definition of the real numbers and continuous function.