# (revised+re-post)Upper and Lower sums & Riemann sums

1. Oct 29, 2009

### Call my name

http://img156.imageshack.us/i/17818455.jpg/
http://img215.imageshack.us/i/53355598.jpg/
http://img509.imageshack.us/i/11493310.jpg/

If you look at the above, I have underlined the problem that I am having.

So, my first question is, where are these inequalities coming from? If you do have other questions involving such approach, please show me.

My other question is from the explanation of Riemann sum, I do not understand the sign "llPll
" thing and I am having trouble understanding Riemann sum and how it really is related to the upper and lower sums.

Thank you for your attention.

Last edited by a moderator: Apr 24, 2017
2. Oct 29, 2009

### Call my name

Last edited by a moderator: Apr 24, 2017
3. Oct 29, 2009

### lanedance

you can write those reasonable easy with tex, click on it below

1)
$$x_{i-1} \leq \frac{x_{i-1} +x_{i}}{2} \leq x_{i}$$

comes pretty easy as by defintion of your partition
$$x_{i-1} \leq x_{i}$$
and probably more actually
$$x_{i-1} < x_{i}$$

split them into two equalities
$$x_{i-1} \leq \frac{x_{i-1} +x_{i}}{2}$$
$$\frac{x_{i-1} +x_{i}}{2} \leq x_{i}$$

multiplying everything by 2 and subtract something & it should be easy to see

2) same thing as before, and teh fact that in this case you know $x_i \geq 1$ , so $x_{i-1}^2 <x_i^2$

3) the last one is the definition of the integral as the limit of the sum when every partition appraches zero

4. Oct 29, 2009

### Call my name

I understand that these inequalities actually do 'work',

but what I do not understand is how you 'approach' these questions.

Like, I understand the concept of upper and lower sums, but how do I come up with the inequalities in the first place?

Do I need to first solve the integral by using the fundamental theorem, and then try to make it look that way by coming up with some inequalities?

How do I in the first place just go 'boom' and come up with the 2nd page's inequality?

Thanks for your attention and hopefully somebody will answer..