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## Tough problem involving algebra

1. The problem statement, all variables and given/known data
Original problem asked me to prove $$\int_a^b \! x \, \mathrm{d} x = \frac{b^2-a^2}{2}$$ using Riemann sums. I've already seen a simpler formula using the left side of the rectangles but I'm curious as to how you would manipulate the formula below by hand to get the answer

2. Relevant equations

$$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}[\frac{(b-a)i}{n}+a][\frac{b-a}{n}]$$

3. The attempt at a solution
pages and pages of algebra leading nowhere!

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You can write that thing into $$I = \lim_{n\rightarrow\infty} \sum_{i=1}^{n} [\frac{(b-a)i}{n}+a](\frac{b-a}{n}) = a(b-a) \lim_{n\rightarrow\infty} \sum_{i=1}^{n} \frac{1}{n} + (b-a)^2 \lim_{n\rightarrow\infty} \sum_{i=1}^{n} \frac{i}{n^2}$$ Now, can you evaluate the sums and then take the limits?

Could you please explain how you got that?   I'm not exactly sure how to evaluate the last limit/sum but does everything else evaluate to just $$a(b-a)+(b-a)^2$$?

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## Tough problem involving algebra

Quote by autodidude 					 				 				Could you please explain how you got that?   I'm not exactly sure how to evaluate the last limit/sum but does everything else evaluate to just $$a(b-a)+(b-a)^2$$?
No, it doesn't. What are $\sum\frac{1}{n}$ and $\sum\frac{1}{n^2}$?

I don't know - the only way I know how to interpret the first one right now is to divide 1 into n parts but then I'm not sure what to do so it becomes an infinitely small number if n->infinity. I guess it would be the same for the second one but it gets smaller faster

Quote by autodidude 					 				 				Could you please explain how you got that?   I'm not exactly sure how to evaluate the last limit/sum but does everything else evaluate to just $$a(b-a)+(b-a)^2$$?
The first sum evaluates to 1 but the second does not. http://www.americanscientist.org/iss...-of-reckoning/ might help.

consider this instead....$I_{n} = \sum_{i=0}^{n} [\frac{(b-a)i}{n}+a](\frac{b-a}{n}) = \frac{a(b-a)}{n} \sum_{i=0}^{n} 1 + \frac{(b-a)^2}{n^2} \sum_{i=0}^{n} i$ now we know adding 1 n times... is n, how about adding numbers from 1 to n? know any relevant formulas for that? once you simplify the expression to only relying on n, take the limit of $I_n$ as n goes to infinity you have to remember, your sums arent over n, they are over i, so n can be take out front

I'm quite sure the formula is $$\frac{n(n+1)}{2}$$  I tried subbing that in and after some algebra, I've got this:  $$\frac{(b-a)^2}{2}+\frac{b^2-a^2}{2n}$$  Am I on the right track?  Also, when you write the sigmas out without the limits, does n just become some arbitrary integer? And is there any difference between writing a sigma with infinity on top and writing sigma with an n but with lim(h->infinity)?

Quote by autodidude 					 				 				I'm quite sure the formula is $$\frac{n(n+1)}{2}$$  I tried subbing that in and after some algebra, I've got this:  $$\frac{(b-a)^2}{2}+\frac{b^2-a^2}{2n}$$  Am I on the right track?
Yeah except you of course should have b2-a2, not (b-a)2. Then just take limit n→∞.

Quote by autodidude 					 				 				Also, when you write the sigmas out without the limits, does n just become some arbitrary integer? And is there any difference between writing a sigma with infinity on top and writing sigma with an n but with lim(h->infinity)?
Infinite sums are always defined as that limit so there's no worries there. You want to find the n:th partial sum, and then take the limit.

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