# Tough problem involving algebra

1. Jul 19, 2012

### autodidude

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!

2. Jul 19, 2012

### clamtrox

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?

3. Jul 24, 2012

### 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 $$a(b-a)+(b-a)^2$$?

4. Jul 24, 2012

### HallsofIvy

Staff Emeritus
No, it doesn't. What are $\sum\frac{1}{n}$ and $\sum\frac{1}{n^2}$?

5. Jul 24, 2012

### autodidude

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

6. Jul 24, 2012

### clamtrox

7. Jul 24, 2012

### tt2348

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

8. Jul 25, 2012

### 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?

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)?

9. Jul 25, 2012

### clamtrox

Yeah except you of course should have b2-a2, not (b-a)2. Then just take limit n→∞.

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.