Tough problem involving algebra

autodidude

1. The problem statement, all variables and given/known data
Original problem asked me to prove [tex]\int_a^b \! x \, \mathrm{d} x = \frac{b^2-a^2}{2}[/tex] 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

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


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

clamtrox

You can write that thing into
[tex] 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}
[/tex]
Now, can you evaluate the sums and then take the limits?

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 [tex]a(b-a)+(b-a)^2[/tex]?

HallsofIvy

No, it doesn't. What are [itex]\sum\frac{1}{n}[/itex] and [itex]\sum\frac{1}{n^2}[/itex]?

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

clamtrox

The first sum evaluates to 1 but the second does not. http://www.americanscientist.org/iss...-of-reckoning/ might help.

tt2348

consider this instead....[itex]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[/itex]
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 [itex]I_n[/itex] 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

autodidude

I'm quite sure the formula is [tex]\frac{n(n+1)}{2}[/tex]

I tried subbing that in and after some algebra, I've got this:

[tex]\frac{(b-a)^2}{2}+\frac{b^2-a^2}{2n}[/tex]

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)?

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.