Can someone explain this Riemann sum?

Can someone explain this Riemann sum?

"Evaluate the integral by computing the limit of riemann sums"

Interval: 0-1 (bottom to top, I dont know how to do the integral symbol)

2x dx

first thing I do is figure out the change in x, which is 1/n (1-0)/n

next thing, I figure out x sub i, which is i/n (dxi)

then my book has me write out the formula, bear with me

(1/n)(summation symbol i=1 with the upperbound as n)(2(i/n))

then, seemingly through magic, it becomes (2/n^2)(n(n+1))/2

I say magic, because it just pretends like its immediately obvious whats going on. I know that im supposed to use that summation formula, no problem understanding that, but I dont understand how it got to 2/n^2

Also, how do I know which summation formula to use?

Re: Can someone explain this Riemann sum?

$\displaystyle \sum_{i = 1}^{n} \frac{i}{n} = \frac{1}{n}+ \frac{2}{n} + \frac{3}{n} + ... + \frac{n-2}{n} + \frac{n-1}{n} + \frac{n}{n} $

if n is even add the last one with the first one then add the second first with the second last and so on

that will give you $\displaystyle \frac{n+1}{n} $ times $\displaystyle \frac{n}{2} $

$\displaystyle \frac{1}{n}+ \frac{2}{n} + \frac{3}{n} + ... + \frac{n-2}{n} + \frac{n-1}{n} + \frac{n}{n} = \frac{n+1}{n} \times \frac{n}{2} $

if n is odd can you think about how we can find this sum ?? it is the same why ?

you can try that with limited terms like

1+2+3+4+...+10 for example and see

or 1+2+...+11 try

Re: Can someone explain this Riemann sum?

hi the riemnann sum is $\displaystyle \text{Limit}\left[\frac{1}{n}\sum _{i=1}^{n-1} \frac{i}{n},n\to \infty \right]$

and the result it can be calculate as $\displaystyle \int_0^1 x \, dx$ equal to 0.5

chao