# How to do a Riemann Sum

I really don't understand how to "do" a Riemann sum. All of Newton's notation is really annoying. I think if someone can do a good worked example for me without the jargon that my textbook gives me, then that would be great. They don't look difficult I just can't understand the notation I think.

So an example of one would be the sum of $\frac{1}{n+1} +\frac{1}{n+2}... +\frac{1}{n+n}$

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What textbook are you using? What jargon do you want to avoid? – Aryabhata Mar 10 '12 at 20:51
James Stewart 6E Calculus. Like the whole xi* stuff and x+x2. Just make sure you show everything I guess and don't assume much knowledge from me pretty much. – Tessa Danger Bamkin Mar 10 '12 at 21:02
It's very hard for me to figure out what it is you are asking or what you are having trouble with; most of the time, one doesn't actually compute Riemann sums, but instead translates the problem into a problem of integration (which can be done by methods other than taking limits of sums). What do you mean by '"do" a Riemann sum"? Setting it up in an application problem? Actually computing it? Can you post one of the problems you are having difficulty with, quoted verbatim, and explain your difficulties with that in mind? – Arturo Magidin Mar 10 '12 at 22:32
Do you want to evaluate a definite integral by expressing it as a limit of a Riemann sum or to express a limit as a definite integral? – Américo Tavares Mar 10 '12 at 23:03
I think i want to know how to evaluate a definite untegral as a limit of a riemann sum. – Tessa Danger Bamkin Mar 11 '12 at 19:29

Let's start with $${1\over n+1}+{1\over n+2}+\cdots+{1\over n+n}={1\over n}\left({n\over n+1}+{n\over n+2}+\cdots+{n\over n+n}\right)$$ The right side can be written as $${1\over n}\left({1\over1+(1/n)}+{1\over1+(2/n)}+\cdots+{1\over1+(n/n)}\right)$$ Now let $f(x)=1/(1+x)$. Then the sum is $$(1/n)(f(1/n)+f(2/n)+\cdots+f(n/n))$$ So you are evaluating the function $f$ at $n$ evenly-spaced points between zero and one, namely, at $1/n,2/n,\dots,n/n$, and then taking the average of all those function values by adding them up and dividing by $n$. That means that the sum you started with is a Riemann sum for the integral of that function $f$ between zero and one: $$\int_0^1{1\over1+x}\,dx$$
"So you are evaluating the function $f$ at $n$ evenly-spaced points between zero and one...." – Gerry Myerson Mar 11 '12 at 23:43
It doesn't have to be zero and one, but look at the places where we are evaluating $f$: $1/n$, $2/n$, ..., $n/n$. These are all between zero and one, and as $n$ increases they get as close as you'd like to every point between zero and one. A different sum might lead to different places, and different limits on the integral. – Gerry Myerson Mar 12 '12 at 22:07