# Thread: Interesting (reverse?) Riemann Sum Problem

1. ## Interesting (reverse?) Riemann Sum Problem

Hi,

I'm working on this question and having trouble coming up with a solution, mainly due to the cos. I have no idea how to type sums in here so I am just going to post two screenshots of the original problem.

Basically all I have to do is solve the Reimann sum that is already provided, but it's way too complicated to work out so I have to convert it back to the integral then solve the integral. I'm having trouble finding what $f$ should be in the original integral. If I set it as cos(t) I almost get the right Riemann sum but there is still a (4pi)/(21) leftover from the delta x part of the Riemann sum, and I don't know how to change this because it's all doomed to be "trapped" inside the cos.

Help!

Thank you!

2. Originally Posted by metaname90
Hi,

I'm working on this question and having trouble coming up with a solution, mainly due to the cos. I have no idea how to type sums in here so I am just going to post two screenshots of the original problem.

Basically all I have to do is solve the Reimann sum that is already provided, but it's way too complicated to work out so I have to convert it back to the integral then solve the integral. I'm having trouble finding what $f$ should be in the original integral. If I set it as cos(t) I almost get the right Riemann sum but there is still a (4pi)/(21) leftover from the delta x part of the Riemann sum, and I don't know how to change this because it's all doomed to be "trapped" inside the cos.

Help!

Thank you!

Well, put $\displaystyle{f(x):=\cos x\Longrightarrow \frac{1}{n}\sum\limits^n_{k=1}\cos\left(\frac{4k\p i}{21n}\right)}$ is just the Riemann

sum of the function in the interval $\displaystyle{[0,\frac{4\pi}{21}]$ partitioned in n equal subintervals and taking the rightmost

end of each subinterval to evaluate the function on it (why can we do this?)

Tonio

3. Alright, so that's what I originally tried, and you get sin(4pi/21) as the solution (integral on that interval of cos(x)), but that is the wrong solution. And when I actually work it out I dont' see how cos(x) can be f(x), or entirely, there must be something else. Here is my working below. The part I was talking about struggling with was that leftover 4pi/21:

4. Originally Posted by metaname90
Alright, so that's what I originally tried, and you get sin(4pi/21) as the solution (integral on that interval of cos(x)), but that is the wrong solution. And when I actually work it out I dont' see how cos(x) can be f(x), or entirely, there must be something else. Here is my working below. The part I was talking about struggling with was that leftover 4pi/21:

Ok, that looks right...now just take the constant factor $\displaystyle{\frac{4\pi}{21}}$ out of the sum

and of the limit and what's left is the integral of cosine over the given interval.

Tonio

### riemann sum backward

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