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I am trying to convert an integral to a Riemann sum like this:

$$ \int_a^b f(x) \,dx = \lim_{n\to \infty} \sum_{k=1}^n f(x_i)\Delta x $$

Where, $\Delta x = \frac{b-a}{n} $ and $x_i = a + i \Delta x$.

My attempt:

$$ \int_1^{n+1} f(x)\, dx = \lim_{n\to \infty} \sum_{k=1}^n f(1+k)$$

Since $\Delta x = 1$ and $ x_i = 1+i$

I do believe this is wrong though. How do I take into account the upper bound $n+1$?

Thank you.

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    $\begingroup$ This is not right. Since you have an $n$ that is on the left-hand side, you can't take $\lim_{n \rightarrow \infty}$. You'll have to use a different variable for the limit. $\endgroup$ – Jair Taylor May 20 '18 at 19:28
  • $\begingroup$ So if I change n to p on the right hand side it is ok? My hope was that since there is a "n+1" the expression on the right hand side would be something different. $\endgroup$ – What Gives May 20 '18 at 20:02
  • $\begingroup$ There's no need to replace b with n+1. It should have been left as b. The summation in the final line is also incorrect; $\Delta x$ and n change as n changes. $\endgroup$ – NicNic8 May 20 '18 at 20:03
  • $\begingroup$ @NicNic8 The integral that I am working with looks like that, it is not something I have constructed on my own. $\endgroup$ – What Gives May 20 '18 at 20:04
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    $\begingroup$ @WhatGives No, that's not enough. Your computation for $\Delta x$ will also change. If you are using $p$ for the number of subintervals then your $\Delta x$ would become $(b-a)/p$. $\endgroup$ – Jair Taylor May 20 '18 at 20:07
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In this case you've got $\Delta x = \frac{n}{k}$ and $x_i = 1 + \frac{in}{k}$ so that $$\int_1^{n+1} f(x) \, dx = \lim_{k\to\infty} \sum_{i=1}^k f(1 + \frac{in}{k})\frac{n}{k}.$$

Note that $n$ is a constant in this example, so you need to choose another letter, say $k$, to represent the variable that determines the number of rectangles in your approximation. Then the variable ``i'' is the `dummy variable' that tells which rectangle to refer to when computing the sum.

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  • $\begingroup$ Thank you for the answer but this was really not the result I was expecting. Is it even possible to expand this for a few terms? Can the constant n be anything? If k-> infinity it would just seems like the limit would be 0 since the last term would be *n/inf? $\endgroup$ – What Gives May 20 '18 at 21:35
  • $\begingroup$ @WhatGives the number of terms of the form $f(x) $ become greater. Think about the average, defined as $\frac{1}{n}\sum_{k=1}^n f(k)$. As $n \to \infty$ averages don't always tend to zero. In this case case we are just taking values of $f$ at $1+\epsilon$ for small epsilon. $\endgroup$ – Brevan Ellefsen May 20 '18 at 21:39
  • $\begingroup$ It is really hard for me to understand this, but thank you for the effort. Maybe I just need to sit with it for a while. $\endgroup$ – What Gives May 20 '18 at 22:10
  • $\begingroup$ @WhatGives I challenge you to take $f(x) = x$, $n =2$, and $k = 4$ (forget about the limit for the moment). Then I would ask you to explicitly write down each term of the Riemann sum. This may help you a little bit. $\endgroup$ – treble May 21 '18 at 0:33

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