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I have the integral $$\int_1^3\sin(x)\,\mathrm dx$$ and I want to convert it to a Riemann Sum. I understand that the first thing I should do is set the limit as $n\to\infty$. I am stumped as what I should do next. Can someone walk me through the steps of converting integrals to Riemann Sums?

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  • $\begingroup$ Perhaps review this and the examples: khanacademy.org/math/ap-calculus-ab/… $\endgroup$ – Moo May 21 '18 at 12:45
  • $\begingroup$ Welcome to Maths SX! I suppose you mean for the integral from $1$ to $3$? $\endgroup$ – Bernard May 21 '18 at 12:46
  • $\begingroup$ @Bernard yes, I'm not quite familiar with mathematical formatting $\endgroup$ – kydd May 21 '18 at 12:54
  • $\begingroup$ Can you show us an attempt at least, so then we can help explain where you went wrong (or right)? $\endgroup$ – Andrew Li May 21 '18 at 12:59
  • $\begingroup$ I started out by finding delta x, which should be (3-1)/n, and that is the width of each rectangle. Then, I need to find the height of each rectangle, but I don't know how to do that. I think it's something like adding on to the first x value on my interval, but I'm not sure. $\endgroup$ – kydd May 21 '18 at 13:06
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You have to introduce the subdivision points. As the intercal is $[1,3]$, with length $2$, we obtain: $$x_0=1,\:x_1=1+\frac2n,\dots,\: x_i=1+i\cdot \frac 2n,\dots,\: x_n=1+n\cdot \frac 2n=3$$

Then, the Riemann sums associated with this subdivision are \begin{align} s_n&=\sum_{i=0}^{n-1}\sin x_i\,\cdot\frac2n=\frac2n\sum_{i=0}^{n-1}\sin\Bigl(1+\frac{2i}n\Bigr),\\[1ex] S_n&=\frac2n\sum_{i=1}^{n}\sin\Bigl(1+\frac{2i}n\Bigr). \end{align}

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