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Calculate the upper sums $U_n$ and lower sums $L_n$, on a regular partition of the intervals, for the following integral: $$\int_1^2 \lfloor x\rfloor dx$$

$$\Delta x=\frac{1}{n}$$

And then I'm unsure as where to go as I have gotten this far

$$\Sigma=\lfloor x_i \rfloor \Delta x$$ $$\Sigma=\left\lfloor 1+\frac{I}{n} \right\rfloor \frac{1}{n}$$

Does the floor function work as a it does for a normal function? Can someone please help

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  • $\begingroup$ The floor function is almost constant on this interval. $\lfloor x \rfloor = 1$ for $1 \le x < 2$, while $\lfloor 2 \rfloor = 2$. $\endgroup$ – Paul Sinclair Apr 18 '16 at 14:09
  • $\begingroup$ would you be able to show me some workings of how you would work through this question as I understand the floor function but am confused on the workings to find the answer $\endgroup$ – jarrod Apr 20 '16 at 7:45
  • $\begingroup$ more specifically the upper sum $\endgroup$ – jarrod Apr 21 '16 at 0:20
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Sorry to take so long. I've been busy. On the interval in question, $$\lfloor x \rfloor = \begin{cases}1, & 1 \le x < 2\\2, & x = 2\end{cases}$$

Given $n$, the intervals are $\left[1 + \frac {k-1}n,1 + \frac{k}n\right], k = 1, \ldots, n$. So the sums are $$L_n = \sum_{k=1}^{n}\left\lfloor 1 + \frac {k-1}n \right\rfloor\frac1n = \left(\sum_{k=1}^{n}1\right)\frac1n = 1\\U_n= \sum_{k=1}^{n}\left\lfloor 1 + \frac {k}n \right\rfloor\frac1n = \left(\sum_{k=1}^{n-1}1 + 2\right)\frac1n = (n + 1)\frac1n = 1 + \frac1n$$

For the upper sum, the values of $k$ between $1$ and $n-1$ give values inside the floor function that are strictly less than $2$, so the floor function gives a value of $1$. But when $k = n$, the value inside the floor function is $2$, so the floor function gives a value of $2$, so the final numerator in the sum is $2$.

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