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

  • $\begingroup$ 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

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 $2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.