Find the upper and lower sum of an integral with a floor I'm having some trouble and looking for some help with a problem i'm trying to solve. Without the floor function it would be easy but the floor has made it a bit trickier:
Find the upper and lower sum: 
$$
\int_1^2 \lfloor x+1 \rfloor dx
$$
As a start i can say $$\Delta x=1/n$$  and $$x_i=1+i/n$$
$$\\$$ $$\\$$ 
EDIT: Thanks for the comments, although I think I might have explained it poorly! I am trying to find the upper and lower bound equation, over a regular partition of the intervals: $$U_N\quad and\quad L_N$$ 
So for example the upper bound equation for: $$\int_1^2 f(x) \quad where \ f(x)=x$$ $$\Delta x =(b-a)/n=1/n \quad and \quad x_i=1+i/n$$ $$Hence \quad U_L =\Delta x \sum_{i=1}^n f(x_i)$$ $$=1/n[\sum_{i=1}^n 1 + i/n]=\frac1n[n+\frac1n(\frac{n^2}{2}+\frac{n}{2})]$$ $$=\frac1n[n+\frac{n}{2}+\frac12]=\frac32+\frac{1}{2n}$$ 
$$Thus:\quad U_L= \frac32+\frac{1}{2n}$$
And that would be the upper limit, the lower is just when X_i = 1+(i-1)/n.
My trouble is doing all this for a floor function.
Thanks for your help!
 A: 
We consider the function 
  \begin{align*}
&f:\mathbb{R}\rightarrow\mathbb{R}\\
&f(x)=\lfloor x+1\rfloor
\end{align*}
  and the definite integral
  \begin{align*}
\int_1^2 f(x) dx=\int_1^2 \lfloor x+1\rfloor    dx
\end{align*}



To calculate lower and upper sums we take  a partition  $\mathcal{P}=\{x_0,x_1,\ldots,x_n\}$ of the interval $[1,2]$ with 
  $$1=x_0<x_1<\ldots<x_{n-1}<x_n=2$$
  The lower sum $L(\mathcal{P})$ and upper sum $U(\mathcal{P})$   of $f$ with respect to the partition $\mathcal{P}$ are  given as
  \begin{align*}
L(\mathcal{P})&=\sum_{k=1}^nf(x_{k-1})(x_k-x_{k-1})=\sum_{k=1}^n\lfloor x_{k-1}+1\rfloor(x_k-x_{k-1})\\
U(\mathcal{P})&=\sum_{k=1}^nf(x_k)(x_k-x_{k-1})=\sum_{k=1}^n\lfloor x_k+1\rfloor(x_k-x_{k-1})\\
\end{align*}

We know the floor function  $\lfloor x\rfloor$ fulfills
$$\lfloor x\rfloor \leq x < \lfloor x\rfloor+1$$
with points of discontinuity at the integers. Therefore
\begin{align*}
f(x)=\lfloor x+1\rfloor=
\begin{cases}
2&\qquad 1\leq x <2\\
3&\qquad x=2
\end{cases}
\end{align*}
It's often convenient to represent the floor function by Iverson brackets
\begin{align*}
\lfloor x\rfloor=\sum_{j\geq 0}[1\leq j \leq x]
\end{align*}
This way we get rid of the floor symbols $\lfloor$ and $\rfloor$ and can manipulate sums instead.

Let's calculate the lower sum $L(\mathcal{P})$:
  \begin{align*}
L(\mathcal{P})&=\sum_{k=1}^n\lfloor x_{k-1}+1\rfloor(x_k-x_{k-1})\\
&=\sum_{k=1}^n\sum_{j\geq 0}[1 \leq j \leq x_{k-1}+1](x_k-x_{k-1})\tag{1}\\
&=\sum_{k=1}^n2(x_k-x_{k-1})\tag{2}\\
&=2x_n-2x_0\tag{3}\\
&=2
\end{align*}

Comment:


*

*In (1) we observe, that $j$ takes always the values $1$ and $2$, since $1\leq x_{k-1}<2$ for $k=1,\ldots,n$ and so the inner sum is $2$.

*In (2) we do a little telescoping, leaving only elements $x_0$ and $x_n$.

*In (3) we only have to cope with the endpoints of the interval $[1,2]$.

And now the upper sum $U(\mathcal{P})$:
  \begin{align*}
U(\mathcal{P})&=\sum_{k=1}^n\lfloor x_{k}+1\rfloor(x_k-x_{k-1})\\
&=\sum_{k=1}^n\sum_{j\geq 0}[1 \leq j \leq x_{k}+1](x_k-x_{k-1})\\
&=\sum_{k=1}^{n-1}\sum_{j\geq 0}[1 \leq j \leq x_{k}+1](x_k-x_{k-1})\\
&\qquad+\sum_{j\geq 0}[1 \leq j \leq x_{n}+1](x_n-x_{n-1})\tag{4}\\
&=\sum_{k=1}^{n-1}2(x_k-x_{k-1})+3(x_n-x_{n-1})\\
&=2x_{n-1}-2x_0+3x_n-3x_{n-1}\\
&=3x_n-x_{n-1}-2x_0\\
&=4-x_{n-1}
\end{align*}

Comment:


*

*In (4) we split the last summand with $k=n$, since the inner sum consists of three summands in that case.



Note: Since the mesh of the partition $\displaystyle{\max_{1\leq k \leq n}}|x_{k}-x_{k-1}|$ tends to zero with growing $n$, we see that
$$\lim_{n\rightarrow \infty}x_{n-1}=x_n=2$$
and therefore the upper sums tend to
$$\lim_{n\rightarrow\infty}(4-x_{n-1})=2$$
the same value as the lower sums.
Note: Since the value of an integral don't change, when we change the function value of $f$ at finite many points, we obtain
\begin{align*}
\int_1^2 f(x) dx=\int_1^2 \lfloor x+1\rfloor    dx=2\int_1^2 dx=\left. 2x \right|_1^2=2
\end{align*}
