[Math] Find the upper and lower sum of an integral with a floor

calculusceiling-and-floor-functionsintegrationsummation

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!

Best Answer

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*}

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