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*}
This sum actually has a surprisingly nice upper and lower bound. It turns out
$$\frac{x(x+1)}{2} \le \sum_{n\le x}\sigma(n) \le x^2$$
You can verify this for yourself with a python script such as this one
Now let's discuss how I arrived at these bounds:
We can reframe this problem to make it a little simpler to think about. Instead of adding up $\sigma(i)$ for every $i$ from $1$ to $x$, we can instead try to find out how many times each factor(from $1$ to $x$) is counted, and use this to find our answer.
For example in the case of $x=9$ we know:
$1$ will be counted $9$ times (at $1,2,3,4,5,6,7,8,9$)
$2$ will be counted $4$ times (at $2,4,6,8$)
$3$ will be counted $3$ times (at $3,6,9$)
$4$ will be counted $2$ times (at $4,8$)
...
In general each factor $k$ is counted $\lfloor \frac{x}{k} \rfloor$ times.
This means we can rewrite our sum as
$$\sum_{k=1}^x k \cdot \lfloor \frac{x}{k} \rfloor$$
We can set $ \lfloor \frac{x}{k} \rfloor=1$ to get our lower bound of $\frac{x(x+1)}{2}$
And we can set $ \lfloor \frac{x}{k} \rfloor= \frac{x}{k}$ to get our upper bound of $x^2$
Best Answer
No, because$$\sum_{i=1}^n\frac1{1+i^3}\not=\int_1^n\frac1{1+x^3}\,\mathrm dx.$$However,$$\int_i^{i+1}\frac1{1+x^3}\,\mathrm dx\leqslant\frac1{1+i^3}\leqslant\int_{i-1}^i\frac1{1+x^3}\,\mathrm dx,$$and you can use this to solve your problem.