More generally,
suppose
$f(x) > 0$ and
$f'(x) < 0$
(like for
$1/x^3$)
and you want to get
lower and upper estimates for
$\int_a^b f(x) dx$
divided into $n$ intervals.
The points are
$a+hi$ for
$i = 0$ to $n$,
where $h$ is the
width of the interval.
Since we want
$a+hn = b$,
$h = \frac{b-a}{n}
$.
The two sums we are interested in are
$S
=h\sum_{i=0}^{n-1} f(a+ih)
$
and
$T
=h\sum_{i=1}^{n} f(a+ih)
$.
$S$ is the sum from
the points at the
left side of each interval,
and
$T$ is the sum from
the points at the
right side of each interval,
For the case when
$f'(x) < 0$,
$hf(a+(i-1)h)
> \int_{a+(i-1)h}^{a+ih} f(x) dx
> hf(a+ih)
$,
with the inequalities reversed
if $f'(x) > 0$.
To see this,
note that the integral
is the area under the function,
and this is between
the areas of the rectangles
with base $h$
and hight of the function at the endpoints of the interval.
The total area
from $a$ to $b$
is
$I
=\int_a^b f(x) dx
=\sum_{i=1}^n \int_{a+(i-1)h}^{a+ih} f(x) dx
$.
Using the first inequality,
$I
<\sum_{i=1}^n hf(a+(i-1)h)
=h\sum_{i=1}^n f(a+(i-1)h)
=h\sum_{i=0}^{n-1} f(a+ih)
=S
$.
Using the second inequality,
$I
>\sum_{i=1}^n hf(a+ih)
=h\sum_{i=1}^{n} f(a+ih)
=T
$.
The difference between
these bounds is
$S-T
=h\sum_{i=0}^{n-1} f(a+ih)-h\sum_{i=1}^{n} f(a+ih)
=h(f(a)-f(b))
=\frac{b-a}{n}(f(a)-f(b))
$.
If you want this
to be less that $c$,
then
$\frac{b-a}{n}(f(a)-f(b))
< c
$
or
$n
> \frac{(b-a)(f(a)-f(b))}{c}
$.
If $f'(x) > 0$,
the same argument
with the signs reversed gives
$n
> \frac{(b-a)(f(b)-f(a))}{c}
$.
Therefore,
if $f'(x)$
is of constant sign,
$n
> \frac{(b-a)|f(b)-f(a)|}{c}
$
will work.
If $c = 1/m$
(a usual condition),
then
$n
> m(b-a)|f(b)-f(a)|
$.
Note that this is a
sufficient value of $n$ -
it will probably not be the
lowest possible.
A simpler estimate,
which is also sufficient,
is
$n
> m(b-a)|\max(f(b), f(a))|
$.
For your function,
$f(x) = 1/x^3$,
$a=1$, and $b=17$,
this gives
$n
>m(17-1)(1)
=16m
$.
If $m=100$,
$n > 1600$
will do.
Suppose that $f\colon[-1,1]\longrightarrow\mathbb R$ is defined by $f(x)=x^2$. Now, take $x_0=-1$, $x_1=-\frac13$, $x_2=\frac13$, and $x_3=1$. Then:
- in $[x_0,x_1]$, the minimum of $f$ is $\frac19$ and the maximum is $1$;
- in $[x_1,x_2]$, the minimum of $f$ is $0$ and the maximum is $\frac19$;
- in $[x_2,x_3]$, the minimum of $f$ is $\frac19$ and the maximum is $1$.
Therefore, if $P=\{x_0,x_1,x_2\}$, then the lower sum of $f$ with respect to $P$ is$$\frac23\times\frac19+\frac23\times0+\frac23\times\frac{1}{9}=\frac4{27}$$and the upper sum is$$\frac23\times1+\frac23\times\frac19+\frac23\times\frac19=\frac{38}{27}.$$
Best Answer
Let us first show that $\ln 2 \lt 1$. We have $\ln 2=\int_1^2\frac{dt}{t}$.
Divide the interval from $1$ to $2$ into $1$ part. The upper Riemann sum is the width of the interval, times the value of the function $\frac{1}{t}$ at $t=1$. This upper Riemann sum is $1$, and is clearly bigger than the integral.
Now let us show that $\ln 4 \gt 1$. We have $\ln 4=\int_1^4\frac{dt}{t}$. Divide the interval from $1$ to $4$ into $3$ equal parts, and find the corresponding lower Riemann sum. The minimum of $\frac{1}{t}$ on the first part is $\frac{1}{2}$. For the other two parts, the minima are $\frac{1}{3}$ and $\frac{1}{4}$. So the lower Riemann sum is $\ge \frac{1}{2}+\frac{1}{3}+\frac{1}{4}$. This is already $\gt 1$, so the integral is $\gt 1$.
We could alternately note that $\ln 4=2\ln 2$. Then we can divide the interval from $1$ to $2$ into $1$ part, and note that our function value is $\frac{1}{2}$ at the right endpoint, and $\frac{1}{t}$ is decreasing. So the lower Riemann sum is $\frac{1}{2}$, and therefore $\ln 2 \gt \frac{1}{2}$, so $2\ln 2 \gt 1$. But this is not quite in the spirit of the game, since we have used a property of $\ln$, namely $\ln 4=2\ln 2$. This would first have to be established from the definition of $\ln$ as an integral.
Remark: We were a little casual about concluding that the integral from $1$ to $2$ is less than $1$. In principle we only showed that it is $\le 1$. If you are very fussy, you can divide the interval $[1,2]$ into two equal parts. Then the upper sum is $\frac{1}{2}\left(\frac{1}{1}+\frac{1}{3/2}\right)\lt 1$.
The above calculations will only make full sense if one draws a picture of $y=\frac{1}{t}$, and visually identifies the upper and lower sums mentioned. They are all either the area of a rectangle, or the sum of the areas of a small number of rectangles.