Modify Figure I.3 so that the ordinate at each point is instead of
.
- Show that the outer and inner sums are given by
- Use the inequalities
to show
for all
, and prove that
is the only number with this property.
- What number is in place of
if the ordinates at each point are given by
?
- Similar to the previous exercise we compute the sums
and
by summing up rectangles lying below and above the curve
, respectively. Each rectangle has width
since we are dividing the interval
into
equal segments. For
the height of each rectangle is the value of
on the left edge of the interval, and for
the height of each rectangle is the value of
on the right edge of the interval. Therefore, we have
and,
- Proof. Next, we start with the given inequality
and multiply each term by
to obtain
Therefore, for all
,
Now, to show that
is the only number that lies between
and
for all
, we suppose
is any such number, i.e.,
. We show that
and
both lead to contradictions, which means
.
Suppose
, then
But this cannot hold for all
since the positive integers are unbounded above.
On the other hand, suppose
, then
Again, this is a cannot hold for all positive integers
. Therefore, both
and
lead to contradictions, so we must have
- Using parts (a) and (b) of this exercise and part (e) of the previous exercise we conclude that the area
under the curve
is given by
Thank you very much. Your solutions help a lot.
Just a little detail: at part b., when analysing
,
equals
, not
.
Thanks for another one! Fixed.