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
- 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