 .
The following graphs show the lower and upper sums, respectively:
So, now we need to compute the lower and upper sums. For the lower sums we add up the area of all of the blue rectangle. Each rectangle has width (since we have divided the interval into equal pieces) and its height is the value of the curve on the left side of the rectangle (since the curve is drawn so that the left corner rests on the curve). Therefore, we have
For the upper sums the width of each rectangle is still , but the height is now given by the function value on the right corner. So we have,
Now, we are given the identity
and so, following the example in Apostol,
Now, we multiply each term in the inequality by so the sums on the far left and far right become our and above, and we can simplify the term in the middle. Thus, for all we have
Now, we must show that if for all then . We accomplish this by showing that and both lead to contradictions.
Suppose . Then, since we can compute,
However, this cannot hold for all since the term on the right is a constant (depending on ), so we can always choose some that is larger since the positive integers are unbounded above.
Next, suppose . Then,
But, again, this cannot be true for all positive integers since the positive integers are unbounded above. Therefore, both and lead to contradictions. Hence, we must have
 .
The following graphs show the lower and upper sums, respectively:
As in part (a), we need to compute the lower and upper sums. For the lower sums we add up the area of all of the blue rectangle. Each rectangle has width (since we have divided the interval into equal pieces) and its height is the value of the curve on the left side of the rectangle (since the curve is drawn so that the left corner rests on the curve). Therefore, we have
For the upper sums the width of each rectangle is still , but the height is now given by the function value on the right corner. So we have,
Now, we are given the identity
and so, following the example in Apostol,
Now, we multiply each term in the inequality by so the sums on the far left and far right become our and above, and we can simplify the term in the middle. Thus, for all we have
Now, we must show that if for all then . We accomplish this by showing that and both lead to contradictions.
Suppose . Then, since we can compute,
However, this cannot hold for all since the term on the right is a constant (depending on ), so we can always choose some that is larger since the positive integers are unbounded above.
Next, suppose . Then,
But, again, this cannot be true for all positive integers since the positive integers are unbounded above. Therefore, both and lead to contradictions. Hence, we must have

The following graphs show the lower and upper sums, respectively:
This will be very similar to parts (a) and (b) except we have the factor of instead of 2 or 3. First, we need to compute the lower and upper sums. For the lower sums we add up the area of all of the blue rectangle. Each rectangle has width (since we have divided the interval into equal pieces) and its height is the value of the curve on the left side of the rectangle (since the curve is drawn so that the left corner rests on the curve). Therefore, we have
For the upper sums the width of each rectangle is still , but the height is now given by the function value on the right corner. So we have,
Now, we are given the identity
and so, following the example in Apostol,
Now, we multiply each term in the inequality by so the sums on the far left and far right become our and above, and we can simplify the term in the middle. Thus, for all we have
Now, we must show that if for all then . We accomplish this by showing that and both lead to contradictions.
Suppose . Then, since we can compute,
However, this cannot hold for all since the term on the right is a constant (depending on ), so we can always choose some that is larger since the positive integers are unbounded above.
Next, suppose . Then,
But, again, this cannot be true for all positive integers since the positive integers are unbounded above. Therefore, both and lead to contradictions. Hence, we must have
 .
The following graphs show the lower and upper sums, respectively:
Again, we need to compute the lower and upper sums. For the lower sums we add up the area of all of the blue rectangle. Each rectangle has width (since we have divided the interval into equal pieces) and its height is the value of the curve on the left side of the rectangle (since the curve is drawn so that the left corner rests on the curve). Therefore, we have
For the upper sums the width of each rectangle is still , but the height is now given by the function value on the right corner. So we have,
Now, we are given the identity
and so, following the example in Apostol,
Next, to get the inequalities we want (with ) it takes a bit more work than in parts (a) – (c) since we now have the term to deal with. First, we multiply each term in the inequality by to obtain
Now, we must show that if for all then . We accomplish this by showing that and both lead to contradictions.
Suppose . Then, since we can compute,
However, this cannot hold for all since the term on the right is a constant (depending on ), so we can always choose some that is larger since the positive integers are unbounded above.
Next, suppose . Then,
But, again, this cannot be true for all positive integers since the positive integers are unbounded above. Therefore, both and lead to contradictions. Hence, we must have
 .
Finally, we follow the same methods as in parts (a) – (d) to compute the lower and upper sums,
Then, similar to part (d), we have
Therefore,
Now, we must show that if for all then . We accomplish this by showing that and both lead to contradictions.
Suppose . Then, since we can compute,
However, this cannot hold for all since the term on the right is a constant (depending on ), so we can always choose some that is larger since the positive integers are unbounded above.
Next, suppose . Then,
But, again, this cannot be true for all positive integers since the positive integers are unbounded above. Therefore, both and lead to contradictions. Hence, we must have