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# Calculate the area under higher power curves

Given the inequality

(valid for all integers ), generalize the results of Exercise I.4.2 to calculate for any integer .

Using the same method as in the previous two exercise, here and here, we first calculate the lower and upper sums by adding up the areas of the rectangles below and above the curve, respectively.

Then, starting with the given inequality

we multiply each term by to obtain

Following the same method as in the previous exercise we then obtain that if the ordinate at each is given by the area is given by

# Use the method of exhaustion to calculate the area under more general curves

Modify Figure I.3 so that the ordinate at each point is instead of .

1. Show that the outer and inner sums are given by

2. Use the inequalities

to show for all , and prove that is the only number with this property.

3. What number is in place of if the ordinates at each point are given by ?

1. 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,

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

3. 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

# Use the method of exhaustion to calculate the area under the following curves.

Calculate the areas under the following curves using the method of exhaustion:

1. ;
2. ;
3. ;
4. ;
5. .

1. .
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

2. .
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

3. 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

4. .
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

5. .

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