Calculus Of One Real Variable By Pheng Kim Ving 
9.2 
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1. Partitions 
Let's partition or subdivide the interval [a, b] in Fig. 1.1 into 5 subintervals using 6
points x_{0}, x_{1}, x_{2}, x_{3}, x_{4}, and x_{5} where
x_{0} = a, x_{5} = b, and x_{0} < x_{1} < x_{2} < x_{3} < x_{4} < x_{5}. We say that we've made a partition
of order 5 of the interval [a, b]. This
partition is determined by selecting points x_{0} at a, x_{5} at b,
and the remaining points such that x_{0} < x_{1} < x_{2} < x_{3} < x_{4} < x_{5}.
So we say that this partition is the ordered set {x_{0}, x_{1}, x_{2}, x_{3}, x_{4}, x_{5}}.
Fig. 1.1 A partition of [a, b]. 
A partition P of order n of the interval [a, b] is an ordered set P = {x_{0}, x_{1}, x_{2}, ..., x_{n}} such that:
a = x_{0} < x_{1} < x_{2} < ... < x_{n} = b,
Fig. 1.2 A Regular Partition Of Order 5 Of 
There are n
left endpoints (because there are n
subintervals); their subscripts run from 0 to n 1. There are also n
right endpoints (again because there are n
subintervals); their subscripts however run from 1 to n. The right endpoints
of the 1st, 2nd, ..., (n
1)th subintervals are the left endpoints of the 2nd, 3rd, ..., nth subintervals respectively.
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2. Areas 
We wish to compute the area of the plane region under the graph
of a function and over a closed finite interval, the
function being continuous and nonnegative on the interval. We take an example.
Let's find the area A of the plane region
bounded by the line y = f(x)
= (1/2)x + 1, the xaxis,
the vertical line x =
1, and the vertical line x
= 3. The region is a
trapezoid, colored in Fig. 2.1.
Using geometry, the area is A = (1/2)(b_{1} + b_{2})h,
where b_{1} and b_{2} are the bases and h
is the height. Now, b_{1} = f(1)
=
(1/2)(1) + 1 = 1/2, b_{2 }= f(3)
= (1/2)(3) + 1 = 5/2, and h = 3 (1) = 4. So A
= (1/2)(1/2 + 5/2)(4) = 6 square units.
Fig. 2.1
Area A Of Trapezoid.

Using calculus, we proceed as follows.
Fig. 2.2
Using Regular Partition Of Order 4 And
Left Endpoints.

First let's set up a regular partition of order 4 of the
interval [1, 3]. See Fig. 2.2. The length of each subinterval is (3
(1))/4 = 1. The 4 left endpoints of the
subintervals are:
Fig. 2.3
Using Regular Partition Of Order 4 And
Right Endpoints.

Next let's use the same regular partition of order 4 but
this time let's employ the right endpoints of the subintervals. The
length of each subinterval is (3 (1))/4 = 1. See Fig. 2.3. The 4 right
endpoints of the subintervals are:
x_{1} = 0, x_{2} = 1, x_{3} = 2, and x_{4} = 3,
Refer to Figs. 2.2 and 2.3 as guides. For any positive integer n we have:

Using Regular Partition Of Order 6 And Left Endpoints
Fig. 2.4
Using Regular Partition Of Order 6 And
Left Endpoints.

Now let's use a regular partition of a greater order, say 6,
and the left endpoints. See Fig. 2.4. The length of each
subinterval is (3 (1))/6 = 2/3. The
7 left endpoints of the subintervals are:
Using Regular Partition Of Order 6 And Right Endpoints
Fig. 2.5
Using Regular Partition Of Order 6 And
Right Endpoints.

Next let's use a regular partition of the same order 6 but
this time let's employ the right endpoints. See Fig. 2.5. The
(upper) sum of the areas of the 6 circumscribed rectangles is:
Using Regular Partition Of Order 8 And Left Endpoints
Fig. 2.6
Using Regular Partition Of Order 8 And
Left Endpoints.

Now let's use a regular partition of a still greater order, say
8, and the left endpoints. See Fig. 2.6. The length of each
subinterval is (3 (1))/8 = 1/2. The
9 endpoints are:
Using Regular Partition Of Order 8 And Right Endpoints
Fig. 2.7
Using Regular Partition Of Order 8 And
Right Endpoints.

Next let's use the same regular partition of order 8 but
this time let's employ the right endpoints. See Fig. 2.7. The
(upper) sum of the areas of the 8 circumscribed rectangles is:
Fig. 2.8
Using Regular Partition Of Order n > 0 And Left Endpoints

Fig. 2.9
Using Regular Partition Of Order n > 0 And Right Endpoints

Now using the right endpoints as shown in Fig. 2.9 we will
get the upper sum of the areas of the n
circumscribed
rectangles. The calculation of this sum is similar to that of the lower sum and
is presented in 9.2
Calculation Of Upper
Riemann Sum For Fig. 2.9. We have:
We have:
Since:
we by definition obtain A = 6 square units, the same value obtained by the geometric approach.
Let A
be the area of the triangle formed by the graph of y = x
+ 3, the xaxis, and
the line x = 2.
a. Find A using a geometric formula.
b. Show that the limits at
infinity of the lower and upper sums exist and are equal to the value of A found in part 1.
a.
Fig. 2.10
Area Of Triangle. 
Let f(x) = y = x
+ 3. If y = 0 then
x + 3 = 0
or x = 3; so
the xintercept
is x = 3. The
base of this triangle, which is
a right triangle, is 3 (2) = 5; its height is f(2) = (2) + 3 = 5. See Fig. 2.10. So:
A = (5/2)(5) = 12.5 square units.
b.
Set up a regular partition of order n where n
is an arbitrary positive integer. See Fig. 2.11. The interval is [2, 3]. The
length of each subinterval is (3 (2))/n
= 5/n. The
endpoints are:
Fig. 2.11 Lower Sum Is Obtained By Choosing Right Endpoints. 
Fig. 2.12 Upper Sum Is Obtained By Choosing Left Endpoints. 
See Fig. 2.12 for the
upper sum. Its calculation is similar to that of the lower sum and is carried
out in 9.2
Calculation
Of Upper Sum For Fig. 2.12. We have:
Therefore the limits at infinity of the lower and upper
sums exist and are equal to the value of A found in part 1.
EOS
Note that here the minimum
function values on the subintervals occur at the right endpoints while the
maximum ones at
the left endpoints.
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3. Areas Of General Regions 
Now, if the geometric approach works
and is simpler, why bother with the calculus approach? Well, because the
geometric approach works only with polygonal regions, ie regions bounded by
straightline segments. If a region has a
curved boundary, then the calculus approach is required, as shown in the
following example.
When we for simplicity say the region under f over [a, b]
we mean the plane region bounded by the graph of f, the
x‑axis, the vertical line x = a,
and the vertical line x
= b.
Let A
be the area of the plane region under y = x^{2} over [1, 4].
See Fig. 3.1. Find A
by treating it as the common limit at
infinity of the lower and upper sums if the limits of these sums exist and are
equal.
Fig. 3.1
Area A.

Note
This region has a curved boundary, which is a piece of the curve y = x^{2}.
Solution
Let f(x)
= y = x^{2}. Let n be an arbitrary positive integer and set up the
regular partition of order n
of [1, 4]. See Figs. 3.2
and 3.3. The length of each subinterval is (4 1)/n = 3/n.
The endpoints are:
Fig. 3.2
Lower Sum.

Fig. 3.3
Upper Sum.

For the upper sum
refer to Fig. 3.3. Its calculation is similar to that of the lower sum. For a look
at its calculation see
9.2
Calculation Of Upper Sum For Fig. 3.3. We have:
EOS
The area of the region under f over [a, b] is the common limit at infinity of the lower and upper sums if the limits at infinity of these sums exist and are equal. 
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4. General Sums 
Consider Figs. 4.1 and 4.2. The regular partition is of
order 3. The minimum of f
over [x_{0}, x_{1}] is attained at the left
endpoint x_{0}, while those
over [x_{1}, x_{2}] and [x_{2}, x_{3}] are attained at
the right endpoints x_{2} and x_{3} respectively. The
Fig. 4.1
For This Function Selecting Only Left endpoints Produces A General Sum. 
Fig. 4.2
And Selecting Only Right Endpoints Produces A General Sum Also. 
maximum of f over [x_{0}, x_{1}] is attained at the right endpoint x_{1}, while that over [x_{1}, x_{2}] is attained at c, which is neither
the left endpoint x_{1} nor the right
endpoint x_{2}, and that over
[x_{2}, x_{3}] is attained at the left
endpoint x_{2}. The same situation
is true for infinitely many regular partitions. So selecting only the left
endpoints won't yield only the minimum or only the
maximum on every subinterval and thus will produce a sum thats not a lower
sum or an upper sum. Similarly for
selecting only the right endpoints. Each selection produces a general sum. A general
sum is a sum that's lower or upper
or neither.
In Example 3.1 we found the limits
of both the lower and upper sums; they exist and are equal; we concluded that
the
area is equal to that common limit. Is it enough to use just 1 sum, either the
lower or the upper one, and if its limit (at
infinity) exists then conclude that the area is equal to that limit? The answer
is yes for that function y
= x^{2}. Is it ok to
use
a general sum, and if its limit exists then conclude that the area is equal to
that limit? The answer again is yes for that
function y = x^{2}. There's a theorem, presented
in Section
9.3 Theorem 5.1, from which we deduce that if a function f is
nonnegative and continuous on [a,
b] then every sum has the same
limit and the area under f
over [a, b] is equal to
that limit. To lighten our work we start using that theorem now. The function y = x^{2} is nonnegative and continuous
everywhere and thus on [1, 4]. For a nonnegative and continuous function any 1
sum is ok and enough.
A sum is produced by selecting a set of points of the
subintervals. The set of left endpoints produces a sum that may be
a lower one or an upper one or neither. Similarly for the set of right
endpoints. We may select another set of points, eg
the set of midpoints of the subintervals, as shown in Fig. 4.3. They're
labelled c_{1}, c_{2},
, c_{n}. The point c_{i} is
the midpoint
of the ith
subinterval [x_{i}_{1}, x_{i}].
Remark that their subscripts run from 1 to n.
The subscript of the midpoint or generally of
any nonendpoint is the same as that of the right endpoint. The rectangle over
[x_{i}_{1}, x_{i}]
has height f(c_{i})
where f(x) = x^{2}
for all x. In Fig.
4.3 this set produces a general sum that's neither a lower one nor an upper one,
but that also has a limit
that's also equal to the area of the region under f over [1, 4] because f is nonnegative and continuous on [1, 4]. We
can
use any 1 sum to find that area, which means that we can select any 1 set of
points of the subintervals and the limit of
the sum produced by it will be the area. For a nonnegative and continuous
function any 1 set of points is ok and enough.
In this tutorial for convenience we'll mostly select either the set of left
endpoints or the set of right endpoints.
The above discussions apply also to nonnegative and
continuous functions that are increasing on some parts of [a, b]
and decreasing on others in addition to ones that are either increasing or
decreasing on the entire [a,
b]. The area of the
region under a function nonnegative and continuous over a closed finite
interval is the limit at infinity of a lower sum or
of an upper sum or of a general sum if that limit exists.
Fig. 4.3
Here Selecting Midpoints Produces A
General Sum.

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Each of the sums discussed so far in this section is called
a Riemann sum, after the 19th century German mathematician
Georg Friedrich Bernhard Riemann, who developed a general theory of such sums.
Let f be a
function that's nonnegative
and continuous on [a,
b]. See Figs. 5.1, 5.2, and
5.3. Let n be an
arbitrary positive integer and set up a regular partition
of order n of [a, b]. Since f
is continuous on [a, b] it's continuous on every
subinterval, which is finite and closed. By
Section
1.2.2 Theorem 2.1 on every subinterval it attains a minimum and a maximum
on that subinterval.
Fig. 5.1
Lower Riemann Sum.

Fig. 5.2
Upper Riemann Sum.

Fig. 5.3
General Riemann Sum.

Fig. 5.4

Definitions
5.1 Riemann Sums
Note that a general Riemann sum of order n may be any sum of order n (choosing any set of
points of the 
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6. Approximations Of Areas By Riemann Sums 
Example 6.1
Approximate the area under y = f(x) = x^{2} over [1, 4] by the lower Riemann sum of order 6 on a regular partition.
Solution
Let A
be the required area. The common length of the subintervals is (4 1)/6 =
1/2. See Fig. 6.2. Since f
is increasing
and continuous a lower Riemann sum is obtained by selecting the left endpoints
of the subintervals. We have:
Fig. 6.2
Approximation Of Area.

EOS
Recall that in Example 3.1 we found the exact value of A to be 21 square units.
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7. Areas As Limits Of Riemann Sums 
Compute the area of the plane region under y = x^{2} 3x + 3 over [1, 3].
Solution
Fig. 7.1 Computation Of Area. 
Let f(x) =
y = x^{2} 3x
+ 3, which is nonnegative and continuous on [1, 3]. Let n
be an arbitrary positive integer. Let's
use the regular partition of order n and the right endpoints. See
Fig. 7.1. We'll get the general Riemann Sum R_{n}
of order
n.
The length of each subinterval is (3 (1))/n = 4/n.
The right endpoints are:
EOS
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Fig. 8.1 Limit Of Riemann Sum Is Less Than Actual Area Under f Over [a, b]. 
Given a regular partition of order n where n
is a given positive integer, how many Riemann sums of order n are possible?
Well, in general, many, actually infinitely many, because there are infinitely
many ways to select the points of the
subintervals.
Problems & Solutions 
1. Approximate
the area of the plane region under y = e^{x}
over [2, 2] by the lower Riemann sum of order 4 on a regular
partition. Use a calculator to
express your answer in decimal format rounded to 2 decimal places.
Solution
Let f(x) =
y = e^{x}.
The length of each subinterval is (2 (2))/4 = 1. Since f is increasing the lower
Riemann sum is
obtained by using the left endpoints. They are 2, 2 + 1 = 1, 1 + 1 = 0,
and 0 + 1 = 1. The required area is:
2. Repeat
Problem & Solution 1, replacing the lower
Riemann sum of order 4 by that of order 8. Of the two answers,
which one do you think is more
accurate? Will
the accuracy improve still if the order of the regular partition increases
further? If the order approaches infinity, what will
the approximation approach?
Solution
Let f(x) = y = e^{x}. The length of each subinterval is (2 (2))/8 = 1/2. The left endpoints of the subintervals are:
2,
2 + 1/2 = 3/2,
3/2 + 1/2 = 1,
1 + 1/2 = 1/2,
1/2 + 1/2 = 0,
0 + 1/2 = 1/2,
1/2 + 1/2 = 1, and
1 + 1/2 = 3/2.
The area is:
The answer in this Problem & Solution is more accurate.
Yes, if the order of the regular partition increases further, the
accuracy will improve still. If the order approaches infinity, the
approximation will approach the actual area.
3. Consider
the area A of the plane region under y
= e^{x}
over [a, b] where a
< b. Let n be a
positive integer and R_{n}
the
approximation of A
by the Riemann sum with the regular partition of order n
and using the right endpoints of the
subintervals.
a. Prove that:
(Thus A = e^{a} e^{b} square units.)
Solution
4. Find the area of the plane region under y = x^{3} + 1 over the interval [1, 1].
Solution
Let f(x) =
y = x^{3} + 1. Let's use
the Riemann Sum R_{n}
with the regular partition of order n, where n
is a positive integer,
and using the left endpoints. The length of each subinterval is (1 (1))/n
= 2/n. The left endpoints are:
Solution
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