Calculus Of One Real Variable – By Pheng Kim Ving Chapter 9: The Integral – Section 9.2: Areas And Riemann Sums 9.2 Areas And Riemann Sums

 1. Partitions

Let's partition or sub-divide the interval [a, b] in Fig. 1.1 into 5 sub-intervals using 6 points x0, x1, x2, x3, x4, and x5 where
x0 = a, x5 = b, and x0 < x1 < x2 < x3 < x4 < x5. We say that we've made a partition of order  5 of the interval [a, b]. This
partition is determined by selecting points x0 at a, x5 at b, and the remaining points such that x0 < x1 < x2 < x3 < x4 < x5.
So we say that this partition is the ordered set {x0, x1, x2, x3, x4, x5}. Fig. 1.1   A partition of [a, b].

A partition P of order n of the interval [a, b] is an ordered set P = {x0, x1, x2, ..., xn} such that:

a = x0 < x1 < x2 < ... < xn = b,  Fig. 1.2   A Regular Partition Of Order 5 Of [a, b]. There are n left endpoints (because there are n sub-intervals); their subscripts run from 0 to n – 1. There are also n
right endpoints (again because there are n sub-intervals); their subscripts however run from 1 to n. The right endpoints
of the 1st, 2nd, ..., (n – 1)th sub-intervals are the left endpoints of the 2nd, 3rd, ..., nth sub-intervals respectively.

 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 non-negative 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 x-axis, 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)(b1 + b2)h, where b1 and b2 are the bases and h is the height. Now, b1 = f(–1) =
(1/2)(–1) + 1 = 1/2, b2 = 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. # Area A Of Trapezoid.

Using calculus, we proceed as follows.

### Using Regular Partition Of Order 4 And Left Endpoints # 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 sub-interval is (3 –
(–1))/4  = 1. The 4 left endpoints of the sub-intervals are: #### Using Regular Partition Of Order 4 And Right Endpoints # 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 sub-intervals. The
length of each sub-interval is (3 – (–1))/4 = 1. See Fig. 2.3. The 4 right endpoints of the sub-intervals are:

x1 = 0,     x2 = 1,     x3 = 2,     and     x4 = 3, ### Remark 2.1 – Sums Of Areas Of Rectangles For Any Positive Integer n

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 # 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
sub-interval is (3 – (–1))/6  = 2/3. The 7 left endpoints of the sub-intervals are: Using Regular Partition Of Order 6 And Right Endpoints # 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 # 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
sub-interval is (3 – (–1))/8  = 1/2. The 9 endpoints are: Using Regular Partition Of Order 8 And Right Endpoints # 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: #### General Case   # Using Regular Partition Of Order n > 0 And Left Endpoints # 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: #### Approximations Of Areas #### Defining Areas

We have: ### Obtaining The Area

Since: we by definition obtain A = 6 square units, the same value obtained by the geometric approach.

### Example 2.1

Let A be the area of the triangle formed by the graph of y = – x + 3, the x-axis, 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.

##### Solution

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 x-intercept 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 sub-interval 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 sub-intervals occur at the right endpoints while the maximum ones at
the left endpoints.

 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 straight-line 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
xaxis, the vertical line x = a, and the vertical line x = b”.

## Example 3.1

Let A be the area of the plane region under y = x2 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. # Area A.

Note

This region has a curved boundary, which is a piece of the curve y = x2.

Solution

Let f(x) = y = x2. 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 sub-interval is (4 – 1)/n = 3/n. The endpoints are:  # Lower Sum. # 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.

 4. General Sums

### General Sums

Consider Figs. 4.1 and 4.2. The regular partition is of order 3. The minimum of f over [x0, x1] is attained at the left
endpoint x0, while those over [x1, x2] and [x2, x3] are attained at the right endpoints x2 and x3 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 [x0, x1] is attained at the right endpoint x1, while that over [x1, x2] is attained at c, which is neither
the left endpoint x1 nor the right endpoint x2, and that over [x2, x3] is attained at the left endpoint x2. 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 sub-interval and thus will produce a sum that’s 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.

### Any One Sum Is Ok And Enough

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 = x2. 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 = x2. There's a theorem, presented in Section 9.3 Theorem 5.1, from which we deduce that if a function f is
non-negative 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 = x2 is non-negative and continuous
everywhere and thus on [1, 4]. For a non-negative and continuous function any 1 sum is ok and enough.

### Any One Set Of Points Is Ok And Enough

A sum is produced by selecting a set of points of the sub-intervals. 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 sub-intervals, as shown in Fig. 4.3. They're labelled c1, c2, …, cn. The point ci is the midpoint
of the ith sub-interval [xi-1, xi]. Remark that their subscripts run from 1 to n. The subscript of the midpoint or generally of
any non-endpoint is the same as that of the right endpoint. The rectangle over [xi-1, xi] has height f(ci) where f(x) = x2
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 non-negative 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 sub-intervals and the limit of
the sum produced by it will be the area. For a non-negative 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.

### Area As Limit Of A Sum

The above discussions apply also to non-negative 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 non-negative 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. # Here Selecting Midpoints Produces A General Sum.

 5. Riemann Sums

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 non-negative
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 sub-interval, which is finite and closed. By
Section 1.2.2 Theorem 2.1 on every sub-interval it attains a minimum and a maximum on that sub-interval.  # Lower Riemann Sum. # Upper Riemann Sum. # General Riemann Sum.  # 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 sub-intervals) including the lower or upper one. 6. Approximations Of Areas By Riemann Sums Example 6.1

Approximate the area under y = f(x) = x2 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 sub-intervals 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 sub-intervals. We have:  # Approximation Of Area.

EOS

Recall that in Example 3.1 we found the exact value of A to be 21 square units.

 7. Areas As Limits Of Riemann Sums #### Example 7.1

Compute the area of the plane region under y = x2 – 3x + 3 over [–1, 3].

Solution # Fig. 7.1

Computation Of Area.

Let f(x) = y = x2 – 3x + 3, which is non-negative 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 Rn of order
n. The length of each sub-interval is (3 – (–1))/n = 4/n. The right endpoints are: EOS  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
sub-intervals.

 Problems & Solutions

1. Approximate the area of the plane region under y = ex 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 = ex. The length of each sub-interval 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 = ex. 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 = ex over [a, b] where a < b. Let n be a positive integer and Rn the
approximation of A by the Riemann sum with the regular partition of order n and using the right endpoints of the
sub-intervals.

a.  Prove that: (Thus A = ea eb square units.)

Solution   4.  Find the area of the plane region under y = x3 + 1 over the interval [–1, 1].

Solution Let f(x) = y = x3 + 1. Let's use the Riemann Sum Rn with the regular partition of order n, where n is a positive integer,
and using the left endpoints. The length of each sub-interval is (1 – (–1))/n = 2/n. The left endpoints are:   Solution  