Lar calc10 ch04_sec2

Integration
Copyright © Cengage Learning. All rights reserved.

Area
Copyright © Cengage Learning. All rights reserved.

3
Objectives
 Use sigma notation to write and evaluate a sum.
 Understand the concept of area.
 Approximate the area of a plane region.
 Find the area of a plane region using limits.

5
Sigma Notation
This section begins by introducing a concise notation for
sums. This notation is called sigma notation because it
uses the uppercase Greek letter sigma, written as

6
Example 1 – Examples of Sigma Notation
From parts (a) and (b), notice that the same sum can be
represented in different ways using sigma notation.

7
Sigma Notation
The following properties of summation can be derived
using the Associative and Commutative Properties of
Addition and the Distributive Property of Addition over
Multiplication. (In the first property, k is a constant.)

8
Sigma Notation
The following theorem lists some useful formulas for sums
of powers.

9
Example 2 – Evaluating a Sum
Solution:
Applying Theorem 4.2, you can write

Example 2 – Solution cont’d
10
Now you can evaluate the sum by substituting the
appropriate values of n, as shown in the table.

12
Area
In Euclidean geometry, the simplest type of plane region is
a rectangle.
Although people often say that the formula for the area of a
rectangle is A = bh, it is actually more proper to say that
this is the definition of the area of a rectangle.
From this definition, you can develop formulas for the areas
of many other plane regions.

13
From this definition, you can develop formulas for the ares
of many other plane regions.
For example, to determine the area of a triangle, you can
form a rectangle whose area is twice that of the triangle, as
shown in Figure 4.5.
Figure 4.5
Area

14
Once you know how to find the area of a triangle, you can
determine the area of any polygon by subdividing the
polygon into triangular regions, as shown in Figure 4.6.
Figure 4.6
Area

15
Area
Finding the areas of regions other than polygons is more
difficult. The ancient Greeks were able to determine
formulas for the areas of some general regions (principally
those bounded by conics) by the exhaustion method.
The clearest description of this method was given by
Archimedes. Essentially, the method is a limiting process in
which the area is squeezed between two polygons—one
inscribed in the region and one circumscribed about the
region.

16
For instance, in Figure 4.7 the area of a circular region is
approximated by an n-sided inscribed polygon and an
n-sided circumscribed polygon.
Figure 4.7
Area

17
Area
For each value of n, the area of the inscribed polygon is
less than the area of the circle, and the area of the
circumscribed polygon is greater than the area of the circle.
Moreover, as n increases, the areas of both polygons
become better and better approximations of the area of the
circle.

18
The Area of a Plane Region

19
Example 3 – Approximating the Area of a Plane Region
Use the five rectangles in Figure 4.8(a) and (b) to find two
approximations of the area of the region lying between the
graph of f(x) = –x2 + 5 and the x-axis between x = 0
and x = 2.
Figure 4.8

20
Example 3(a) – Solution
The right endpoints of the five intervals are
where i = 1, 2, 3, 4, 5.
The width of each rectangle is , and the height of each
rectangle can be obtained by evaluating f at the right
endpoint of each interval.

21
Example 3(a) – Solution
The sum of the areas of the five rectangles is
Because each of the five rectangles lies inside the
parabolic region, you can conclude that the area of the
parabolic region is greater than 6.48.
cont’d

22
Example 3(b) – Solution
The left endpoints of the five intervals are (i – 1),
where i = 1, 2, 3, 4, 5.
The width of each rectangle is , and the height of each
rectangle can be obtained by evaluating f at the left
endpoint of each interval. So, the sum is
cont’d

23
Example 3(b) – Solution
Because the parabolic region lies within the union of the
five rectangular regions, you can conclude that the area of
the parabolic region is less than 8.08.
By combining the results in parts (a) and (b), you can
conclude that 6.48 < (Area of region) < 8.08.
cont’d

25
Upper and Lower Sums
Consider a plane region bounded above by the graph of a
nonnegative, continuous function y = f (x), as shown in
Figure 4.9.
The region is bounded below by the x-axis, and the left and
right boundaries of the region are the vertical lines x = a
and x = b.
Figure 4.9

26
To approximate the area of the region, begin by subdividing
the interval [a, b] into n subintervals, each of width
as shown in Figure 4.10.
Figure 4.10

27
The endpoints of the intervals are as follows.
Because f is continuous, the Extreme Value Theorem
guarantees the existence of a minimum and a maximum
value of f (x) in each subinterval.
f (mi) = Minimum value of f (x) in ith subinterval
f (Mi) = Maximum value of f (x) in ith subinterval

28
Next, define an inscribed rectangle lying inside the ith
subregion and a circumscribed rectangle extending
outside the ith subregion. The height of the ith inscribed
rectangle is f(mi) and the height of the ith circumscribed
rectangle is f(Mi).
For each i, the area of the inscribed rectangle is less than
or equal to the area of the circumscribed rectangle.

29
The sum of the areas of the inscribed rectangles is called a
lower sum, and the sum of the areas of the circumscribed
rectangles is called an upper sum.

30
From Figure 4.11, you can see that the lower sum s(n) is
less than or equal to the upper sum S(n).
Moreover, the actual area of the region lies between these
two sums.
Figure 4.11

31
Example 4 – Finding Upper and Lower Sums for a Region
Find the upper and lower sums for the region bounded by
the graph of f(x) = x2 and the x-axis between x = 0 and
x = 2
Solution:
To begin, partition the interval [0, 2] into n subintervals,
each of width

32
Example 4 – Solution
Figure 4.12 shows the endpoints of the subintervals and
several inscribed and circumscribed rectangles.
cont’d
Figure 4.12

33
Because f is increasing on the interval [0, 2], the minimum
value on each subinterval occurs at the left endpoint, and
the maximum value occurs at the right endpoint.
Using the left endpoints, the lower sum is
cont’d

34

35
Using the right endpoints, the upper sum is
cont’d

36
The next theorem shows that the equivalence of the limits (as
n  ∞) of the upper and lower sums is not mere coincidence.
It is true for all functions that are continuous and nonnegative
on the closed interval [a, b].

37
In Theorem 4.3, the same limit is attained for both the
minimum value f(mi) and the maximum value f(Mi).
So, it follows from the Squeeze Theorem that the choice of
x in the ith subinterval does not affect the limit.
This means that you are free to choose an arbitrary x-value
in the ith subinterval, as in the following definition of the
area of a region in the plane.

39
Example 5 – Finding Area by the Limit Definition
Find the area of the region bounded by the graph f(x) = x3,
the x-axis, and the vertical lines x = 0 and x = 1 as shown
in Figure 4.14.
Figure 4.14

40
Begin by noting that f is continuous and nonnegative on the
interval [0, 1]. Next, partition the interval [0, 1] into
n subintervals, each of width Δx = 1/n.
According to the definition of area, you can choose any
x-value in the ith subinterval.
For this example, the right endpoints ci = i/n are convenient.

41
The area of the region is

Lar calc10 ch04_sec2

More Related Content

What's hot (20)

Similar to Lar calc10 ch04_sec2 (20)

More from Institute of Applied Technology (20)

Recently uploaded (20)

Lar calc10 ch04_sec2