Area/Sigma Notation

Area/Sigma Notation

Objective: To define area for plane regions with curvilinear boundaries. To use Sigma Notation to find areas.

The Area Problem

• Formulas for the areas of polygons are well known. However, the problem of finding formulas for regions with curved boundaries caused difficulties for early mathematicians.

The Area Problem

• Formulas for the areas of polygons are well known. However, the problem of finding formulas for regions with curved boundaries caused difficulties for early mathematicians.

• The first real progress was made by Archimedes who obtained areas of regions with curved bounds by the method of exhaustion.

The Area Problem

• This method, when applied to a circle, consists of inscribing a succession of regular polygons in the circle and allowing the number of sides to increase indefinitely. As the number of sides increases, the polygons tend to “exhaust” the region inside the circle, and the area of the polygons become better and better approximations of the exact area.

The Rectangle Method

• We will now use Archimedes’ method of exhaustion with rectangles to find the area under a curve in the following way:


• We will now use Archimedes’ method of exhaustion with rectangles in the following way:

• Divide the interval [a, b] into n equal subintervals, and over each subinterval construct a rectangle that extends from the x-axis to any point on the curve that is above the subinterval.


• For each n, the total area of the rectangles can be viewed as an approximation to the exact area under the curve over the interval [a, b]. As n increases these approximations will get better and better and will approach the exact area as a limit.

Area as a Limit; Sigma Notation

• The notation we will use is called sigma notation or summation notation because it uses the uppercase Greek letter (sigma) to denote various kinds of sums. To illustrate how this notation works, consider the sum in which each term is of the form k2, where k is one of the integers from 1-5. In Sigma notation this can be written as

which is read “the summation of k2 from 1 to 5”.

22222 54321

5

1

2

k

k

Sigma Notation

• In general, we can look at sigma notation like this:

Example 1

• Let’s look at some examples of sigma notation.

8

4

3

k

k

5

1

2k

k

5

0

)12(k

k

5

0

)12()1(k

k k

Other Sums

• Here are two other ideas we need to know1. If the upper and lower limits of summation are the

same, we evaluate that number once in the function.

2. If the function we are evaluating is a constant, we add that number to itself n times, or Cn.

2

2

33 2k

k

5

1

10)5(2222222i

n

k

CnC1

Changing the limits of Summation

• A sum can be written in more than one way using Sigma Notation with different limits of summation. For example, these three are all the same.

10864225

1

i

i

4

0

22j

j

7

3

42k

k

Example 2

• Express the following in sigma notation so that the lower limit of summation is 0 rather than 3.

7

3

25k

k

Example 2

• Express the following in sigma notation so that the lower limit of summation is 0 rather than 3.

• We need to subtract 3 from the upper and lower limits of summation. If you subtract 3 from the limits, you must add 3 to k in the function. Always do the opposite.

7

3

25k

k

Changing Limits

• Be careful… you add/subtract from k, not the exponent. For example, change the lower limit of summation from 3 to 1.

5

1

3227

23

1)2(27

3

12 333k

k

k

k

k

k

Properties of Sums

• Theorem 6.4.1

a)

b)

c)

n

kk

n

kk acca

11

n

kk

n

kkk

n

kk baba

111

)(

n

kk

n

kkk

n

kk baba

111

)(

Summation Formulas

• Theorem 6.4.2

a)

b)

c)

2

)1(...21

1

nnnk

n

k

6

)12)(1(...21 222

1

2

nnnnk

n

k

2

1

3333

2

)1(...21

n

k

nnnk

Example 3

• Evaluate

30

1

)1(k

kk

Example 3

• Evaluate

30

1

)1(k

kk

30

1

30

1

30

1

2230

1

)1(k k kk

kkkkkk

Example 3

• Evaluate

30

1

)1(k

kk

30

1

30

1

30

1

2230

1

)1(k k kk

kkkkkk

99202

)31(30

6

)61)(31(30

Definition of Area

• We will now turn to the problem of giving a precise definition of what is meant by “area under a curve.” Specifically, suppose that the function f is continuous and nonnegative on the interval [a,b], and let R denote the region bounded below by the x-axis, bounded on the sides by the vertical lines x = a and x = b, and bounded above by the curve y = f(x).

Definition of Area

• Divide the interval [a, b] into n equal subintervals by inserting n – 1 equally spaced points between a and b and denote those points by 121 ,..., nxxx

Definition of Area

• Each of these subintervals has width (b – a)/n, which is customarily denoted by

n

abx

Definition of Area

• Over each interval construct a rectangle whose height is the value of f at an arbitrarily selected point in the subinterval. Thus, if denote the points selected in the subintervals, then the rectangles will have heights and areas

**2

*1 ,..., nxxx

)(),...(),( **2

*1 nxfxfxf

xxfxxfxxf n )(,...)(,)( **2

*1

Definition of Area

• This can be expressed more compactly in sigma notation as: (k = # of rectangle)

• We will repeat the process using more and more subdivisions, and define the area of R to be the “limit” of the areas of the approximating regions Rn as n increases without bound. That is, we define the area A as

xxfAn

kk

)(1

*

xxfAn

kk

n

)(lim1

*

Definition of Area

• Definition 6.4.3 (Area under a curve)• If the function f is continuous on [a, b] and if f(x) > 0

for all x in [a, b], then the area under the curve y = f(x) over the interval [a, b] is defined by

xxfAn

kk

n

)(lim1

*

Points

• The values of can be chosen arbitrarily, so it is conceivable that different choices of these values might produce different values of A. Were this to happen, then the definition of area would not be acceptable. This does not happen. We will get the same area regardless of which points we choose.

**2

*1 ,..., nxxx

Points

• The three ways we will look at this is:

1. The left endpoint of each subinterval2. The right endpoint of each subinterval3. The midpoint of each subinterval

Points

• To be more specific, suppose that the interval [a, b] is divided into n equal parts of length by the points and let x0 = a and xn = b. Then

for k = 0, 1, 2,…,n

nabx /)( *1

*2

*1 ,..., nxxx

xkaxk

Points

• We will look at each point as:

• Left endpoint

• Right endpoint

• Midpoint

xkaxx kk )1(1*

xkaxx kk 1*

xkaxx kk )2/1(1*

Example 5

• Use the definition of area using the right endpoint of each subinterval to find the area between the graph of f(x) = x2 and the interval [0, 1].

Example 5


• We will use the same problem solving process for each point we choose.

1. Find x

Example 5



1. Find x nnx

101

Example 5



1. Find 2. Find

x nnx

101

*kx

Example 5



1. Find 2. Find

x nnx

101

*kx

n

k

nkxk 1

0*

Example 5



1. Find 2. Find

x nnx

101

*kx

n

k

nkxk 1

0*

n

k

n

kk

n n

k

nn

kxxf

13

22

1

* 1)(lim

Example 5



1. Find 2. Find

x nnx

101

*kx

n

k

nkxk 1

0*

6

)12)(1(11)(lim

31

3

22

1

* nnn

nn

k

nn

kxxf

n

k

n

kk

n

Example 5



1. Find 2. Find

x nnx

101

*kx

n

k

nkxk 1

0*

3

1

6

2

6

)12)(1(1lim

3

3

3

n

nnnn

nn

Theorem 6.4.4

• Here are a few limits that you may or may not use. They can make the end of the problems easier.

n

kn n

a1

111

lim)

n

kn

kn

b1

2 2

11lim)

n

kn

kn

c1

23 3

11lim)

n

kn

kn

d1

34 4

11lim)

Example 6

• Use the definition of area using the midpoint to find the area under the curve f(x) = 9 – x2 over the interval [0, 3].

Example 6


1. Find x nnx

303

Example 6


1. Find2. Find

x nnx

303

*kx

nn

k

nkxk 2

333)2/1(0*

Example 6


1. Find2. Find

x nnx

303

*kx

nn

k

nkxk 2

333)2/1(0*

nnn

kxxf

n

kk

n

3

2

339)(lim

2

1

*

Example 6


1. Find2. Find

x nnx

303

*kx

nn

k

nkxk 2

333)2/1(0*

nnn

k

n

k

nnn

kxxf

n

kk

n

3

4

9

2

1899

3

2

339)(lim

222

22

1

*

Example 6


1. Find2. Find

x nnx

303

*kx

nn

k

nkxk 2

333)2/1(0*

333

2

222

2

4

27

2

5427273

4

9

2

1899lim

nn

k

n

k

nnnn

k

n

kn

18003

2727

Example 6

• Let’s quickly look at this as a right endpoint and compare the two answers.

nnn

kxxf

n

kk

n

3

2

339)(lim

2

1

*

nn

kxxf

n

kk

n

339)(lim

2

1

*

Example 6

• Let’s quickly look at this as a right endpoint and compare the two answers.

3

22272733

9limn

k

nnn

kn

333

2

222

2

4

27

2

5427273

4

9

2

1899lim

nn

k

n

k

nnnn

k

n

kn

Net Signed Area

• In our definition of area, we assumed that f was continuous and nonnegative over the interval [a, b].

• If f is both negative and positive over the interval, our definition no longer represents the area between the curve y = f(x) and the interval [a, b]; rather, it represents a difference of areas- the area above the x-axis minus the area below the x-axis. We call this the net signed area.

Definition 6.4.5

• Net Signed Area- If the function f is continuous on [a, b] then the net signed area A between y = f(x) and the interval [a, b] is defined by

xxfAn

kk

n

)(lim1

*

Example 7

• Use our definition of area with the left endpoint to find the net signed area between the graph of

y = x – 1 and the interval [0, 2]

Example 7

• Use our definition of area with the left endpoint to find the net signed area between the graph of

y = x – 1 and the interval [0, 2]1. Find

2. Find

nnx

202

x

*kx nn

k

nkxk

222)1(0*

nnn

k

nnn

kxxf

n

kk

n

24421

22)(lim

221

*

022

Example 7• Again, let’s compare this to the right endpoint.• Using the right endpoint, n

kxk

20*

nnn

k

nnn

kxxf

n

kk

n

24421

22)(lim

221

*

022

nn

k

nn

kxxf

n

kk

n

2421

2)(lim

21

*

Homework

• Page 384• 3-15 odd• 27, 39-51 odd

Documents

Area/Sigma Notation