4.2 Area

Sigma Notation

Summation ExamplesExample:

5

1i

i

Example:

3

1

2

j

j

Example:

4

1

3k

1 2 3 4 5 15

1 4 9 14

3 3 33 12

Example:

3

1

52i

i 7 9 11 27

Example 1

12 2

0

kn

n

k

More Summation Examples

112 2

n 12

2 2

n 13

2 2

n 10

2 2

n

12 2 nn

......

i

n

ii xxf

1

)( 11)( xxf 22 )( xxf 33)( xxf

......nn xxf )(

Theorem 4.2 Summation Rules

n

iika

1

n

iiak

1

n

iii ba

1

)(

n

iia

1

n

iib

1

n

i

c1

cn

n

i

i1 2

)1( nn

n

i

i1

2

6

)12)(1( nnn

n

i

i1

3

4

)1( 22 nn

Theorem 4.2 Summation Rules

Example 2 Evaluate the summation

100

1

)92(i

i

2100

1i

i

100

1

9i

2

)101(1002 9100

100,10 900 000,11

Solution

100

1

)92(i

i

Examples

Example 3 Compute

4

1

2)13(i

i

4

1

2 )169(i

ii

4

1

29i

i

4

1

6i

i

4

1

1i

6

)9)(5(49

2

)5(46 )1)(4(

270 60 4 214

4

1

2)13(i

i

Solution

Examples

Example 4 Evaluate the summation for n = 100 and 10000

1

22

1n

k n

k

Solution

1

22

1n

k n

k

1

22

)1(1 n

k

kn

n

k

kn 1

2

1

Note that we change (shift) the upper and lower bound

2

)1(12

nn

n n

n

2

1

For n = 100

200

101

2

1

n

n For n = 10000

20000

10001

2

1

n

n

Examples

Summation and LimitsExample 5 Find the limit for

n

n

i nni

1

2 )1()2( lim

n

n

i ni

n 1

2)2(1

lim

n

n

i nii

nn 12

2

)44(1

lim

n

nnn

n

nn

nn

n 6

)12)(1(1

2

)1(44

1 lim

2

n

n

n

n

n

n 121

6

1)1(24 lim

Continued…

n n

n

n

n

n

n 121

6

1)1(24 lim

n nnn

12

11

6

11124 lim

)2)(1(61(2)(1)4

319

316

Area

wlA hbA 2

12rA

2

2)( xxf

x

y

Lower ApproximationUsing 4 inscribed rectangles of equal width

Lower approximation =(sum of the rectangles)

4

91

4

10

4

2

4

14

2

14

7

2

2)( xxf

The total number of inscribed rectangles

x

y

Using 4 circumscribed rectangles of equal width

Upper approximation =(sum of the rectangles)

4

4

91

4

1

4

2

4

30

2

1

4

15

2

2)( xxf

Upper Approximation

The total number of circumscribed rectangles

Continued…

4

91

4

10

2

1

4

14

2

1

4

7

L

4

4

91

4

1

2

1

4

30

2

1

4

15

U

L A U

4

7 A 4

15The average of the lower and upper approximations is

2

LU

2

415

47

2

422

4

11

A is approximately 4

11

Upper and Lower Sums

The procedure we just used can be generalized to the methodology to calculate the area of a plane region. We begin with

subdividing the interval [a, b] into n subintervals, each of equal width x = (b – a)/n. The endpoints of the intervals are

babaxnax

xax

xax

axax

n

)()(

)(2

)(1

)(0

2

1

0

Upper and Lower Sums

Because the function f(x) is continuous, the Extreme Value Theorem guarantees the existence of a minimum and a maximum value of f(x) in each subinterval.

We know that the height of the i-th inscribed rectangle is f(mi) and that of circumscribed rectangle is f(Mi).

)(min)(],[ 1

xfmfii xxx

i

)(max)(],[ 1

xfMfii xxx

i

Upper and Lower Sums

The i-th regional area Ai is bounded by the inscribed and circumscribed rectangles.

We know that the relationship among the Lower Sum, area of the region, and the Upper Sum is

xMfAxmf iii )()(

n

ii

n

ii

n

ii xMfAxmf

111

)()(

)()()()(111

nSxMfAAxmfnsn

ii

n

ii

n

ii

Theorem 4.3 Limits of the Upper and Lower Sums

x

y

2

2)( xxf

length = 2 – 0 = 2

xnn

202

n = # of rectangles

A

n

n

i nin

f1

22 lim

n

n

i

inn 1

22

42 lim

n

n

i

inn 1

22

42 lim

Exact Area Using the Limit

in

M i

2

A

n

n

i nin

f1

22 lim

n

n

i

inn 1

22

42 lim

n

n

i

inn 1

22

42 lim

n

nnn

nn 6

)12)(1(42 lim

2

n n

n

n

n 121

3

4 lim

n nn

12

11

3

4 lim

)2)(1(3

4

3

8

Exact Area Using the Limit

Definition of the Area of a Region in the Plane

a b

n

abx

Area =

n

n

i n

abi

n

abaf

1

lim

height x base

In General - Finding Area Using the Limit

Or, xi , the i-th right endpoint

Regular Right-Endpoint Formula

RR-EFintervals are regular in length

squaring from right endpt of rect.

Example 6 Find the area under the graph of 5] [1, interval on the 64)( 2 xxxf

1 5

a = 1b = 5

n

ab

nn

415

in

aba i

n

41

A =

n

n

i nin

f1

441 lim

n

n

i

in

inn 1

2

64

144

14

lim

n

n

i

in

in

inn 1

22

616

4168

14

lim

n

n

i nin

fA1

441 lim

n

n

i

in

inn 1

2

64

144

14

lim

n

n

i

in

in

inn 1

22

616

4168

14

lim

n

n

i

in

inn 1

22

38164

lim

nn

nn

n

nnn

nn3

2

)1(8

6

)12)(1(164 lim

2

Regular Right-Endpoint Formula

nn

nn

n

nnn

nn3

2

)1(8

6

)12)(1(164 lim

2

n n

nnn

n12

)1(16)12)(1(

3

32 lim

2

n n

n

n

n

n

n12

116

121

3

32 lim

12)1(16)2)(1(3

32

n nnn12

1116

12

11

3

32 lim

Continued

3

52

HomeworkPg. 267 1, 7, 11, 15, 21, 31, 33, 41, 23-29 odd, 39, 43