CHAPTER 4SECTION 4.2

AREA

Sigma (summation) notation Sigma (summation) notation REVIEWREVIEW5

1

1 2 3 4 5k

k

In this case k is the index of summation

The lower and upper bounds of summation are 1 and 5

63 3 3 3 3 3 3

1

1 2 3 4 5 6

i

i

In this case i is the index of summation

The lower and upper bounds of summation are 1 and 6

63

1

1 8 27 64 125 216

i

i

Sigma notationSigma notation

4

1

1 2 3 4 163

1 2 3 4 5 60k

k

k

3

1 1 2 2 3 31

( ) ( ) ( ) ( )k kk

f x x f x x f x x f x x

Sigma Summation Sigma Summation NotationNotation

Practice with Practice with Summation NotationSummation Notation

= 3080= 3080

Practice with Summation Practice with Summation NotationNotation

Numerical Problems can be done with the Numerical Problems can be done with the TI83+/84 as was done in PreCalc AlgebraTI83+/84 as was done in PreCalc Algebra

Sum is in LIST, MATHSum is in LIST, MATH

Seq is on LIST, OPSSeq is on LIST, OPS

Area Under a Curve by Limit DefinitionArea Under a Curve by Limit Definition

The area under a curve can be approximated by the sum of rectangles. The figure on the left shows inscribed rectangles while the figure on the right shows circumscribed rectangles

This gives the upper sum.

This gives the lower sum.

211

8y x 0 4x

Left endpoint approximation:

Approximate area: 1 1 1 31 1 1 2 5 5.75

8 2 8 4

(too low)

Approximate area: 1 1 1 31 1 2 3 7 7.75

8 2 8 4

211

8y x 0 4x

Right endpoint approximation:

(too high)

Averaging the right and left endpoint approximations:

7.75 5.756.75

2

(closer to the actual value)

Approximating definite integrals:different choices for the sample points

• If xi* is chosen to be the left endpoint of the interval, then xi* = xi-1 and we have

• If xi* is chosen to be the right endpoint of the interval, then xi* = xi and we have

• Ln and Rn are called the left endpoint approximation and right endpoint approximation , respectively.

xxfRdxxfb

a

n

iin

)()(1

xxfLdxxfb

a

n

iin

)()(

11

xxfdxxfb

a

n

ii

)()(1

*

1.031251.28125

1.781252.53125

211

8y x 0 4x

Can also apply midpoint approximation:choose the midpoint of the subinterval as the sample point.

Approximate area: 6.625

The midpoint rule gives a closer approximation than the trapezoidal rule, but in the opposite direction.

Midpoint rule

],[ ofmidpoint )( and

where

)]()()([)(

112

1

21

iiiii

nn

b

a

xxxxx

n

abx

xfxfxfxMdxxf

Approximating the Area of a Plane Region

5

1

2

3

4

2/5 4/5 6/5 8/5 2

f(x) = -x2 + 5

2/5 4/5 6/5 8/5 2

f(x) = -x2 + 5

x

y

x

y

To approximate the area under each curve, you must sum the area of each rectangle.

See next slide

a. b.

5

4

3

2

1

a. The right endpoints, Mi, of the intervals are 2i/5, where i = 1, 2, 3, 4, 5. The width of each rectangle is 2/5 and the height of each rectangle can be obtained by evaluating f at the right endpoint of each interval.

[0, 2/5], [2/5, 4/5], [4/5, 6/5], [6/5, 8/5], [8/5, 10/5]

Evaluate f(x) at the right endpoints of each of these intervals.

The sum of the area of the five rectangles is

Height Width

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.

5

1

5

1

2

48.625

162

5

25

5

2

5

2

5

2

i i

iif

Approximating the Area of a Plane Region for b (con’t)

b. The left endpoints of the five intervals are 2/5(i _ 1), where i = 1, 2, 3,

4, 5. The width of each rectangle is 2/5, and the height of each

rectangle can be found by evaluating f at the left endpoint of each

interval.

Height Width

Because the parabolic region lies within the union of the five rectangular

region, that the area of the parabolic region is less than 8.08.

6.48 < Area of region < 8.08

5

1

5

1

2

08.825

202

5

25

5

22

5

2

5

22

i i

iif

ON CALCULATOR

2 2( (( ((2 2) / 5) 5) , ,1,5) 8.08

5sum seq x x

2 2( (( ((2 / 5) 5) , ,1,5) 6.48

5sum seq x x

In general for the upper sum S(n) and Lower sum s(n), you use the following for curves f(x) bound between x=a and x=b.

i1

1

b-a( ) ( ) , where M = a + i( x) (the right endpoint)

n

b-a( ) ( ) , where x = m a + (i-1)( x) (the left endpoint)

n i

n

ii

n

ii

S n f M x x and

s n f m x and

Finding Upper and Lower Sums for a RegionFind 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 Begin by partitioning the interval [0, 2] into n subintervals, each of length

nnn

abx

202

1 2 1 2

f(x) = x2f(x) = x2

1

2

3

4 4

3

2

1

A. B.

Left endpoints Right endpoints

n

i

nimi

12210

n

i

niM i

220

2

233

3

1i1i1i

23

1i

23

2

1i1i

3

44

3

832

3

4

2

12

6

1218

128

128212

212

is sumlower theendpoints,left the UsingA.

nnnnn

n

nnnnnn

n

iin

iinnn

i

nn

ifxmfns

nnn

nn

i

nn

i

2

233

31

23

1

2

11

3

44

3

832

3

4

6

12188

2222

is sumupper theendpoints,right the UsingB.

nnnnn

n

nnn

ni

n

nn

i

nn

ifxMfnS

n

i

n

i

n

i

n

ii

2 2

Notice that for any value of the lower sum is less than

8 4 4 8 4 4(or equal to) the upper sum

3 3 3 3The difference between these two sums lessens as increases.

If you take the limit as

n

s n S nn n n n

n

8 , both the upper and lower sum approaches

3This leads us to the Limit of the Lower and Upper Sums Theorem

n

Limit of the Lower and Upper Sums

1 1

Let be continuous and nonnegative on the interval [ ]. The limits as

of both the lower and upper sums exist and are equal to each other. That is,

lim lim limn n

i in n n

i i

f a, b n

f m x f M x

where and and are the minimum and maximum

values of on the th subinterval.i i

S n

x b a n f m f M

f i

Definition of the Area of a Region in the Plane

11

Let be continuous and nonnegative on the interval [ ]. The area of

the region bounded by the graph of , the -axis, and the vertical lines

and is

Area lim , n

i in

i

f a, b

f x

x a x b

f c x x

where

ic x

x b a n

This gives the lower sum.

If the width of each of nn rectangles is x, and the height is the minimumminimum value of f in the rectangle, f(Mi), then the area is the limit of the area of the rectangles as n

Area Under a Curve by Limit DefinitionArea Under a Curve by Limit Definition

Area under a curve by limit Area under a curve by limit definitiondefinition

This gives the upper sum.

If the width of each of nn rectangles is x, and the height is the maximummaximum value of f in the rectangle, f(mi), then the area is the limit of the area of the rectangles as n

Area under a curve by limit Area under a curve by limit definitiondefinition

The limit as n The limit as n of the Upper Sum = of the Upper Sum =

The limit as n The limit as n of the Lower Sum = of the Lower Sum =

The area under the curve between x = a and x = b.The area under the curve between x = a and x = b.

Theorem 4.3 Limits of the Lower and Upper Sums

Definition of the Area of a Region in the Plane

VisualizationVisualization

iithth interval interval

f(cf(cii))

Width = Width = ΔΔxx

ccii

Example: Area under a curve by limit Example: Area under a curve by limit definitiondefinition

Find the area of the region bounded by the graph f(x) = 2x – x3 , the x-axis, and the vertical lines x = 0 and x = 1, as shown in the figure.

Area under a curve by limit definitionArea under a curve by limit definition

Why is right, endpoint Why is right, endpoint i/ni/n??

Suppose the interval from 0 Suppose the interval from 0 to 1 is divided into 10 to 1 is divided into 10 subintervals, the endpoint subintervals, the endpoint of the first one is 1/10, of the first one is 1/10, endpoint of the second one endpoint of the second one is 2/10 … so the right is 2/10 … so the right endpoint of the endpoint of the iithth is is ii/10/10..

Visualization againVisualization again

iithth interval interval

f(cf(cii))

Width = Width = ΔΔx=x=

cci i = i/n= i/n

b a

n

1

3

1

lim ( )

1lim 2

n

in

i

n

ni

Area f c x

i i

n n n

32 4

1 1

2 1lim

n n

ni i

i in n

Find the area of the region bounded by the graph Find the area of the region bounded by the graph f(x) = 2x – xf(x) = 2x – x33 on [0, 1] on [0, 1]

2 2

2 4

2 ( 1) 1 ( 1)lim

2 4n

n n n n

n n

Sum of all the rectanglesSum of all the rectangles

Right endpoint Right endpoint

Sub Sub for for x in f(x)x in f(x)

Use rules of summationUse rules of summation

ii n

c

ii n

c

…continued1

3

1

lim ( )

1lim 2

n

ini

n

ni

Area f c x

i i

n n n

32 4

1 1

2 2lim

n n

ni i

i in n

2

2

( 1) ( 1)lim

4n

n n

n n

2 2

2 4

2 ( 1) 1 ( 1)lim

2 4n

n n n n

n n

2

1 1 1 1lim 1

4 2 4n n n n

1 3

14 4

Foil & SimplifyFoil & Simplify

2

2

1 2 1lim 1

4n

n n

n n

The area of the region bounded by the graph f(x) = 2x – x3 , the x-axis, and the vertical lines x = 0 and x = 1, as shown in the figure = .75

0.750.75

Practice with LimitsPractice with Limits

2

2

1 ( )lim

2n

n n

n

1 1lim 1

2n n

2

1 ( 1)lim

2n

n n

n

1 11 0

2 2

Multiply Multiply outout

SeparateSeparate