Approximation of Integrals Yvonne a. Greenbaun Chap7_Sec7

8/9/2019 Approximation of Integrals Yvonne a. Greenbaun Chap7_Sec7

1/102

7TECHNIQUES OF INTEGRATIONTECHNIQUES OF INTEGRATION


2/102

There are two situations in which it is

impossible to find the exact value of

a definite integral.

TECHNIQUES OF INTEGRATION


3/102


The first situation arises from the fact that,

in order to evaluate using the

Fundamental Theorem of Calculus (FTC),

we need to know an antiderivative of f .

( )b

a f x dx∫


4/102


owever, sometimes, it is difficult, or

even impossible, to find an antiderivative

(!ection 7.").

For example, it is impossible to evaluate

the following integrals exactl#$

21 1 3

0 11 xe dx x dx

−+∫ ∫


5/102


6/102


&n both cases, we need to find

approximate values of definite

integrals.


7/102

7.7

Approximate Integration

&n this section, we will learn$

ow to find approximate values

of definite integrals.



8/102

APPROXIMATE INTEGRATION

'e alread# know one method for

approximate integration.

ecall that the definite integral is defined as

a limit of iemann sums.

!o, an# iemann sum could be used as

an approximation to the integral.


9/102


&f we divide a, b* into n subintervals

of e+ual length x - (b a)/n, we have$

where x i 0 is an# point in the i th subinterval x i 12, x i *.

1

( ) ( *)

nb

ia

i

f x dx f x x=

≈ ∆∑∫


10/102

Ln APPROXIMATION

&f x i 0 is chosen to be the left endpoint of

the interval, then x i 0 - x i 12 and we have$

The approximation Ln is called the left endpointapproximation.

1

1( ) ( )

nb

n ia

i f x dx L f x x−

=≈ = ∆∑∫

E!ation "


11/102

&f f ( x ) 3 4, the integral represents an area

and %+uation 2 represents an approximation

of this area b# the rectangles shown here.

Ln APPROXIMATION


12/102

&f we choose x i 0 to be the right endpoint,

x i 0 - x i and we have$

The approximation R n

is called right endpoint

approximation.

E!ation #

1

( ) ( )nb

n ia

i

f x dx R f x x=

≈ = ∆∑∫

R n APPROXIMATION


13/102


&n !ection ".5, we also considered the case

where x i 0 is chosen to be the midpoint

of the subinterval x i 12, x i *.i x


14/102

M n APPROXIMATION

The figure shows

the midpoint

approximation M n.


15/102

M n APPROXIMATION

M n appears to be better

than either Ln or R n.


16/102

THE MI$POINT RU%E

where

and

1 2

( )

[ ( ) ( ) ... ( )]

b

na

n

f x dx M

x f x f x f x≈= ∆ + + +∫

b a x

n

−∆ =

11 12 ( ) midpoint of [ , ]i i i i i x x x x x− −= + =


17/102

TRAPE&OI$A% RU%E

6nother approximationcalled the

Trape8oidal uleresults from averaging

the approximations in %+uations 2 and 5,

as follows.


18/102

TRAPE&OI$A% RU%E

[

]

[

]

11 1

1

1

0 1 1 2

1

0 1 2

1

1( ) ( ) ( )

2

( ( ) ( ))2

( ( ) ( )) ( ( ) ( ))2

... ( ( ) ( ))

( ) 2 ( ) 2 ( )2

... 2 ( ) ( )

n nb

i iai i

n

i i

i

n n

n n

f x dx f x x f x x

x f x f x

x f x f x f x f x

f x f x

x f x f x f x

f x f x

−= =

−=

−

−

≈ ∆ + ∆ ∆

= +

∆= + + +

+ + +

∆= + +

+ + +

∑ ∑∫ ∑


19/102

THE TRAPE&OI$A% RU%E

where x - (b a)/n and x i - a 9 i x

[

]

0 1 2

1

( )

( ) 2 ( ) 2 ( )2

... 2 ( ) ( )

b

na

n n

f x dx T x

f x f x f x

f x f x−

≈∆

= + +

+ + +

∫


20/102

TRAPE&OI$A% RU%E

The reason for the name can be seen

from the figure, which illustrates the case

f ( x ) 3 4.


21/102

TRAPE&OI$A% RU%E

The area of the trape8oid that lies above

the i th subinterval is$

&f we add the areas of

all these trape8oids,we get the right side of

the Trape8oidal ule.

11

( ) ( )[ ( ) ( )]

2 2

i ii i

f x f x x x f x f x− −

+ ∆ ∆ = + ÷


22/102


6pproximate the integral

with n - ", using$

a. Trape8oidal ule

b. :idpoint ule

Examp'e "2

1

(1/ ) x dx

∫


23/102


'ith n - ", a - 2 and b - 5,

we have$ x - (5 2)/" - 4.5

!o, the Trape8oidal ule gives$

2

51

1 0.2[ (1) 2 (1.2) 2 (1.4)

2

2 (1.6) 2 (1.8) (2)]

1 2 2 2 2 10.1

1 1.2 1.4 1.6 1.8 2

0.695635

dx T f f f x

f f f

≈ = + +

+ + +

= + + + + + ÷

≈

∫

Examp'e " a


24/102


The approximation is illustrated

here.

Examp'e " a


25/102


The midpoints of the five subintervals

are$ 2.2, 2.;, 2.", 2.7, 2.<

Examp'e " (


26/102


!o, the :idpoint ule gives$

2

1

1[ (1.1) (1.3) (1.5)

(1.7) (1.9)]

1 1 1 1 1 1

5 1.1 1.3 1.5 1.7 1.90.691908

dx x f f f x

f f

≈ ∆ + +

+ +

= + + + + ÷

≈

∫

Examp'e " (


27/102


&n %xample 2, we deliberatel# chose

an integral whose value can be computed

explicitl# so that we can see how accurate

the Trape8oidal and :idpoint ules are.

=# the FTC,

2 2

11

1 ln ] ln 2 0.693147...dx x x

= = =∫


28/102

APPROXIMATION ERROR

The error in using an approximation is

defined as the amount that needs to be

added to the approximation to make it

exact.


29/102


From the values in %xample 2, we see that

the errors in the Trape8oidal and :idpoint

ule approximations for n - " are$

E T > 4.445?@@

E M > 4.4425;<


30/102


&n general, we have$

( )

( )

b

T na

b

M na

E f x dx T

E f x dx M

= −

= −

∫

∫


31/102


The tables show the results of calculations

similar to those in %xample 2. owever, these are for n - ", 24, and 54 and for

the left and right endpoint approximations and also

the Trape8oidal and :idpoint ules.


32/102


'e can make several observations

from these tables.


33/102

&n all the methods. we get more accurate

approximations when we increase n. owever, ver# large values of n result in so man#

arithmetic operations that we have to beware ofaccumulated round1off error.

O)SER*ATION "


34/102

O)SER*ATION #

The errors in the left and right endpoint

approximations are$ Apposite in sign

6ppear to decrease b# a factor of about 5

when we double the value of n


35/102

O)SER*ATION +

The Trape8oidal and :idpoint ules

are much more accurate than the endpoint

approximations.


36/102

O)SER*ATION ,

The errors in the Trape8oidal and :idpoint

ules are$ Apposite in sign

6ppear to decrease b# a factor of about ?

when we double the value of n


37/102

O)SER*ATION -

The si8e of the error in the :idpoint ule

is about half that in the Trape8oidal ule.


38/102

MI$POINT RU%E *S. TRAPE&OI$A% RU%E

The figure shows wh# we can usuall# expect

the :idpoint ule to be more accurate than

the Trape8oidal ule.


39/102


The area of a t#pical rectangle in

the :idpoint ule is the same as the area

of the trape8oid ABCD whose upper side is

tangent to the graph at P .


40/102


The area of this trape8oid is closer to

the area under the graph than is the area

of that used in the Trape8oidal ule.


41/102


The midpoint error (shaded red) is smaller

than the trape8oidal error (shaded blue).


42/102

O)SER*ATIONS

These observations are corroborated

in the following error estimateswhich

are proved in books on numerical

anal#sis.


43/102

O)SER*ATIONS

Botice that Abservation ? corresponds

to the n5 in each denominator because$

(5n)5 - ?n5


44/102


That the estimates depend on the si8e of

the second derivative is not surprising if #ou

look at the figure.

f’’ ( x ) measures how much

the graph is curved.

ecall that f’’ ( x ) measures

how fast the slope of y - f ( x )

changes.


45/102

ERROR )OUN$S

!uppose f’’ ( x ) D K for a D x D b.

&f E T and E M are the errors in

the Trape8oidal and :idpoint ules,

then

3 3

2 2( ) ( )and12 24

T M K b a K b a E E

n n− −≤ ≤

Etimate +


46/102

ERROR )OUN$S

Eets appl# this error estimate to the

Trape8oidal ule approximation in %xample 2.

&f f ( x ) - 2/ x , then f’ ( x ) - 12/ x 5 and f’’ ( x ) - 5/ x ;.

6s 2 D x D 5, we have 2/ x ≤ 2G

so,

3 32 2''( ) 2

1 f x

x= ≤ =


47/102

ERROR )OUN$S

!o, taking K - 5, a - 2, b - 5, and n - "

in the error estimate (;), we see$

3

22(2 1) 112(5) 150

0.006667

T E −≤ =

≈


48/102

ERROR )OUN$S

Comparing this estimate with the actual error

of about 4.445?@@, we see that it can happen

that the actual error is substantiall# less than

the upper bound for the error given b# (;).


49/102

ERROR ESTIMATES

ow large should we take n in order to

guarantee that the Trape8oidal and :idpoint

ule approximations for are

accurate to within 4.4442H

Examp'e #

2

1(1/ ) x dx∫


50/102

ERROR ESTIMATES

'e saw in the preceding calculation

that f’’ ( x ) D 5 for 2 D x D 5

!o, we can take K - 5, a - 2, and b - 5 in (;).

Examp'e #


51/102

ERROR ESTIMATES

6ccurac# to within 4.4442 means that

the si8e of the error should be less than

4.4442

Therefore, we choose n so that$3

2

2(1)0.0001

12n<

Examp'e #


52/102

ERROR ESTIMATES

!olving the ine+ualit# for n,

we get

or

Thus, n = ?2 will ensure

the desired accurac#.

2 2

12(0.0001)n >

140.8

0.0006n > ≈

Examp'e #


53/102

ERROR ESTIMATES

&ts +uite possible that a lower value

for n would suffice.

owever, ?2 is the smallest value for whichthe error1bound formula can guarantee us accurac#

to within 4.4442

Examp'e #


54/102

ERROR ESTIMATES

For the same accurac# with the :idpoint ule,

we choose n so that$

This gives$

3

2

2(1)

0.000124n <

129

0.0012n > ≈

Examp'e #


55/102

ERROR ESTIMATES

a. Ise the :idpoint ule with n - 24 to

approximate the integral

b. Jive an upper bound for the error involved

in this approximation.

21

0. xe dx∫

Examp'e +


56/102

ERROR ESTIMATES

6s a - 4, b - 2, and n - 24, the :idpoint ule

gives$

21

0

0.0025 0.0225 0.0625 0.1225 0.2025

0.3025 0.4225 0.5625 0.7225 0.9025

[ (0.05) (0.15) ... (0.85) (0.95)]

0.1

1.460393

xe dx

x f f f f

e e e e e

e e e e e

≈ ∆ + + + +

+ + + +

= + + + + + ≈

∫

Examp'e + a


57/102


58/102

ERROR ESTIMATES

6s f ( x ) - e x 2 , we have$

f’ ( x ) - 5 xe x 2 and f’’ ( x ) - (5 9 ? x 5)e x 2

6lso, since 4 D x D 2, we have x 5 D 2.

ence, 4 D f’’ ( x ) - (5 9 ? x 5) e x 2 D Ke

Examp'e + (


59/102

ERROR ESTIMATES

Taking K - Ke, a - 4, b - 2, and n - 24 in

the error estimate (;), we see that an upper

bound for the error is$3

2

6 (1)

24(10) 400

0.007

e e

=

≈

Examp'e + (


60/102

ERROR ESTIMATES

%rror estimates give upper bounds

for the error.

The# are theoretical, worst1case scenarios.

The actual error in this case turns out to be

about 4.445;


61/102


6nother rule for approximate integration

results from using parabolas instead

of straight line segments to approximate

a curve.


62/102


6s before, we divide a, b* into n subintervals

of e+ual length h - x - (b a)/n.

owever, this time, we assume n is an even number.


63/102


Then, on each consecutive pair of intervals,

we approximate the curve y - f ( x ) 3 4 b#

a parabola, as shown.


64/102


&f y i - f ( x i ), then P i ( x i , y i ) is the point on

the curve l#ing above x i .

6 t#pical parabola passes through

three consecutive points$ P i , P i 92, P i 95


65/102


To simplif# our calculations, we first

consider the case where$

x 4 - 1h, x 2 - 4, x 5 - h


66/102


'e know that the e+uation of

the parabola through P 4, P 2, and P 5

is of the form

y = Ax 5 + Bx + C


67/102


Therefore, the area under the parabola

from x - 1 h to x - h is$

2 2

0

3

0

3

2

( ) 2 ( )

23

23

(2 6 )

3

h h

h

h

Ax Bx C dx Ax C dx

x A Cx

h A Ch

h Ah C

−+ + = +

= +

= + ÷

= +

∫ ∫

O G O


68/102


owever, as the parabola passes through

P 4(1 h, y 4), P 2(4, y 2), and P 5(h, y 5), we have$

y 4 - A( h)5

9 B(1 h) 9 C - Ah5

Bh 9 C y 2 - C

y 5 - Ah5 9 Bh 9 C



69/102


Therefore,

y 4 9 ?y 2 9 y 5 - 5 Ah5 9 KC

!o, we can rewrite the area under

the parabola as$

0 1 2( 4 )

3

h y y y+ +



70/102


Bow, b# shifting this parabola

hori8ontall#, we do not change

the area under it.



71/102


This means that the area under the parabola

through P 4, P 2, and P 5 from x - x 4 to x - x 5

is still$0 1 2( 4 )

3

h y y y+ +



72/102


!imilarl#, the area under the parabola

through P 5, P ;, and P ? from x - x 5 to x - x ?

is$2 3 4( 4 )

3

h y y y+ +



73/102


Thus, if we compute the areas under all

the parabolas and add the results, we get$

0 1 2 2 3 4

2 1

0 1 2 3 4

2 1

( ) ( 4 ) ( 4 )

3 3

... ( 4 )3

( 4 2 4 23

... 2 4 )

− −

− −

≈ + + + + +

+ + + +

= + + + +

+ + + +

∫ b

a

n n n

n n n

h h f x dx y y y y y y

h y y y

h y y y y y

y y y



74/102


Though we have derived this approximation

for the case in which f ( x ) 3 4, it is a

reasonable approximation for an# continuous

function f .

Bote the pattern of coefficients$

2, ?, 5, ?, 5, ?, 5, . . . , ?, 5, ?, 2

SIMPSON/S RU%E


75/102

SIMPSON/S RU%E

This is called !impsons uleafter

the %nglish the %nglish mathematician

Thomas !impson (272427K2).

SIMPSON/S RU%E R '


76/102

SIMPSON/S RU%E

where n is even and x - (b a)/n.

R!'e

0 1 2 3

2 1

( )

[ ( ) 4 ( ) 2 ( ) 4 ( )3

... 2 ( ) 4 ( ) ( )]

b

na

n n n

f x dx S

x f x f x f x f x

f x f x f x− −

≈

∆= + + +

+ + + +

∫

SIMPSON/S RU%E Examp'e ,


77/102

SIMPSON/S RU%E

Ise !impsons ule

with n - 24 to approximate

2

1(1/ ) x dx∫

Examp'e ,

SIMPSON/S RU%E Examp'e ,


78/102

SIMPSON/S RU%E

Lutting f ( x ) - 2/ x , n - 24, and x - 4.2 in

!impsons ule, we obtain$

2

101

1[ (1) 4 (1.1) 2 (1.2) 4 (1.3)

3

... 2 (1.8) 4 (1.9) (2)]

1 4 2 4 2 4 2 4

0.1 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7

2 4 13

1.8 1.9 2

0.693150

xdx S f f f f

x

f f f

∆≈ = + + +

+ + + +

+ + + + + + + ÷

= ÷ ÷+ + + ÷

≈

∫

Examp'e ,

SIMPSON/S RU%E


79/102

SIMPSON S RU%E

&n %xample ?, notice that !impsons ule

gives a much better approximation

(S24 > 4.K


80/102

SIMPSON S RU%E

&t turns out that the approximations in

!impsons ule are weighted averages of

those in the Trape8oidal and :idpoint ules$

ecall that E T and E M usuall# have opposite signsand E M is about half the si8e of E T |.

1 22 3 3n n n

S T M = +

SIMPSON/S RU%E


81/102

SIMPSON S RU%E

&n man# applications of calculus, we need to

evaluate an integral even if no explicit formula

is known for y as a function of x .

6 function ma# be given graphicall# or as

a table of values of collected data.

SIMPSON/S RU%E


82/102

SIMPSON S RU%E

&f there is evidence that the values are not

changing rapidl#, then the Trape8oidal ule

or !impsons ule can still be used to find

an approximate value for .b

a y dx∫

SIMPSON/S RU%E Examp'e -


83/102

SIMPSON S RU%E

The figure shows data traffic on the link from

the I.!. to !'&TC, the !wiss academic and

research network, on Februar# 24, 2


84/102

SIMPSON S RU%E

Ise !impsons ule to estimate the total

amount of data transmitted on the link up to

noon on that da#.

Examp'e -



85/102

SIMPSON S RU%E

!ince we want the units to be consistent

and D(t ) is measured in :b/s, we convert

the units for t from hours to seconds.

Examp'e -



86/102

SIMPSON S RU%E

&f we let A(t ) be the amount of data (in :b)

transmitted b# time t , where t is measured in

seconds, then A’ (t ) - D(t ).

!o, b# the Bet Change Theorem (!ection ".?),

the total amount of data transmitted b# noon

(when t - 25 x K45 - ?;,544) is$

43,200

0(43, 200) ( ) A D t dt = ∫

Examp'e -



87/102

SIMPSON S RU%E

'e estimate the values of D(t ) at hourl#

intervals from the graph and compile them

here.

Examp'e -



88/102

SIMPSON S RU%E

Then, we use !impsons ule

with n - 25 and t - ;K44 to estimate

the integral, as follows.

Examp'e -



89/102

SIMPSON S RU%E

43,2000

( )

[ (0) 4 (3600) 2 (7200)3

... 4 (39, 600) (43, 200)]

3600[3.2 4(2.7) 2(1.9) 4(1.7)

3

2(1.3) 4(1.0) 2(1.1) 4(1.3)

2(2.8) 4(5.7) 2(7.1) 4(7.7) 7.9] 143,880

A t dt

t D D D

D D

∆≈ + +

+ + +

≈ + + +

+ + + +

+ + + + + =

∫

Examp'e -

The total amount of data transmitted up to noon

is 2??,444 :bs, or 2?? gigabits.

SIMPSON/S RU%E *S. MI$POINT RU%E


90/102

SIMPSON S RU%E *S. MI$POINT RU%E

The table shows how !impsons ule

compares with the :idpoint ule for

the integral , whose true value

is about 4.K


91/102

SIMPSON S RU%E

This table shows how the error E s

in !impsons ule decreases b#

a factor of about 2K when n is doubled.

SIMPSON/S RU%E


92/102

SIMPSON S RU%E

That is consistent with the appearance of n?

in the denominator of the following error

estimate for !impsons ule.

&t is similar to the estimates given in (;)

for the Trape8oidal and :idpoint ules.

owever, it uses the fourth derivative of f .

ERROR )OUN$ 0SIMPSON/S RU%E1 Etimate ,


93/102

ERROR )OUN$ 0SIMPSON S RU%E1

!uppose that f (?)( x ) D K for a D x D b.

&f E s is the error involved in using

!impsons ule, then 5

4

( )

180 s

K b a E

n

−≤

Etimate ,

ERROR )OUN$ 0SIMPSON/S RU%E1 Examp'e 2


94/102

0 1

ow large should we take n to guarantee

that the !impsons ule approximation

for is accurate to within 4.4442H2

1(1/ ) x dx∫

Examp'e 2



95/102

0 1

&f f ( x ) - 2/ x , then f (?)( x ) - 5?/ x ".

!ince x 3 2, we have 2/ x D 2, and so

Thus, we can take K - 5? in (?).

(4)

5

24( ) 24 f x

x= ≤

a p e 2



96/102

0 1

!o, for an error less than 4.4442, we should

choose n so that$

This gives

or

5

4

24(1)0.0001

180n<

4 24

180(0.0001)n >

41 6.04

0.00075n > ≈

p



97/102

0 1

Therefore, n - @ (n must be even)

gives the desired accurac#.

Compare this with %xample 5, where we obtained

n - ?2 for the Trape8oidal ule and n - 5< for

the :idpoint ule.

p



98/102

0 1

a. Ise !impsons ule with n - 24

to approximate the integral .

b. %stimate the error involved in thisapproximation.

21

0

xe dx∫

p

ERROR )OUN$ 0SIMPSON/S RU%E1 Examp'e 7 a


99/102

0 1

&f n -24, then x - 4.2 and the rule gives$

21

0

0 0.01 0.04 0.09 0.16

0.25 0.36 0.49 0.64 0.81 1

[ (0) 4 (0.1) 2 (0.2) ...3

2 (0.8) 4 (0.9) (1)]0.1

[ 4 2 4 23

4 2 4 2 4 ]

1.462681

x xe f f f

f f f

e e e e e

e e e e e e

∆≈ + + +

+ + += + + + +

+ + + + + +≈

∫

p

ERROR )OUN$ 0SIMPSON/S RU%E1 Examp'e 7 (


100/102

0 1

The fourth derivative of f ( x ) - e x 5 is$

f (?)( x ) - (25 9 ?@ x 5 9 2K x ?)e x 5

!o, since 4 D x D 2, we have$

4 D f (?)( x ) D (25 9 ?@ 92K)e2 - 7Ke

p



101/102

Lutting K - 7Ke, a - 4, b - 2, and n - 24

in (?), we see that the error is at most$

Compare this with %xample ;.

5

4

76 (1)0.000115

180(10)

e

≈

p



102/102

Thus, correct to three decimal places,

we have$

21

0 1.463 x

e dx ≈∫

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Approximation of Integrals Yvonne a. Greenbaun Chap7_Sec7