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lent form of equation.

INTERPOLATION

Let y^ = A(x-b){x-c)(x-d) . .

. {x-k)

+B(x-a)(x-c)(x-d) . .

. (x-k)

+C{x-a){x-b)(x-d) . .

. (x-k)

+D{x-a){x-b){x-c) . .

. (x-k)

+ etc. to n terms.

105

Each one of the n terms on the right-hand side of the equation

lacks one of the factors x â€” a, x â€” b, x â€” c, xâ€”d, . . . xâ€”k,

and each is affected with an arbitrary constant. The expression

on the right-hand side of the equation is a rational integral

function of x.

Letting x = a gives

and

ya = A{a â€” b)(aâ€”c)(aâ€”d) . . . aâ€”k,

ya

(a â€” b)(a â€” c){a â€” d) . . . a â€” k'

Letting x = b gives

B =

Jb

{b-a){b-c){b-d) . . . {b-k)'

Proceeding in the same way we obtain values for all of the

constants and, finally,

{x â€” b){x â€” c){x â€” d) . . . {x â€” k)

yx=ya

+yi

{a â€” b){a â€” c){a â€” d)

{x â€” a){x â€” c)(x â€” d)

+yc

{b-a){b-c){b-d) .

{x â€” a){x â€” b)(x â€” d) .

+yd

{câ€”a){câ€”b){câ€”d) .

{x â€” a){x â€” b){x â€” c) .

{d-a){d-b){d-c) .

{x â€” a)(x â€” b)(x â€” c) .

'^^\k-a){k-b){k-c) .

. {a-k)

. {x â€” k)

. (b-k)

. {x â€” k)

. ic-k)

. {x â€” k)

. {d-k)

(2)

106

EMPIRICAL FORMULAS

This is called Lagrange's theorem for interpolation.

I. Apply formula (2) to the data given under formula XIX

for finding the value of y corresponding to it: = 0.425. Select

two values on either side of the value required,

6 = .40, yi> =

X in the formula must be taken as o

>' = (-695)

+(.757)

= .744.

i(-i)(-t)

(-i)(-2)(-3)

a)(Â§)(-f)

730,

757,

780.

5-

â– f(.73o)

(2)(i)(-i)

+ (.780)

t(-|)(-f)

(l)(-l)(-2)

(3)(2)(i)

2. Required an approximate value of log 212 from the fol-

lowing data:

log 210 = 2.3222193,

log 211 = 2.3242825,

log 213 = 2.3283796,

log 2 14 = 2.3304138.

(-i)(-3)(-4)^^ ^^ ^\i)(-2)(-3)

log 212 = (2.3222193)-

+ (2.3283796)

(2)(l)(-2)

+ (2.3304138)

(2)(l)(-l )

(4)(3)(i)'

(3)(2)(-i)

= 2.326359.

This is correct to the last figure.

In case the values given are periodic it is better to use a

formula involving circular functions. In Chapter V the approxi-

mate values of the constants in formula XX were derived. This

formula could be used as an interpolation formula. But on

account of the work involved in determining the constants it is

INTERPOLATION

107

much more convenient to use an equivalent one which does not

necessitate the determination of constants.* The equivalent

formula given by Gauss is

y^=y

^-y\

+3^-

+ etc.

sin^(x â€” 6) sin |(x â€” c) . .

. sm\{x-k)

"sin J(a â€” 5) sin^(a â€” c) . .

sin |(a: â€” a) sin Â§(x â€” c) .

. sin ^{a â€” k)

. sin ^{x â€” k)

sin Â§(6 â€” a) sin \{b â€” c) .

sin \{x â€” (i) sin \{x â€” 'b) .

. sin^(x â€” ^)

" sin \{c â€” a) sin \{c â€” h) .

. . dn^{c-k)

(3)

It is evident that the value of ja is obtained from this formula

by putting x = a. The value of Jd is obtained by putting x = b,

and jc by putting x = c.

The proof that (3) is equivalent to XX need not be given

here.

Let it be required to find an approximate value of y cor-

responding to it: = 42Â° from the values given.

From (3)

>' = (io.i)

X

y

30Â°

10. 1

40"

9.8

50"

8.5

sin iÂ°sin (^4Â°)

sin(-sÂ°)sin(-ioÂ°)

f(9-8)

+ (8.5)

sin 6Â° sin ( â€” 4Â°)

sin 5Â° sin (-5Â°)

sin 6Â° sin 1Â°

sin 10Â° sin 5Â°

- rrn T^i:^II5)C^698) , . (.1 045) (.0698)

(.1045) (.01 75)

(.i736)(.o872)

= 9.618.

+ (8.5)

* Trigometrische Interpolation, Encyklopadie der Mathematischen

Wissenchaften, Vol. II, pt. I, pp. 642-693.

108 EMPIRICAL FORMULAS

A better result would have been obtained by using four sets

of values.

Differentiation of Tabulated Functions

It is frequently desirable to obtain the first and second

derivatives of a tabulated function to a closer approximation

than graphical methods will yield. For that purpose we will

derive differentiation formulas from (i) and (2). From

^ x{x-i){x-2){x-s) ^^^^^^

k

By differentiating it follows that

, ^ , 2:1:â€” 1 .0 , sx^ â€” 6x+2^^

yx=Ayo-\ â€” â€”A^yo+^ â€” I A^yo

^ 4^-i2.M-...-6 ^,^^^ (^)

Differentiating again

yJ'=A^yo+(x-i)A^yo+(ix^-x+H)A'yo+ ..... . . (s)

As an illustration let it be required to find the first and

second derivatives of the function given in the table below and

determine whether the series of observations is periodic*

The consecutive daily observations of a function being

0.099833, 0.208460, 0.314566, 0.416871, 0.514136, 0.605186,

0.688921, 0.764329, show that the function is periodic and deter-

mine its period.

* Interpolation and Numerical Integration, by David Gibb.

INTERPOLATION

109

From the given observations the following table may be

ttpn !

written:

y=f(x)

0-099833

0.208460

0.314566

0.416871

0.514136

6 0.605186

7 0.688921

8 0.764329

0.108627

0.106106

0.102305

(-)o.097265

(-)o.o9io5o

(-)o.o83735

(-)o.o754o8

From (4)

y\= .108627

.001260

. 109887

.000437

. 109450

â€” .000427

â€” .000010

-.000437

y3= .102305 -.000392

.002520 â€” .000019

.104825

.000411

. 104414

.000411

.002521

.003801

.005040

.006215

.007315

.008327

^2

A'

â€” .001280

( + ) -.001239

(+)-.ooii75

(+) â€” .001100

(+) â€” .001012

A*

.000041

.000064

.000075

.000088

.106106

.001900

. 108006

â€” .000429

.107577

.097265

.003108

.100373

â€” .000389

.099984

â€” .000413

â€” .000016

â€” .000429

.000367

.000022

.000389

For the remaining first derivatives the order must be reversed

and the resulting sign changed.

^5=- .097265 .002520 ^6= â€” .091050 .003108

,000010 .000413 â€” .000016 .000392

097275

002933

002933

094342

â€” .091066

â€¢003500

.087566

â€¢003500

110

EMPIRICAL FORMULAS

/7=". 083735

,000019

083754

004025

,003658

.000367

004025

â€” .006279

.000038

â€” .006241

y'7=-. 007315

â€” .001100

â€” .008415

.000069

â€” . 008346

y8=-.o754o8

â€” .000022

.079729

From (5)

y'i = -. 002521

.001318

.001280

.000038

.001318

.001175

.000069

.001244

y'2

â€” .001203

/';,= -. 005040

.001244

y'4

-.003796

y"r, = - . 005040

-.001239

y\

-075430

.004501

.070929

y 8

003801

001298

002503

006215

001181

005034

006215

001175

007390

000059

007331

008327

001012

009339

000081

009258

.004164

,000337

004501

.001239

,000059

001298

OOIIOO

,000081

,001181

X

y

y

y"

yi

y

I

.099833

. 109450

â€” .001203

â€” .0121

2

. 208460

.107577

â€” .002503

â€” .0120

3

.314566

. 104414

-.003796

â€” .0121

4

.416871

.099984

-.005034

â€” .0121

INTERPOLATION

111

x

y

/

/'

y

5

.514136

â€¢094342

.006241

â€” .0121

6

.605186

.087566

â€¢007331

â€” .0121

7

.688921

.079729

.008346

â€” .0121

8

.764329

.070929 â€”

.009258

â€” .0121

Since â€” is very nearly constant and equal to â€”.0121, the

y

corresponding differential equation is

y'^-{-.oi2iy = o,

whose solution is

;y = ^ coso.iix+^sino.iix.

This shows that y is a, period function of x, and its period is

27r J

, or 57.12 days.

o.ii

Convenient formulas for the first and second derivatives may

also be obtained by differentiating Lagrange's formula for inter-

polation. Using five points the formula is

_ (x â€” b){xâ€”c)(xâ€”d)(x â€” e) (x â€” a)(xâ€”c)(xâ€”d)(xâ€”e)

^"~^" (a-b){a-c)(a-d)(a-ey^' {b-a){b-c){b-d){h-e)

{x â€” a){x â€” b){x â€” d){x â€” e) {x â€” a){x â€” b){x â€” c){x â€” e)

â– ^^^ {c-a){c-b){c-d){c-e) "^^' {d-a){d-b){d-c){d-e)

{x â€” a){x â€” b){x â€” c){x â€” d) . .

'^^' {e-a){e-b){e-c){e-d) ^^^

Selecting the points at equal intervals and letting

eâ€”d = dâ€”c = c â€” b = b â€” a = hj

and differentiation

ya=-^l-2sya+4^yb-3^yc-\-i6ya-sye],

y'h = â€” rl- 3>'a-io>+i8>'c- 63;^+ y^,

\2h

112

EMPIRICAL FORMULAS

yc = ^[ ya- ^yi> + 8>'d-3'e],

yd = ^[- >'a+ 6>-i83;c+io>'d+3>'e],

Differentiating again

y"a = j;^[35>'a - 1043^6+ 1 i4>'c - 56>'d+ 1 1}'.],

/'&= â€” jrh^^ya- 2o>'ft+ 63^0+ 4yd-ye],

12nr

y'c = ^[->'a+ l6>- 3OJC+ l63;d->'e],

yd = â€” pl-^^ 43^*+ 63'c- 20>'d+II>],

3''e = -^[ii3'a- 56>'6+ii4>'c-i04>'d+35>'J-

The results of applying these formulas to the function given

are expressed in the table below.

X

y

/

r

I

â€¢099833

.109451

.001203

2

. 208460

â€¢ 107583

002524

3

.314566

.104415 -

.003804

4

.416871

.099986 -

.005045

5

â€¢514136

â€¢094347

.006221

6

.605186

.087568 -

007322

7

.688921

â€¢079733

008334

8

.764329

^A r ^ ? 1

.070929

009258

1 1

These results agree fairly well with those previously obtained.

It is probable that the formulas derived from the interpolation

formula give the most satisfactory results.

INTERPOLATION

113

As another application let us find the maximum or minimum

value of a function having given three values near the critical

point.

Let ya, jb, and jc be three values of a function of x near

its maximum or minimum corresponding to the values of Xy

a, b, and c respectively.

From (2)

_ (x â€” b)(xâ€”c) (x â€” a)(x â€” c) (x â€” a)(x â€” b)

^'"^" (a-b){a-cy^\b-a){b-cy^' {c-a){c-b)'

Equating to zero the first derivative with respect to x

2x â€” a â€” b

2X â€” b â€” c

y'^ 7 â€” TVT â€” ^+>

2x â€” a-

^yc

{a-b){a-c) ' '" {b-a){b-c) ' â– " {c-a){c-b)

_ yajb^ - c^) +>(c^ - a^) -\-ycia^ - b^)

=0;

(6)

2[ya{b-c)-\-yb{c-a)-\-yc{a-b)]

This is equivalent to drawing the parabola

y=A+Bx-\-Cx^

through the three points and determining its maximum or

minimum.

From the table of values

6.0

6.5

7.0

10.05

10.14

10.10

the abscissa of the maximum point is found from (6).

(io.o5)(-6.75) + (io-i4)(i3) + (iaio)(-6.25) ^

2[(io.o5)(-.5) + (io.i4)(i) + (io.io)(-.5) ''â€¢^^''

3; = 10.1424.

CHAPTER VIII

NUMERICAL INTEGRATION

Areas

An area bounded by the curve, y=J{x), the axis of x, and

two given ordinates is represented by the definite integral

r

ydx,

where the ordinates are taken at :r = a and x = n. It may be

said that the definite integral represents the area under the

curve, or that the area under the curve represents the value of

the definite integral.

If a function is given by its graph, it is possible, by means

of the planimeter, to find roughly the area bounded by the curve,

two given ordinates and the x â€” axis, or, what amounts to the

same thing, the area enclosed by a curve. This method is used

in finding the area of the indicator diagrams of steam, gas or

oil engines, and various other diagrams. The approximations

in these cases are close enough to satisfy the requirements.

If, however, considerable accuracy is sought, or whenever

the function is defined by a table of numerical values another

method must be employed.

Mechanical Quadrature or Numerical Integration is the method

of evaluating the definite integral of a function when the func-

tion is given by a series of numerical values. Even when the

function is defined by an analytical expression but which can-

not be integrated in terms of known functions by the method

of the integral calculus, numerical integration must be resorted

to for its evaluation.

The formulas employed in numerical integration are derived

from those established for interpolation.

114

NUMERICAL INTEGRATION 115

In interpolation it was found that the order of differences

which must be taken into account depends upon the rapidity

with which the differences decrease as the order increases.

This is also true of numerical integration. It is the same as

saying that if the series employed does not converge the process

will give unsatisfactory results. An illustration will be given

later.

Formulas for numerical integration will be derived from (i)

of Chapter VII.

In this formula it was assumed that the ordinates are given

at equal intervals.

, ^ .x(xâ€”l).o , x{xâ€”l){x â€” 2) .o

yx=yo-{-xAyo+^ -A^yo+ â€” , ^A^yo

11 11

^ x(x-i){x-2)(x-s) ^. ^ x{x-i)(x-2)(x-s){x-4) ^.^

^ x(x-i){x-2){x-s){x-4){x-s) ^, J ^j>j

[6

Integrating the right-hand member,

\ ydx=yQ I dx-\-Ayo\ xdx-\ â€” p- | x{x â€” i)dx

Jo Jo \2_J

+^Vx{x-l){x-2)dx

b Jo

^^ C\{^^-^){0c-2){x-^)dx

k Jo

H-^ \x{x-i){x-2){x-i){x-4)dx

15 Jo

\ x{x-i){x-2){x-^){x-4){x-s)dx-\- . . .

=nyo-\ â€” Ayo+ -r^+ n^-^n^ ) - -

2 \3 2/ |2 V4 / |3

â– "|6

\5 2 3 /

A^o

k

116 EMPIRICAL FORMULAS

\6 4 3 / |5

\7 2 ' 4 3 / P

The data given in any particular problem will enable us

to compute the successive differences of ^'o up to A">'o. On the

assumption that all succeeding differences are so small as to

be negligible the above formula gives an approximate value

of the integral. It is only necessary to assign particular values

tow.

Let w = 2, then

I yxdx = 2yo+2Ayo+iA^yoj

Ayo=yi-yo,

A^yo = Ayi - Ayo = y2-yi-yi +yo,

=y2-2yi+yo.

Substituting these values in the above integral it becomes

I y2dx = 2yo-]-2yi-2yo+iy2-iyi+^yo,

_yo+4yi+y2

This is equivalent to assuming that the curve coincides with

a parabola of the second degree.

If the common distance between the ordinates is h, the

value becomes

'2h

ydx=U(yo+4yi+y2) (7)

X

IfÂ»=3

ydx = syo+^Ayo+iA^yo+i^^yoj

Ayo=yi-yo,

A^yo = Ayi-Ayo=y2- 2yi +yoy

i

NUMERICAL INTEGRATION 117

=yz-2>y2+2>yi-yo'

Substituting these values in the equations,

1 ydx = 7^yo-\-^yi-^yQ+ly2-hi+ho-\-h^-h2+hi-hoj

= l>'0 + ljl+f3'2+|>'3,

= l(3'o+33'i+3>'2+3'3).

If the common distance between the ordinates is h the

formula becomes

â€¢3ft

X

= iKyQ-\-^yi-{-2>y2+y^) (8)

This is equivalent to assuming that the curve coincides with

a parabola of the third degree.

If there are five equidistant ordinates, h representing the

distance between successive ordinates

/â€¢4ft

Jo ^

yd^ - 14(3^0+^4) +64(3^1 +^3) +24y2 ^^^ ... (9)

b 45

If the area is divided into six parts bounded by seven equi-

distant ordinates the integral becomes

i

6

ydx = 6>'o + 1 ^^yo + 2 7 A^^yo + 24 A^^yo + ^-L^yo

+ffA5>'o+AVA63;o.

Sine 3 the last coefficient, 1^, differs but slightly from t^

and by the assumption that A^^'o is small the error will be slight

if the last coefficient is replaced by tq-

Doing this and replacing

^yohyyi-yo,

^^yohyy2-2yi+yQ,

A^yo by yd-;^y2+^yi -yo,

118

EMPIRICAL FORMULAS

A*yo by yA-4ys+6y2 -4yi+yo,

A^yo by y5-5>'4H-ioy3-io>'2+53'i-3'o,

A^^'o by / - 6>'5 -f- 1 5>'4 - 2oy3 + 1 53'2 - 6yi +yo,

gives the formula

X

ydx = TV/b'o+>'2+>'4+3'6+5(>'i +>'5) +6>'3]. . (lo)

The application of these

formulas is illustrated by

finding the area in Fig. 27.

Fig. 27.

By (7)

A=ih(yo+4yi + 2y2-\-4y3 + 2y4:+4y5+2ye+4y7+2y8+4y9

-\-2yio+4yn+yi2)'

By (8)

^=l^(yo+3>'i+3>'2 + 2y3+3>'4+3>'5+2y6+3>'7+33'8+2y9

+3>'io+33'ii+>'i2).

By (9)

A=-hhli4(yo+2y^-[-2y8+yi2)+64(yi-\-y3+y5+y7+y9+yii)

+ 24(y2+yQ-\-yio)].

By (10)

A=^hlyo+y2+y4.+2y6-\-y8+yio-\-yi2+5(yi+y5+y7+yii)

+6(^3 +^9)].. .

I. A rough comparison of the approximations by the use

of these formulas will be obtained by finding the value of

'ax

Jn

â€” . The value of this definite integral is log 13 = 2.565. It

I X

is also equal to the area under the curve

I

y=-

X

NUMERICAL INTEGRATION 119

from x = i to a: -=13. Dividing the area up into 12 strips of

unit width by 13 ordinates the corresponding values of x and

y are

X

I

2

3

4

5

6

7

8

9

10

II

12

13

y

I

1

2

i

i

i

i

1

T

1

i

A

tV

A

tV

By (7)

^ =Mi+2+i+i+f +1 +f +Ht+f +A+i+Ty

= 2.578, error .5%;

By (8), ^=2.585, error .8%;

By (9), ^ = 2.573, error .3%;

By (10), A = 2.572, error .3%.

2. The accuracy of the approximation is much increased

by taking the ordinates nearer together, as is shown by the

following evaluation of

^^ dx^

h 1+^

X'

The value of this integral is equal to the area under the

curve

I

i-\-x

from x=o tG x = i. Dividing the area into twelve parts by

dx

thirteen equidistant ordinates the value of | is found to be

Jo I ' "

By (7), 0.69314866, error 0.00000148;

By (8), 0.69315046, error 0.00000328;

By (9), 0.69314725, error 0.00000007;

By (10), 0.69314722, error 0.00000004.

The correct value is, of course, loge 2, which is 0.69314718.

Formulas (7) and (8) are Simpson's Rules, (10) is Weddle's

Rule.

120

EMPIRICAL FORMULAS

3. Apply the above formulas to the area of that part of the

semi-ellipse included between the two perpendiculars erected at

the middle points of the semi-major axes. Let this area be

divided into twelve parts by equidistant ordinates.

Since the equation of the ellipse is

these ordinates are

Ws b, ^V7^ b, iVs b, iVYs b, iVJ~s b, tVVm3 b, b,

t^jVh3 ^, WYs b, \y/Vs b, JVS 6, ^V^ b, iV3 b.

By (7), .4=0.9566099^6;

By (8), ^=0.956608006;

By (9), ^=0.9566x1406;

By (10), A =0.9566x1406.

The correct value to seven

places is 0.9566x1506.

In the application of these

formulas it is highly desirable

to avoid large dififerences among

the ordinates. For that reason

the formulas do not give so

good results when applied to

the quadrant of the ellipse.

4. The area under the curve,

Fig. 28, determined by the

following sets of values:

yi

3.0

2.5

2.0

1.5

y

â€” ,

>v

/

Y

N

^

/

/

0.5

A

X I

A .6 .8 1.0 1.2 Â»

Fig. 28.

) .2 .4

x.o

1.2

1.5

2.2

2.7

2.6

2.3

y \ i-o

is by (7)

^ =i - 5-(i-o+6.o+4.4+xo.8+5. 2+9.2 + 2. x) = 2.58,

and by (8),

.4 =|-i(x.o+4.5+6.6+5.4+7.8+6.9+2.x) = 2.5725.

2.1

NUMERICAL INTEGRATION

121

X1.2

ydx.

The area found is therefore the approximate value of this

integral

5. Find the area under the curve determined by the points

2.8 3.2 3.6 4.0 4.6 4.8

XI 1.5 1.9 2.3

5-0

y\o .40 1.08 1.82 2.06 2.20 2.30 2.25 2.00 1.80 1.5

The points located by the above sets of values are plotted

in Fig. 29 and a smooth curve drawn through them. The area

y

^'T~^=T^

2 â– - Jr\

L ~^

"2

^

/

A

V

J

-Ai

~l

1

* f

t

/

/

^

y"

â€” 1 â€” 1 â€” 1 1 X

12345

Fig. 29.

is divided into strips each having a width of .4. Rectangles

are formed with the same area as the corresponding strips.

The eye is a very good judge of the position of the upper bound-

ary of each rectangle. Adding the lengths of these rectangles

and multiplying the sum by .4 the area is found to be 6.644.

By Simpson's Rule, formula (7), are found

for h = .2,

^=6.639,

^=6.645.

The graphical determination of areas can be made to yield

a close approximation by taking narrow strips, and where the

points are given at irregular intervals the area can be found

more rapidly than by the application of Simpson's Rules.

122 EMPIRICAL FORMULAS

6. A gas expands from volume 2 to volume 10, so that its

pressure p and volume v satisfy the equation pv = ioo. Find

the average pressure between v = 2 and 7; = 10.

The average pressure is equal to the work done divided by

8. The work is equal to the area under the curve p = ^^ from

V

v = 2 to z; = 10, which is

n^ 100 r 1 ^Â°

I â€” dv = 100 log V = 160.044.

J2 V L J2

That this area represents the work done in expanding the

volume from 2 to 10 becomes evident in the following way.

Let s represent the surface inclosing the gas, ps will then be

the total pressure on that surface. The element of work will

then be

dW = psdn,

when dn lepresents the element along the normal.

PF = j psdn.

But

sdn = dv,

and

W^j'pdv.

This is the equation above. The average pressure over the

change of volume from 2 to 10 is

160.944 -^ 8 = 20. 1x8.

7. Find the mean value of sin^ x from x = o to ii; = 27r. Plot

the curve y = sin^ x by the following values of x and y :

TT

12

TT

6

TT

4

TT

3

12

TT

2

.0670

.2500

.5000

.7500

â€¢9330

I. 0000

11

12

27r

3

31

4

6

IITT

12

9330 -7500 .5000 .2500 .0670

NUMERICAL INTEGRATION

123

X

IT

12

77r

6

4

47r

3

12

31

2

y

o

.0670

.2500

.5000

.7500

â€¢9330

I. 0000

X

IQTT

12

Si

3

Ttt

4

IITT

6

2_3Z

12

27r

â€¢9330 -7500 .5000 .2500 .0670 o

Applying Simpson's Rule, formula (7), the area is found to

be TT. The mean value is the area divided by 27r or .5.

8. A body weighing 100 lb. moves along a straight line

without rotating, so that its velocity v at time t is given by the

following table:

/sec

V ft./sec

1.47 1.58 . 1.67 1.76 1.86

Find the mean value of its kinetic energy from t = i to / = 9.

t

I

3

5

7

9

v^

2.1609

2.4964

2.7889

3.0976

3-4596

Kinetic energy .

3-355

3.876

4-331

4.810

5-372

Plotting kinetic energy to /, the area under the curve is

34.755. This divided by 8 gives the mean kinetic energy as

4-357-

Volumes Ai/

Fig. 30 explains the ap-

plication of the formulas

to the problem of finding

the approximate volume of

an irregular figure. Tiie area

of the sections at right angles

to the axis of x are :

Ai=\k{yi-\-^y5-^y^),

^2=i^(3'6 + 4>'9+3'8),

Az=\k{y2+Ay7+y^)' Fig. 30.

124 EMPIRICAL FORMULAS

If the areas of these sections be looked upon as ordinates,

h being the distance between two adjacent ones, it is evident

that the volume may be represented by the area under the

curve drawn through the extremities of these ordinates.

Substituting the values of A\, A2, and A3 in this equation,

the volume becomes

V = ih[ik{yi-{-4y5+y4)+ik(y6-^4yc)-\-y8)+lHy2-\-4y7-\-y3)]

= iM[yi +>'2 +>'3 +3'4 +4(75 +>'G+>'7 +y8) + 16^9]

In order to apply formulas (8), (9) and (10), the solid would

have to be divided differently, but the method of application

is at once evident from the above and needs no further discussion.

1. The following are values of the area in square feet of the

cross-section of a railway cutting taken at intervals of 6 ft.

How many cubic feet of earth must be removed in making the

cutting between the two end sections given?

91, 95, 100, 102, 98, 90, 79.

These cross-section -areas were obtained by the application

of Simpson's Rules.

By (7),

F = | â€¢ 6(91 -1-380 -f 200 -f 408 + 196 H-36o-f 79) =3428;

By (8),

7 = 1-6(91 + 285+300 -F204-f 294 -f-27o-f79) =3426.8.

2. A is the area of the surface of the water in a reservoir

when full to a depth h.

h ft I 30 25 20 15 10 5 o

^ sq.ft. . 1 26,700 22,400 19,000 16,500 14,000 10,000 5,000

NUMERICAL INTEGRATION 125

- Find (a) the volume of water in the reservoir, (b) the work

done in pumping water out of the reservoir to a height of loo ft.

above the bottom until the remaining water has a depth of

lO ft.

F = f (26, 700 +89, 600 +38,000 +66,000 +28,000+40,000+5,000)

= 488,833 cu. ft.

Work = 'ZÂ£; ( A(ioo â€” h)dh, where ze; = weight of i cu.ft. of water

Jio

= 62.3 lb. The value of this integral will be approximately the

area under the curve determined by the points

30 25 20 15 10

^(100 â€” /?) -i 1,869,000 1,680,000 1,520,000 1,402,500 1,260,000

multiplied by 62.3.

This area is equal to

1(1,869,000+6,920,000+3,040,000+5,610,000 + 1,260,000)

= 31,165,000.

Multiplying this by 62.3 gives the work equal to 1,941,579,500

ft.-lb.

3. When the curve in Fig. 29 revolves about the ::i;-axis,

find the volume generated.

The areas of the cross-sections corresponding to the given

values of x are given in the following table:

I .0 1.2

TT 2.257r 4.847r 7.29^ 6.76^ 5 â€¢ 29^ 4.41^

By (7) F = 5.8627r = 18.416.

By (8) F = 5.8o37r = 18.231.

4. When the curve in Fig. 30 revolves about the x-axis,

find the volume generated from x=i tox = 4.2. From the curve

the following sets of values are obtained :

126 EMPIRICAL FORMULAS

X

i.o 1.2 1.4 1.6

1.8

2.0

2.2

2.4

2.6

y

.11 .29 .53

.87

1.37

1. 71

1.90

2.01

X

.012 .084 .281

2.8 3.0 3.2

â€¢757

3-4

1.877

3.6

2.924

3.8

3.610

4.0

4.040

4.2

y_

2 . 06 2.12 2.2

2.27

2.30

2.28

2.25

2.20

f 4.2444.4944.84 5.153 5.290 5.198 5.062 4.84

INTERPOLATION

Let y^ = A(x-b){x-c)(x-d) . .

. {x-k)

+B(x-a)(x-c)(x-d) . .

. (x-k)

+C{x-a){x-b)(x-d) . .

. (x-k)

+D{x-a){x-b){x-c) . .

. (x-k)

+ etc. to n terms.

105

Each one of the n terms on the right-hand side of the equation

lacks one of the factors x â€” a, x â€” b, x â€” c, xâ€”d, . . . xâ€”k,

and each is affected with an arbitrary constant. The expression

on the right-hand side of the equation is a rational integral

function of x.

Letting x = a gives

and

ya = A{a â€” b)(aâ€”c)(aâ€”d) . . . aâ€”k,

ya

(a â€” b)(a â€” c){a â€” d) . . . a â€” k'

Letting x = b gives

B =

Jb

{b-a){b-c){b-d) . . . {b-k)'

Proceeding in the same way we obtain values for all of the

constants and, finally,

{x â€” b){x â€” c){x â€” d) . . . {x â€” k)

yx=ya

+yi

{a â€” b){a â€” c){a â€” d)

{x â€” a){x â€” c)(x â€” d)

+yc

{b-a){b-c){b-d) .

{x â€” a){x â€” b)(x â€” d) .

+yd

{câ€”a){câ€”b){câ€”d) .

{x â€” a){x â€” b){x â€” c) .

{d-a){d-b){d-c) .

{x â€” a)(x â€” b)(x â€” c) .

'^^\k-a){k-b){k-c) .

. {a-k)

. {x â€” k)

. (b-k)

. {x â€” k)

. ic-k)

. {x â€” k)

. {d-k)

(2)

106

EMPIRICAL FORMULAS

This is called Lagrange's theorem for interpolation.

I. Apply formula (2) to the data given under formula XIX

for finding the value of y corresponding to it: = 0.425. Select

two values on either side of the value required,

6 = .40, yi> =

X in the formula must be taken as o

>' = (-695)

+(.757)

= .744.

i(-i)(-t)

(-i)(-2)(-3)

a)(Â§)(-f)

730,

757,

780.

5-

â– f(.73o)

(2)(i)(-i)

+ (.780)

t(-|)(-f)

(l)(-l)(-2)

(3)(2)(i)

2. Required an approximate value of log 212 from the fol-

lowing data:

log 210 = 2.3222193,

log 211 = 2.3242825,

log 213 = 2.3283796,

log 2 14 = 2.3304138.

(-i)(-3)(-4)^^ ^^ ^\i)(-2)(-3)

log 212 = (2.3222193)-

+ (2.3283796)

(2)(l)(-2)

+ (2.3304138)

(2)(l)(-l )

(4)(3)(i)'

(3)(2)(-i)

= 2.326359.

This is correct to the last figure.

In case the values given are periodic it is better to use a

formula involving circular functions. In Chapter V the approxi-

mate values of the constants in formula XX were derived. This

formula could be used as an interpolation formula. But on

account of the work involved in determining the constants it is

INTERPOLATION

107

much more convenient to use an equivalent one which does not

necessitate the determination of constants.* The equivalent

formula given by Gauss is

y^=y

^-y\

+3^-

+ etc.

sin^(x â€” 6) sin |(x â€” c) . .

. sm\{x-k)

"sin J(a â€” 5) sin^(a â€” c) . .

sin |(a: â€” a) sin Â§(x â€” c) .

. sin ^{a â€” k)

. sin ^{x â€” k)

sin Â§(6 â€” a) sin \{b â€” c) .

sin \{x â€” (i) sin \{x â€” 'b) .

. sin^(x â€” ^)

" sin \{c â€” a) sin \{c â€” h) .

. . dn^{c-k)

(3)

It is evident that the value of ja is obtained from this formula

by putting x = a. The value of Jd is obtained by putting x = b,

and jc by putting x = c.

The proof that (3) is equivalent to XX need not be given

here.

Let it be required to find an approximate value of y cor-

responding to it: = 42Â° from the values given.

From (3)

>' = (io.i)

X

y

30Â°

10. 1

40"

9.8

50"

8.5

sin iÂ°sin (^4Â°)

sin(-sÂ°)sin(-ioÂ°)

f(9-8)

+ (8.5)

sin 6Â° sin ( â€” 4Â°)

sin 5Â° sin (-5Â°)

sin 6Â° sin 1Â°

sin 10Â° sin 5Â°

- rrn T^i:^II5)C^698) , . (.1 045) (.0698)

(.1045) (.01 75)

(.i736)(.o872)

= 9.618.

+ (8.5)

* Trigometrische Interpolation, Encyklopadie der Mathematischen

Wissenchaften, Vol. II, pt. I, pp. 642-693.

108 EMPIRICAL FORMULAS

A better result would have been obtained by using four sets

of values.

Differentiation of Tabulated Functions

It is frequently desirable to obtain the first and second

derivatives of a tabulated function to a closer approximation

than graphical methods will yield. For that purpose we will

derive differentiation formulas from (i) and (2). From

^ x{x-i){x-2){x-s) ^^^^^^

k

By differentiating it follows that

, ^ , 2:1:â€” 1 .0 , sx^ â€” 6x+2^^

yx=Ayo-\ â€” â€”A^yo+^ â€” I A^yo

^ 4^-i2.M-...-6 ^,^^^ (^)

Differentiating again

yJ'=A^yo+(x-i)A^yo+(ix^-x+H)A'yo+ ..... . . (s)

As an illustration let it be required to find the first and

second derivatives of the function given in the table below and

determine whether the series of observations is periodic*

The consecutive daily observations of a function being

0.099833, 0.208460, 0.314566, 0.416871, 0.514136, 0.605186,

0.688921, 0.764329, show that the function is periodic and deter-

mine its period.

* Interpolation and Numerical Integration, by David Gibb.

INTERPOLATION

109

From the given observations the following table may be

ttpn !

written:

y=f(x)

0-099833

0.208460

0.314566

0.416871

0.514136

6 0.605186

7 0.688921

8 0.764329

0.108627

0.106106

0.102305

(-)o.097265

(-)o.o9io5o

(-)o.o83735

(-)o.o754o8

From (4)

y\= .108627

.001260

. 109887

.000437

. 109450

â€” .000427

â€” .000010

-.000437

y3= .102305 -.000392

.002520 â€” .000019

.104825

.000411

. 104414

.000411

.002521

.003801

.005040

.006215

.007315

.008327

^2

A'

â€” .001280

( + ) -.001239

(+)-.ooii75

(+) â€” .001100

(+) â€” .001012

A*

.000041

.000064

.000075

.000088

.106106

.001900

. 108006

â€” .000429

.107577

.097265

.003108

.100373

â€” .000389

.099984

â€” .000413

â€” .000016

â€” .000429

.000367

.000022

.000389

For the remaining first derivatives the order must be reversed

and the resulting sign changed.

^5=- .097265 .002520 ^6= â€” .091050 .003108

,000010 .000413 â€” .000016 .000392

097275

002933

002933

094342

â€” .091066

â€¢003500

.087566

â€¢003500

110

EMPIRICAL FORMULAS

/7=". 083735

,000019

083754

004025

,003658

.000367

004025

â€” .006279

.000038

â€” .006241

y'7=-. 007315

â€” .001100

â€” .008415

.000069

â€” . 008346

y8=-.o754o8

â€” .000022

.079729

From (5)

y'i = -. 002521

.001318

.001280

.000038

.001318

.001175

.000069

.001244

y'2

â€” .001203

/';,= -. 005040

.001244

y'4

-.003796

y"r, = - . 005040

-.001239

y\

-075430

.004501

.070929

y 8

003801

001298

002503

006215

001181

005034

006215

001175

007390

000059

007331

008327

001012

009339

000081

009258

.004164

,000337

004501

.001239

,000059

001298

OOIIOO

,000081

,001181

X

y

y

y"

yi

y

I

.099833

. 109450

â€” .001203

â€” .0121

2

. 208460

.107577

â€” .002503

â€” .0120

3

.314566

. 104414

-.003796

â€” .0121

4

.416871

.099984

-.005034

â€” .0121

INTERPOLATION

111

x

y

/

/'

y

5

.514136

â€¢094342

.006241

â€” .0121

6

.605186

.087566

â€¢007331

â€” .0121

7

.688921

.079729

.008346

â€” .0121

8

.764329

.070929 â€”

.009258

â€” .0121

Since â€” is very nearly constant and equal to â€”.0121, the

y

corresponding differential equation is

y'^-{-.oi2iy = o,

whose solution is

;y = ^ coso.iix+^sino.iix.

This shows that y is a, period function of x, and its period is

27r J

, or 57.12 days.

o.ii

Convenient formulas for the first and second derivatives may

also be obtained by differentiating Lagrange's formula for inter-

polation. Using five points the formula is

_ (x â€” b){xâ€”c)(xâ€”d)(x â€” e) (x â€” a)(xâ€”c)(xâ€”d)(xâ€”e)

^"~^" (a-b){a-c)(a-d)(a-ey^' {b-a){b-c){b-d){h-e)

{x â€” a){x â€” b){x â€” d){x â€” e) {x â€” a){x â€” b){x â€” c){x â€” e)

â– ^^^ {c-a){c-b){c-d){c-e) "^^' {d-a){d-b){d-c){d-e)

{x â€” a){x â€” b){x â€” c){x â€” d) . .

'^^' {e-a){e-b){e-c){e-d) ^^^

Selecting the points at equal intervals and letting

eâ€”d = dâ€”c = c â€” b = b â€” a = hj

and differentiation

ya=-^l-2sya+4^yb-3^yc-\-i6ya-sye],

y'h = â€” rl- 3>'a-io>+i8>'c- 63;^+ y^,

\2h

112

EMPIRICAL FORMULAS

yc = ^[ ya- ^yi> + 8>'d-3'e],

yd = ^[- >'a+ 6>-i83;c+io>'d+3>'e],

Differentiating again

y"a = j;^[35>'a - 1043^6+ 1 i4>'c - 56>'d+ 1 1}'.],

/'&= â€” jrh^^ya- 2o>'ft+ 63^0+ 4yd-ye],

12nr

y'c = ^[->'a+ l6>- 3OJC+ l63;d->'e],

yd = â€” pl-^^ 43^*+ 63'c- 20>'d+II>],

3''e = -^[ii3'a- 56>'6+ii4>'c-i04>'d+35>'J-

The results of applying these formulas to the function given

are expressed in the table below.

X

y

/

r

I

â€¢099833

.109451

.001203

2

. 208460

â€¢ 107583

002524

3

.314566

.104415 -

.003804

4

.416871

.099986 -

.005045

5

â€¢514136

â€¢094347

.006221

6

.605186

.087568 -

007322

7

.688921

â€¢079733

008334

8

.764329

^A r ^ ? 1

.070929

009258

1 1

These results agree fairly well with those previously obtained.

It is probable that the formulas derived from the interpolation

formula give the most satisfactory results.

INTERPOLATION

113

As another application let us find the maximum or minimum

value of a function having given three values near the critical

point.

Let ya, jb, and jc be three values of a function of x near

its maximum or minimum corresponding to the values of Xy

a, b, and c respectively.

From (2)

_ (x â€” b)(xâ€”c) (x â€” a)(x â€” c) (x â€” a)(x â€” b)

^'"^" (a-b){a-cy^\b-a){b-cy^' {c-a){c-b)'

Equating to zero the first derivative with respect to x

2x â€” a â€” b

2X â€” b â€” c

y'^ 7 â€” TVT â€” ^+>

2x â€” a-

^yc

{a-b){a-c) ' '" {b-a){b-c) ' â– " {c-a){c-b)

_ yajb^ - c^) +>(c^ - a^) -\-ycia^ - b^)

=0;

(6)

2[ya{b-c)-\-yb{c-a)-\-yc{a-b)]

This is equivalent to drawing the parabola

y=A+Bx-\-Cx^

through the three points and determining its maximum or

minimum.

From the table of values

6.0

6.5

7.0

10.05

10.14

10.10

the abscissa of the maximum point is found from (6).

(io.o5)(-6.75) + (io-i4)(i3) + (iaio)(-6.25) ^

2[(io.o5)(-.5) + (io.i4)(i) + (io.io)(-.5) ''â€¢^^''

3; = 10.1424.

CHAPTER VIII

NUMERICAL INTEGRATION

Areas

An area bounded by the curve, y=J{x), the axis of x, and

two given ordinates is represented by the definite integral

r

ydx,

where the ordinates are taken at :r = a and x = n. It may be

said that the definite integral represents the area under the

curve, or that the area under the curve represents the value of

the definite integral.

If a function is given by its graph, it is possible, by means

of the planimeter, to find roughly the area bounded by the curve,

two given ordinates and the x â€” axis, or, what amounts to the

same thing, the area enclosed by a curve. This method is used

in finding the area of the indicator diagrams of steam, gas or

oil engines, and various other diagrams. The approximations

in these cases are close enough to satisfy the requirements.

If, however, considerable accuracy is sought, or whenever

the function is defined by a table of numerical values another

method must be employed.

Mechanical Quadrature or Numerical Integration is the method

of evaluating the definite integral of a function when the func-

tion is given by a series of numerical values. Even when the

function is defined by an analytical expression but which can-

not be integrated in terms of known functions by the method

of the integral calculus, numerical integration must be resorted

to for its evaluation.

The formulas employed in numerical integration are derived

from those established for interpolation.

114

NUMERICAL INTEGRATION 115

In interpolation it was found that the order of differences

which must be taken into account depends upon the rapidity

with which the differences decrease as the order increases.

This is also true of numerical integration. It is the same as

saying that if the series employed does not converge the process

will give unsatisfactory results. An illustration will be given

later.

Formulas for numerical integration will be derived from (i)

of Chapter VII.

In this formula it was assumed that the ordinates are given

at equal intervals.

, ^ .x(xâ€”l).o , x{xâ€”l){x â€” 2) .o

yx=yo-{-xAyo+^ -A^yo+ â€” , ^A^yo

11 11

^ x(x-i){x-2)(x-s) ^. ^ x{x-i)(x-2)(x-s){x-4) ^.^

^ x(x-i){x-2){x-s){x-4){x-s) ^, J ^j>j

[6

Integrating the right-hand member,

\ ydx=yQ I dx-\-Ayo\ xdx-\ â€” p- | x{x â€” i)dx

Jo Jo \2_J

+^Vx{x-l){x-2)dx

b Jo

^^ C\{^^-^){0c-2){x-^)dx

k Jo

H-^ \x{x-i){x-2){x-i){x-4)dx

15 Jo

\ x{x-i){x-2){x-^){x-4){x-s)dx-\- . . .

=nyo-\ â€” Ayo+ -r^+ n^-^n^ ) - -

2 \3 2/ |2 V4 / |3

â– "|6

\5 2 3 /

A^o

k

116 EMPIRICAL FORMULAS

\6 4 3 / |5

\7 2 ' 4 3 / P

The data given in any particular problem will enable us

to compute the successive differences of ^'o up to A">'o. On the

assumption that all succeeding differences are so small as to

be negligible the above formula gives an approximate value

of the integral. It is only necessary to assign particular values

tow.

Let w = 2, then

I yxdx = 2yo+2Ayo+iA^yoj

Ayo=yi-yo,

A^yo = Ayi - Ayo = y2-yi-yi +yo,

=y2-2yi+yo.

Substituting these values in the above integral it becomes

I y2dx = 2yo-]-2yi-2yo+iy2-iyi+^yo,

_yo+4yi+y2

This is equivalent to assuming that the curve coincides with

a parabola of the second degree.

If the common distance between the ordinates is h, the

value becomes

'2h

ydx=U(yo+4yi+y2) (7)

X

IfÂ»=3

ydx = syo+^Ayo+iA^yo+i^^yoj

Ayo=yi-yo,

A^yo = Ayi-Ayo=y2- 2yi +yoy

i

NUMERICAL INTEGRATION 117

=yz-2>y2+2>yi-yo'

Substituting these values in the equations,

1 ydx = 7^yo-\-^yi-^yQ+ly2-hi+ho-\-h^-h2+hi-hoj

= l>'0 + ljl+f3'2+|>'3,

= l(3'o+33'i+3>'2+3'3).

If the common distance between the ordinates is h the

formula becomes

â€¢3ft

X

= iKyQ-\-^yi-{-2>y2+y^) (8)

This is equivalent to assuming that the curve coincides with

a parabola of the third degree.

If there are five equidistant ordinates, h representing the

distance between successive ordinates

/â€¢4ft

Jo ^

yd^ - 14(3^0+^4) +64(3^1 +^3) +24y2 ^^^ ... (9)

b 45

If the area is divided into six parts bounded by seven equi-

distant ordinates the integral becomes

i

6

ydx = 6>'o + 1 ^^yo + 2 7 A^^yo + 24 A^^yo + ^-L^yo

+ffA5>'o+AVA63;o.

Sine 3 the last coefficient, 1^, differs but slightly from t^

and by the assumption that A^^'o is small the error will be slight

if the last coefficient is replaced by tq-

Doing this and replacing

^yohyyi-yo,

^^yohyy2-2yi+yQ,

A^yo by yd-;^y2+^yi -yo,

118

EMPIRICAL FORMULAS

A*yo by yA-4ys+6y2 -4yi+yo,

A^yo by y5-5>'4H-ioy3-io>'2+53'i-3'o,

A^^'o by / - 6>'5 -f- 1 5>'4 - 2oy3 + 1 53'2 - 6yi +yo,

gives the formula

X

ydx = TV/b'o+>'2+>'4+3'6+5(>'i +>'5) +6>'3]. . (lo)

The application of these

formulas is illustrated by

finding the area in Fig. 27.

Fig. 27.

By (7)

A=ih(yo+4yi + 2y2-\-4y3 + 2y4:+4y5+2ye+4y7+2y8+4y9

-\-2yio+4yn+yi2)'

By (8)

^=l^(yo+3>'i+3>'2 + 2y3+3>'4+3>'5+2y6+3>'7+33'8+2y9

+3>'io+33'ii+>'i2).

By (9)

A=-hhli4(yo+2y^-[-2y8+yi2)+64(yi-\-y3+y5+y7+y9+yii)

+ 24(y2+yQ-\-yio)].

By (10)

A=^hlyo+y2+y4.+2y6-\-y8+yio-\-yi2+5(yi+y5+y7+yii)

+6(^3 +^9)].. .

I. A rough comparison of the approximations by the use

of these formulas will be obtained by finding the value of

'ax

Jn

â€” . The value of this definite integral is log 13 = 2.565. It

I X

is also equal to the area under the curve

I

y=-

X

NUMERICAL INTEGRATION 119

from x = i to a: -=13. Dividing the area up into 12 strips of

unit width by 13 ordinates the corresponding values of x and

y are

X

I

2

3

4

5

6

7

8

9

10

II

12

13

y

I

1

2

i

i

i

i

1

T

1

i

A

tV

A

tV

By (7)

^ =Mi+2+i+i+f +1 +f +Ht+f +A+i+Ty

= 2.578, error .5%;

By (8), ^=2.585, error .8%;

By (9), ^ = 2.573, error .3%;

By (10), A = 2.572, error .3%.

2. The accuracy of the approximation is much increased

by taking the ordinates nearer together, as is shown by the

following evaluation of

^^ dx^

h 1+^

X'

The value of this integral is equal to the area under the

curve

I

i-\-x

from x=o tG x = i. Dividing the area into twelve parts by

dx

thirteen equidistant ordinates the value of | is found to be

Jo I ' "

By (7), 0.69314866, error 0.00000148;

By (8), 0.69315046, error 0.00000328;

By (9), 0.69314725, error 0.00000007;

By (10), 0.69314722, error 0.00000004.

The correct value is, of course, loge 2, which is 0.69314718.

Formulas (7) and (8) are Simpson's Rules, (10) is Weddle's

Rule.

120

EMPIRICAL FORMULAS

3. Apply the above formulas to the area of that part of the

semi-ellipse included between the two perpendiculars erected at

the middle points of the semi-major axes. Let this area be

divided into twelve parts by equidistant ordinates.

Since the equation of the ellipse is

these ordinates are

Ws b, ^V7^ b, iVs b, iVYs b, iVJ~s b, tVVm3 b, b,

t^jVh3 ^, WYs b, \y/Vs b, JVS 6, ^V^ b, iV3 b.

By (7), .4=0.9566099^6;

By (8), ^=0.956608006;

By (9), ^=0.9566x1406;

By (10), A =0.9566x1406.

The correct value to seven

places is 0.9566x1506.

In the application of these

formulas it is highly desirable

to avoid large dififerences among

the ordinates. For that reason

the formulas do not give so

good results when applied to

the quadrant of the ellipse.

4. The area under the curve,

Fig. 28, determined by the

following sets of values:

yi

3.0

2.5

2.0

1.5

y

â€” ,

>v

/

Y

N

^

/

/

0.5

A

X I

A .6 .8 1.0 1.2 Â»

Fig. 28.

) .2 .4

x.o

1.2

1.5

2.2

2.7

2.6

2.3

y \ i-o

is by (7)

^ =i - 5-(i-o+6.o+4.4+xo.8+5. 2+9.2 + 2. x) = 2.58,

and by (8),

.4 =|-i(x.o+4.5+6.6+5.4+7.8+6.9+2.x) = 2.5725.

2.1

NUMERICAL INTEGRATION

121

X1.2

ydx.

The area found is therefore the approximate value of this

integral

5. Find the area under the curve determined by the points

2.8 3.2 3.6 4.0 4.6 4.8

XI 1.5 1.9 2.3

5-0

y\o .40 1.08 1.82 2.06 2.20 2.30 2.25 2.00 1.80 1.5

The points located by the above sets of values are plotted

in Fig. 29 and a smooth curve drawn through them. The area

y

^'T~^=T^

2 â– - Jr\

L ~^

"2

^

/

A

V

J

-Ai

~l

1

* f

t

/

/

^

y"

â€” 1 â€” 1 â€” 1 1 X

12345

Fig. 29.

is divided into strips each having a width of .4. Rectangles

are formed with the same area as the corresponding strips.

The eye is a very good judge of the position of the upper bound-

ary of each rectangle. Adding the lengths of these rectangles

and multiplying the sum by .4 the area is found to be 6.644.

By Simpson's Rule, formula (7), are found

for h = .2,

^=6.639,

^=6.645.

The graphical determination of areas can be made to yield

a close approximation by taking narrow strips, and where the

points are given at irregular intervals the area can be found

more rapidly than by the application of Simpson's Rules.

122 EMPIRICAL FORMULAS

6. A gas expands from volume 2 to volume 10, so that its

pressure p and volume v satisfy the equation pv = ioo. Find

the average pressure between v = 2 and 7; = 10.

The average pressure is equal to the work done divided by

8. The work is equal to the area under the curve p = ^^ from

V

v = 2 to z; = 10, which is

n^ 100 r 1 ^Â°

I â€” dv = 100 log V = 160.044.

J2 V L J2

That this area represents the work done in expanding the

volume from 2 to 10 becomes evident in the following way.

Let s represent the surface inclosing the gas, ps will then be

the total pressure on that surface. The element of work will

then be

dW = psdn,

when dn lepresents the element along the normal.

PF = j psdn.

But

sdn = dv,

and

W^j'pdv.

This is the equation above. The average pressure over the

change of volume from 2 to 10 is

160.944 -^ 8 = 20. 1x8.

7. Find the mean value of sin^ x from x = o to ii; = 27r. Plot

the curve y = sin^ x by the following values of x and y :

TT

12

TT

6

TT

4

TT

3

12

TT

2

.0670

.2500

.5000

.7500

â€¢9330

I. 0000

11

12

27r

3

31

4

6

IITT

12

9330 -7500 .5000 .2500 .0670

NUMERICAL INTEGRATION

123

X

IT

12

77r

6

4

47r

3

12

31

2

y

o

.0670

.2500

.5000

.7500

â€¢9330

I. 0000

X

IQTT

12

Si

3

Ttt

4

IITT

6

2_3Z

12

27r

â€¢9330 -7500 .5000 .2500 .0670 o

Applying Simpson's Rule, formula (7), the area is found to

be TT. The mean value is the area divided by 27r or .5.

8. A body weighing 100 lb. moves along a straight line

without rotating, so that its velocity v at time t is given by the

following table:

/sec

V ft./sec

1.47 1.58 . 1.67 1.76 1.86

Find the mean value of its kinetic energy from t = i to / = 9.

t

I

3

5

7

9

v^

2.1609

2.4964

2.7889

3.0976

3-4596

Kinetic energy .

3-355

3.876

4-331

4.810

5-372

Plotting kinetic energy to /, the area under the curve is

34.755. This divided by 8 gives the mean kinetic energy as

4-357-

Volumes Ai/

Fig. 30 explains the ap-

plication of the formulas

to the problem of finding

the approximate volume of

an irregular figure. Tiie area

of the sections at right angles

to the axis of x are :

Ai=\k{yi-\-^y5-^y^),

^2=i^(3'6 + 4>'9+3'8),

Az=\k{y2+Ay7+y^)' Fig. 30.

124 EMPIRICAL FORMULAS

If the areas of these sections be looked upon as ordinates,

h being the distance between two adjacent ones, it is evident

that the volume may be represented by the area under the

curve drawn through the extremities of these ordinates.

Substituting the values of A\, A2, and A3 in this equation,

the volume becomes

V = ih[ik{yi-{-4y5+y4)+ik(y6-^4yc)-\-y8)+lHy2-\-4y7-\-y3)]

= iM[yi +>'2 +>'3 +3'4 +4(75 +>'G+>'7 +y8) + 16^9]

In order to apply formulas (8), (9) and (10), the solid would

have to be divided differently, but the method of application

is at once evident from the above and needs no further discussion.

1. The following are values of the area in square feet of the

cross-section of a railway cutting taken at intervals of 6 ft.

How many cubic feet of earth must be removed in making the

cutting between the two end sections given?

91, 95, 100, 102, 98, 90, 79.

These cross-section -areas were obtained by the application

of Simpson's Rules.

By (7),

F = | â€¢ 6(91 -1-380 -f 200 -f 408 + 196 H-36o-f 79) =3428;

By (8),

7 = 1-6(91 + 285+300 -F204-f 294 -f-27o-f79) =3426.8.

2. A is the area of the surface of the water in a reservoir

when full to a depth h.

h ft I 30 25 20 15 10 5 o

^ sq.ft. . 1 26,700 22,400 19,000 16,500 14,000 10,000 5,000

NUMERICAL INTEGRATION 125

- Find (a) the volume of water in the reservoir, (b) the work

done in pumping water out of the reservoir to a height of loo ft.

above the bottom until the remaining water has a depth of

lO ft.

F = f (26, 700 +89, 600 +38,000 +66,000 +28,000+40,000+5,000)

= 488,833 cu. ft.

Work = 'ZÂ£; ( A(ioo â€” h)dh, where ze; = weight of i cu.ft. of water

Jio

= 62.3 lb. The value of this integral will be approximately the

area under the curve determined by the points

30 25 20 15 10

^(100 â€” /?) -i 1,869,000 1,680,000 1,520,000 1,402,500 1,260,000

multiplied by 62.3.

This area is equal to

1(1,869,000+6,920,000+3,040,000+5,610,000 + 1,260,000)

= 31,165,000.

Multiplying this by 62.3 gives the work equal to 1,941,579,500

ft.-lb.

3. When the curve in Fig. 29 revolves about the ::i;-axis,

find the volume generated.

The areas of the cross-sections corresponding to the given

values of x are given in the following table:

I .0 1.2

TT 2.257r 4.847r 7.29^ 6.76^ 5 â€¢ 29^ 4.41^

By (7) F = 5.8627r = 18.416.

By (8) F = 5.8o37r = 18.231.

4. When the curve in Fig. 30 revolves about the x-axis,

find the volume generated from x=i tox = 4.2. From the curve

the following sets of values are obtained :

126 EMPIRICAL FORMULAS

X

i.o 1.2 1.4 1.6

1.8

2.0

2.2

2.4

2.6

y

.11 .29 .53

.87

1.37

1. 71

1.90

2.01

X

.012 .084 .281

2.8 3.0 3.2

â€¢757

3-4

1.877

3.6

2.924

3.8

3.610

4.0

4.040

4.2

y_

2 . 06 2.12 2.2

2.27

2.30

2.28

2.25

2.20

f 4.2444.4944.84 5.153 5.290 5.198 5.062 4.84

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