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Numerical Integration

1.1 Introduction

The definite integral

I+f/ a

bf+x/ x

is defined in calculus as a limit of what are called Riemann sums. It is then

proven that

I(f) = F(b) - F(a)

where F+x/ is any antiderivative off+x/; this is the Fundamental Theorem ofCalculus. Many integrals can be evaluated by using this formula, and a

significant portion of most calculus textbooks is devoted to this approach.

Nonetheless, most integrals cannot be evaluated by using (0.0) because most

integrands f+x/ do not have antiderivatives expressible in terms of elementaryfunctions. Examples of such integrals are

0

1x2

x 0

x sin. x 2 x

So other methods are needed for evaluating such integrals.

In the first section of this chapter, we define two of the oldest and most popular

numerical methods for approximating integrals like (0.0): the trapezoidal rule

and Simpson's rule. We also analyze the error in using these methods and then

obtain improvements on them. A third method is Gaussian quadrature which

will be discussed in addition to the two methods stated. This method is more

complicated in its origin than Simpson's and the trapezoidal method, but it is

almost always much superior in accuracy for similar amounts of computation.

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1.2 Trapezoidal Method

The central idea behind most formulas for approximating

I+f/ a

bf+x/ x

is to replace f+x/ by an approximating function whose integral can be evaluated.I this section we will look at methods based on linear and quadratic

interpolation.

1.2.1 Trapezoidal Method

Approximate f+x/ by a linear polynomial

p1+x/ +bx/ f+a/ +xa/ f+b/ba

which interpolates f+x/ at a and b (see Figure 0.0). The integral ofp1+x/ over#a, b' is the area of the shaded trapezoid shown in Figure 0.0; it is given by

yf#x'

p1+x/

a b

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Figure 0.0. An illustration of the trapezoidal rule (0.0).

T1+f/ +ba/ f+a/ f+b/2

.

This approximates the integral I+f/ iff+x/ is almost linear on #a, b'.Example 0.0. Trapezoidal Method I

Approximate the integral

I 0

1 1

1xx

Solution 0.1. The true value of this integral is

I0 0

1 1

x 1x

log+2/

Defining the integrand as

f+x_/ :1

x 1

and applying the interpolation formula to the interval #0, 1' we get the firstapproximation

T1 1

2+1 0/ +f+0/ f+1//

3

4

The approximation deviates from the exact result by

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N#I0 T1'0.0568528

The approximation is exact in the first digit on the right hand side of the radix

point.

To improve the approximation T1+f/ in (0.0) when f+x/ is not a nearly linearfunction on #a, b', break the interval #a, b' into smaller subintervals and apply(0.0) on each subinterval. If the subintervals are small enough, then f+x/ will benearly linear on each one. This idea is illustrated in Figure 0.0

yf+x/

p1+x/

a b

x0 x1 x2

Figure 0.0. An illustration of the trapezoidal rule T2+f/ (5.2.1).Example 0.0. Trapezoidal Method II

Approximate the integral

I 0

1 1

1xx

by using T1+f/ on two subintervals of equal length.Solution 0.2. The true value of this integral is

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I0 0

1 1

x 1x

log

+2

/For two subintervals of equal length

I1 0

1

21

x 1x 1

2

1 1

x 1x

log3

2 log

4

3

Defining the integrand as

f+x_/ : 1x 1

and applying the interpolation formula to the intervals #0, 1 s 2' and #1 s 2, 1' weget the first approximation

T2 1

2

1

2 0 f+0/ f 1

2

1

21

1

2f+1/ f 1

2

17

24

The approximation deviates from the exact result by

N#I1 T2'0.0151862

The error in T2 is about 1 s 4 of that given forT1 compare Example 0.0.We will derive a general formula to simplify the calculations when using several

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subintervals of equal length. Let the number of subintervals be denoted by n,

and let

h ba

n

be the length of each subinterval. The endpoints of the subintervals are given by

xj a j h j 0, 1, , n.

Then break the integral into n subintervals

I+f/ a

bf+x/ x

x0

xnf+x/ x

i0

n1

xi

xi1f+x/ x.

Approximate each subinterval by using a linear interpolating polynomial such as

(0.0) and noting that each subinterval #xi, xi1' has length h. Then

Tn+f/ i0

n1 h

2+f+xi/ f+xi1// h 1

2f+x0/ 1

2f+xn/

i1

n2

f+xi/ .

This is called the trapezoidal numerical integration rule. The subscript n gives

the number of subintervals being used; and the points x0, x1, , xn are called

the numerical integration node points.

The following lines show an implementation of the trapezoidal method in a few

steps using formula (0.0). The implementation is straight forward and does not

use special features of the evaluation.

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trapezoidalMethod#f_, x_, a_, b_, n_' :Block%h, xval, fval,+ determine the step length /h

b a

n;

+ generate the integration nodes /xval Table#a i h, i, 0, n';+ function values are generated /fval Map#+f s. x ! / &, xval';fval317 First#fval'

2;

fval3Length#fval'7 Last

#fval

'2 ;+ sum up the terms according to +5.12/ /N#h Fold#Plus, 0, fval''

)The following line is an application of the function trapezoidalMethod to the

function f+x/ x2 x2

trapezoidalMethod-x2 x2 , x, 0, 1, 210.18932

Before giving some numerical examples ofTn+f/, we would like to discuss thechoice ofn. With a sequence of increasing values ofn, Tn+f/ will usually be anincreasingly accurate approximation ofI+f/. But which sequence of values ofnshould be used? Ifn is doubled repeatedly, then the function values used in

each T2 n+

f

/will include all of the earlier function values used in the preceding

Tn+f/. Thus, the doubling ofn will ensure that all previously computedinformation is used in the new calculation, making the trapezoidal rule less

expensive than it would be otherwise. To illustrate how function values are

reduced when n is doubled, considerT2+f/ and T4+f/.

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T2+f/ h f+x0/2

f+x1/ f+x2/2

with h ba

2, x0 a, x1

ab

2, x2 b.

Also

T4+f/ h f+x0/2

f+x1/ f+x2/ f+x3/ f+x4/2

with h ba

4,

x0 a, x1 3 ab

4, x2

ab

2, x3

a3 b

4, x4 b.

Comparing the two approximations, we observe that f+x1/ and f+x3/ need to beevaluated, as the other function values are known from the lower

approximation. For this and other reasons, all of our examples ofTn+f/ arebased on doubling n.

Example 0.0. Higher Trapezoidal Method

Here we will calculate Tn+f/ using trapezoidal interpolation for three differentfunctions. The functions are:

f+x_/ : x2

for the interval #0, 1'.

g+x_/ : 1x2 1

for the interval #0, 4' and

h+x_/ : 1cos+x/ 2

for the interval #0, 2 '. The number of iterations aren 2, 4, 8, 16, 32, 64, 128, 256.Solution 0.3. The exact values for these functions can be determined by using

the antiderivatives of the functions

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I1 0

1

f+x/x

1

2 erf+1/

I2 0

4

g+x/x

tan1+4/

and

I3 0

2

h+x/x

2

3

The approximation for the different functions and different iteration numbers

follow from the lines below

Tf +trapezoidalMethod+f+x/, x, 0, 1, 1/ &/ s2, 4, 8, 16, 32, 64, 128, 256

0.73137, 0.742984, 0.745866,0.746585, 0.746764, 0.746809, 0.74682, 0.746823

Tg

+trapezoidalMethod

+g

+x

/,

x, 0, 4

, 1

/&

/ s

2, 4, 8, 16, 32, 64, 128, 2561.45882, 1.32941, 1.32525, 1.32567, 1.32578, 1.32581, 1.32582, 1.32582

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Th +trapezoidalMethod+h+x/, x, 0, 2 , 1/ &/ s2, 4, 8, 16, 32, 64, 128, 256

4.18879, 3.66519, 3.62779, 3.6276, 3.6276, 3.6276, 3.6276, 3.6276

The errors with respect to the exact value of the integrals are

1 I1 Tf

0.0154539, 0.00384004, 0.000958518, 0.000239536,0.0000598782, 0.0000149692, 3.74227 106, 9.35566 107

2 I2 Tg

0.133006, 0.0035941, 0.000564261, 0.000144082,0.000036038, 9.01059 106, 2.25272 106, 5.63183 107

3 I3 Th

0.561191, 0.0375927, 0.000192788, 5.12258 109,

8.88178 1016, 8.88178 1016, 0., 8.881781016

These data are collected in the following table for each integral and each

number of used interpolation steps.

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TableForm$Prepend$2, 4, 8, 16, 32, 64, 128, 256, 1, 2, 3,n, "1", "2", "3"((

n 1 2 32 0.0154539 0.133006 0.561191

4 0.00384004 0.0035941 0.0375927

8 0.000958518 0.000564261 0.000192788

16 0.000239536 0.000144082 5.12258 109

32 0.0000598782 0.000036038 8.88178 1016

64 0.0000149692 9.01059 106 8.88178 1016

128 3.74227 106 2.25272 106 0.

256 9.35566 107 5.63183 107 8.88178 1016

From the table it is obvious that the error decreases with increasing n. The third

example converges very rapidly.

1.2.2 Generalized Integration Rules

The generalization of the Trapezoidal and Simpson's rule are known as Newton-

Cotes formulas. The integration (quadrature) formulas by Newton-Cotes can be

derived if the integrand f is replaced by an appropriate interpolation polynomial

pn+x/ so that

a

bf+x/ x

a

bpn+x/ x

For the following calculation we assume an equidistant partition of the interval

#a, b' withxi a i h and i 0, 1, 2, , n

where h

+ba

/ sn and n 0 and n N0. If we assume that the interpolation

polynomial is generated by a Lagrange polynomial of ordern then we have

pn+x/ i0

n

fi Li+x/ with Li+x/ kik0

n xxk

xixk

and fi f+xi/ fori 0, 1, 2, , n. Introducing new variables with x ah t we

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can rewrite the Lagrange polynomial in a simpler form

Li+x/ i+t/ kik0

n t k

i k.

This representation can be used in the integration of the polynomial pn+x/ as

a

bpn+x/ x

i0

n

fia

bLi+x/ x h

i0

n

fi0

ni+t/ t h

i0

n

fii.

The weights i are functions ofn only and do not depend on the function values

off.

Defining the Lagrange polynomials in their reduced representation is straightforward in Mathematica and allows us to use them in the calculation of the

weights. The reduced Lagrange polynomials are defined by

lagrangePolynomial+n_, i_/ : k0

n

If%i k, t ki k

, 1)

This function delivers the corresponding function i+t/ by

lagrangePolynomial+3, 1/1

2+2 t/ +3 t/ t

These functions can be used in the calculations of the coefficients i. The

following calculation lists the weights forn 2 corresponding to the 1 s 3Simpson rule

n 2; Table%0

n

lagrangePolynomial+n, i/t, i, 0, n)

13

,4

3,

1

3!

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The weights in general satisfy the property

i0

n

i n

We can check this relation with our function.

n 4; Fold%Plus, 0, Table%0

n

lagrangePolynomial+n, i/t, i, 0, n))

4

The related approximation of the integral follows next by summing over the

corresponding function values.

n 8;

Factor%Fold%Plus, 0,

Simplify%Table%h f+a h i/ 0

n

lagrangePolynomial+n, i/t, i, 0, n))))

114175

4 h +5888 f+ah/ 928 f+a 2 h/ 10496 f+a 3 h/ 4540 f+a 4 h/ 10496f+a 5 h/ 928 f+a 6 h/ 5888 f+a 7 h/ 989 f+a 8 h/989 f+a//

representing Simpson's basic rule of integration. To get the higher order

approximations we define a relation in Mathematica which delivers the

quadrature formulas.

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integrationRulesLagrange#n_, h_' :.Fold%Plus, 0,

Table%h f#a i h' 0n

lagrangePolynomial#n, i' t,i, 0, n) ss Simplify) ss Factor2

The function integrationRulesLagrange allows us to generate the integration

formulas in a straight forward way. The next line sets up a table of these

formulas up to order 6.

TableForm$i, integrationRulesLagrange#i, h'i16 ,TableHeadings None, n, Formula(

n Formula

11

2h +f+a h/ f+a//

21

3h +4 f+a h/ f+a 2 h/ f+a//

33

8h +3 f+a h/ 3 f+a 2 h/ f+a 3 h/ f+a//

42

45

h

+32 f

+a h

/ 12 f

+a 2 h

/ 32 f

+a 3 h

/ 7 f

+a 4 h

/ 7 f

+a

//5 5288

h +75 f+a h/ 50 f+a 2 h/ 50 f+a 3 h/ 75 f+a 4 h/ 19 f+a 6

1

140h +216 f+a h/ 27 f+a 2 h/ 272 f+a 3 h/ 27 f+a 4 h/ 216 f+

41 f+a 6 h/ 41 f+a//The first formula is known as Trapezoidal rule, the second as Simpson's rule,

the third is called 3 s 8 Simpson rule the forth is the rule established by Milne,the fifth has no specific name, and the sixth is Weddle's rule to approximate

integrals. For larger orders n 6 the function integrationRulesLagrange does

not deliver useful formulas because of the insensitivity of the integral due to

changes in the function f. In principle we can generate these formulas but they

are not drastically improving the accuracy of the integral. This behavior is

discussed in more detail in the following example.

Example 0.0. Error Estimation

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Recall the examples we discussed in the previous section, with

I 0

1 1

1xx ln+2/.

Solution 0.6. Here f+x/ 1 s +1x/, #a, b' #0, 1', and h +ba/ s n. The Tablefrom above will give us the related results and errors.

f+x_/ : 1x 1

TableForm%

Table%i, integrationRulesLagrange i,1

i,

log+2./ integrationRulesLagrange i, 1i

!, i, 1, 10) s.a 0.,

TableHeadings None, "Formula", "Value", "Error")

Formula Value Error

1 0.75 0.0568528

2 0.694444 0.001297263 0.69375 0.000602819

4 0.693175 0.0000274226

5 0.693163 0.0000158485

6 0.693148 8.81695 107

7 0.693148 5.52783 107

8 0.693147 3.39735 108

9 0.693147 2.22241 108

10 0.693147 1.45038 109

It is obvious from this table that the higher order interpolation polynomials

reduce the error of approximation. However, within the range of 6 t0 10 there is

only a marginal improvement of the error.

1.2.3 Error Estimations for Trapezoidal Rule

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