Chapter 7Numerical

Integration

Lecture 8

Integration

Indefinite Integrals

Indefinite Integrals of a function are functions that differ from each other by a constant.

cx

dxx2

2

Definite Integrals

Definite Integrals are numbers.

2

1

2

1

0

1

0

2

x

xdx

for solution form closedNo

for tiveantideriva no is There2

2

b

a

x

x

dxe

e

Why Numerical Integration?

• Very often, the function f(x) to differentiate or the integrand to integrate is too complex to derive exact analytical solutions.

• In most cases in engineering, the function f(x) is only available in a tabulated form with values known only at discrete points.

Numerical Solution

The general form of numerical integration of a function f (x) over some interval [a, b] is a weighted sum of the function values at a finite number (n) of sample points (nodes), referred to as ‘quadrature’:

Numerical Integration

b

af(x)dxArea

One interpretation of the definite integral is

Integral = area under the curve

a b

f(x)

Newton-Cotes IntegrationCommon numerical integration schemeBased on the strategy of replacing a

complicated function or tabulated data with some approximating function that is easy to integrate

nnn

b

a

n

b

a

xaxaaxPxf

dxxPdxxfI

....)( 10

Pn(x) is an nth orderpolynomial

Trapezoidal RuleCorresponds to the case where the

polynomial is a first order

b

a

b

a

dxxPdxxfI 1

0

1

2

3

4

5

0 5 10

x

f(x)

h

F(a)

F(b)

From the trapezoidal rule we can obtain for the total area of (n-1) intervals

0

1

2

3

4

5

0 1 2 3 4 5 6 7 8 9 10

x

f(x)

222

13221 nn xfxfh

xfxfh

xfxfhI

where there are n equally spaced base points.

n

n

x

x

x

x

x

x

dxxfdxxfdxxfI1

3

2

2

)()()(1

Error Estimate in the trapezoidal rule

It can be obtained by integrating the interpolation error we defined in previous chapter for Lagrange polynomial as

0

1

2

3

4

5

0 5 10

x

f(x)

Error

Example

)(''max12 ],[

2 xfhab

Errorbax

5

010

2

1 error ,)sin( thatsohfinddxx

52 102

1

121)('' hErrorxf

)sin()('');cos()(';0; xxfxxfab

52 106

h

0 1

0 11/2

1 12

0 11/21/4 3/4

1 12 2 2

Remark 1: in this example instead of re-computation of some function values when h is changed to h/2 we observe that

Simpson’s Rules

0

1

2

3

4

5

0 5 10

x

f(x)

Simpson’s 1/3 rule can be obtained by passing a parabolic interpolant through three adjacent nodes.

The area under the parabola is

To obtain the total area of (n-1) even intervals we apply the following general Simpson’s 1/3 rule

f(x1), f(xn)Note:

1

f(x2), f(x4), f(x6),..

4f(x3), f(x5), f(x7),..

2

Remark 2: Simpson’s 1/3 rule requires the number of intervals to be even. If this condition is not satisfied, we can integrate over the first (or last) three intervals with Simpson’s 3/8 rule which can be obtained by passing a cubic interpolant through four adjacent nodes, and defined by

The error in the Simpson’s rule is

Because the number of panels is odd, we compute the integral over the first three intervals by Simpson’s 3/8 rule, and use the 1/3 rule for the last two intervals:

Simpson’s 3/8 rule Simpson’s 1/3 rule

Summary

Newton Cotes formulae for Numerical

integration.

Trapezoidal Rule

Simpson’s Rules.

Romberg Integration

Double Integrals

To be continued in

Lecture 9