Numerical Methods for Civil Engineers Lecture 9 Numerical Integration

Numerical Methods for Civil Engineers Lecture 9

Numerical Integration - Basic Ideas - Symbolic vs. vs. Numerical Integration - Trapezoid Rule - Simpson’ Simpson’s Rule - MATLAB quad and quad8 quad8 Functions Mongkol JIRAVACHARADET

SURAN AR EE

INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY

SCHOOL OF CIVIL ENGINEERING

BASIC IDEAS b

I = ∫ f ( x ) dx = Area under curve f ( x ) a

Node f (x) Trapezoidal region

x a

b

Approximated by piecewise-linear: Area = Sum of trapezoidal regions

Symbolic Math Toolbox Symbolic Math Toolboxes incorporate symbolic computation into the numeric environment of MATLAB. Symbolic Integration with MATLAB

>> >> >> >>

I I I I

= = = =

int(f) % Indefinite integral int(f, v) % Designating integration variable int(f, a, b) % Definite integral on close interval int(f, v, a, b)

where f = symbolic expression Variables must be define as symbolic by sym or syms

>> x = sym(‘x’), y = sym(‘y’), z = sym(‘z’) or >> syms x y z

b

Example : I = ∫ ( x 3 − c ) dx a

>> syms x a b c >> I = int(x^3-c,x,a,b)

I =

1/4*b^4-c*b-1/4*a^4+c*a

What is Integration? The integral is equivalent to the area under the curve. f(x) b

I = ∫ f ( x ) dx

I

a

x a

b

Examples of how integration is used to evaluate areas Net force due to wind blowing against the building

Cross-sectional area of a river

Newton-Cotes Formulas Replace a complicated function with a polynomial that is easy to integrate b

b

a

a

I = ∫ f ( x ) dx ≅ ∫ fn ( x ) dx where fn ( x ) = a0 + a1x + L + an x n −1 + an +1x n f(x)

f(x)

x a

b

Approximate area by a straight line

x a

b

Approximate area by a parabola

Trapezoidal Rule 1st degree polynomial = Linear line

f(x) f(b)

f1( x ) = f (a) +

f1(x)

f (b ) − f (a ) ( x − a) b−a

b

I =

f(a)

∫ f1( x ) dx = (b − a ) a

a

f (a ) + f ( b ) 2

x

b

Area of trapezoid = width x average height width = b – a = h

f (a ) + f ( b ) average height = 2

Composite Trapezoidal Rule Improve accuracy by dividing interval into subintervals

f(x) f2

f3

f4

f5

x4

x5

f1

x1

x2

x3

x5

x2

x3

x4

x1

x1

x2

x3

∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx + ∫ f ( x ) dx

+

x5

∫ f ( x ) dx

x4

f(x) There are n segment with equal width: h =

b–a n

x2

x3

xn

x1

x2

x n −1

I = ∫ f ( x ) dx + ∫ f ( x ) dx + L +

∫ f ( x ) dx

f ( x1 ) + f ( x 2 ) f ( x2 ) + f ( x 3 ) +h +L 2 2 f ( x n−1 ) + f ( x n ) +h 2

I =h x1 x1 = a

x2

x3

x4

x5

x6 xn = b

b–a h= n

f +f f +f f +f I = h 1 2 + h 2 3 + L + h n −1 n 2 2 2

n −1  h  I = f1 + 2 ∑ fi + fn   2  i =2 

f(x) f4

MEAN

f5

f3

f2

f6

f1 x1

Substitute h =

x2

b–a n

x3

x4

x5

x6

n −1   f + 2 f + f ∑ i n 1 i =2   I = (b − a ) 2n

width

average height

Example 9-1 Use Trapezoidal rule to numerically integrate f ( x ) = 0.2 + 25 x − 200 x 2 + 675 x 3 − 900 x 4 + 400 x 5 from a = 0 to b = 0.8. Note that the exact value of the integral can be determined analytically to be 1.640533. or using MATLAB’s Symbolic Math Toolbox : >> x = sym(‘x’) >> I=int(0.2+25*x-200*x^2+675*x^3-900*x^4+400*x^5,x,0,0.8) I = 3076/1875 = 1.6405 Single Trapeziod: f(0) = 0.2 and f(0.8) = 0.232

I = (b − a)

f (a ) + f (b ) 2

= (0.8 − 0)

εt =

f(x)

0.2 + 0.232 = 0.1728 2

Error

1.6405 − 0.1728 × 100% = 89 .5% 1.6405

Integral estimate 0

2 Trapeziods: n = 2 (h = 0.4) : f(0) = 0.2

f(0.4) = 2.456

f(0.8) = 0.232

n −1   f1 + 2 ∑ fi + fn  i =2  I = (b − a)  2n

I = (0.8 )

εt =

0.2 + 2(2.456 ) + 0.232 = 1.0688 4

1.6405 − 1.0688 × 100 % = 34.9% 1.6405

0.8

MATLAB’s trapz function Trapezoidal numerical integration >> z = trapz(y) >> z = trapz(x, y)

% Integral of y with unit spacing % Integral of y with respect to x

90o

∫ sin x dx

Example:

0o

>> angle = 0:15:90; >> x = (pi*angle/180); >> y = sin(x); >> z = trapz(x,y) >> z = 0.9943

Example 9-2 Use MATLAB’s trapz function to numerically integrate f ( x ) = 0.2 + 25 x − 200 x 2 + 675 x 3 − 900 x 4 + 400 x 5 from a = 0 to b = 0.8. Note that the exact value of the integral can be determined analytically to be 1.640533. For n = 2 :

>> x = linspace(0,0.8,3); >> fx = 0.2+25*x-200*x.^2+675*x.^3-900*x.^4+400*x.^5; >> I = trapz(x,fx) I = 1.0688 n

I

εt

2 3 4 5 6

1.0688 1.3639 1.4848 1.5399 1.5703

34.9 16.5 9.49 6.13 4.28

Simpson’s Rules Using higher-order polynomials to connect the points f(x)

f(x)

x Simpson’s 1/3 rule Quadratic connecting 3 points

x Simpson’s 3/8 rule Cubic connecting 4 points

Simpson’s 1/3 Rules f(x)

Quadratic connecting 3 points f(x1)

f2(x) f(x2)

f(x0)

x1

x0

I≅

x2

I=

x2

x2

∫ f ( x)dx ≅ ∫ f

x0

2

( x)dx

x0

f 2 ( x) =

( x − x1 )( x − x2 ) f ( x0 ) ( x0 − x1 )( x0 − x2 )

+

( x − x0 )( x − x2 ) f ( x1 ) ( x1 − x0 )( x1 − x2 )

+

( x − x0 )( x − x1 ) f ( x2 ) ( x2 − x0 )( x2 − x1 )

h x −x [ f ( x0 ) + 4 f ( x1 ) + f ( x2 )] , h = 2 0 3 2

Example 9-3 Use Simpson’s 1/3 rule to numerically integrate f ( x ) = 0.2 + 25 x − 200 x 2 + 675 x 3 − 900 x 4 + 400 x 5 from a = 0 to b = 0.8. Note that the exact value of the integral can be determined analytically to be 1.640533. Solution: n = 2 (h = 0.4) : f(0) = 0.2

f(0.4) = 2.456

f(0.8) = 0.232

I =

0.4 (0.2 + 4( 2.456) + 0.232) = 1.3675 3

εt =

1.6405 − 1.3675 × 100 % = 16.6% 1.6405

Composite Simpson’s 1/3 Rules x2

x4

xn

x0

x2

xn − 2

I = ∫ f ( x ) dx + ∫ f ( x ) dx + L + I=

∫ f ( x ) dx

where h =

b−a , n = even number n

h [f0 + 4f1 + f2 ] + h [f2 + 4f3 + f4 ] + L + h [fn−2 + 4fn −1 + fn ] 3 3 3 1 1

1

4

1

4

2

4

1 1

4

1 4

1

1 1

4

2

4

1

2 4

a

4

2

4

1

b

n −1 n −2  (b − a)  I= f ( x ) + 4 f ( x ) + 2 f ( x ) + f ( x )  0 ∑ i j =∑ j n  3n  i =1,3,5 2, 4, 6 

Example 9-4 Use composite Simpson’s 1/3 rule with n = 4 to numerically integrate f ( x ) = 0.2 + 25 x − 200 x 2 + 675 x 3 − 900 x 4 + 400 x 5 from a = 0 to b = 0.8. Note that the exact value of the integral can be determined analytically to be 1.640533. Solution: n = 4 (h = 0.2) : f(0) = 0.2

f(0.2) = 1.288

f(0.4) = 2.456

f(0.6) = 3.464

f(0.8) = 0.232

n −1 n−2  (b − a )  I= f ( x ) + 4 f ( x ) + 2 f ( x ) + f ( x )  0 ∑ i j =∑ j n  3n  i =1,3,5 2, 4,6 

I =

εt =

0.8 ( 0.2 + 4(1.288 + 3.464) + 2( 2.456) + 0.232) = 1.624 3( 4)

1.6405 − 1.624 × 100 % = 1.04% 1.6405

Simpson’s 3/8 Rules Cubic connecting 4 points

b

b

a

a

I = ∫ f ( x)dx ≅ ∫ f 2 ( x)dx f(x)

3h [ f ( x0 ) + 3 f ( x1 ) + 3 f ( x2 ) + f ( x3 )] 8 h = ( x3 − x0 ) / 3 I≅

x0 a or…

x1 I =

x2

x3 b

x

(b − a) [f ( x0 ) + 3 f ( x1) + 3 f ( x2 ) + f ( x3 )] 8

Truncation Errors Interval width

Et

Trapezoid

1 3 h f ′′(ξ ) 12

h =b−a

(b − a )3 f ′′(ξ ) 12

Simpson’s 1/3

1 5 (4) h f (ξ ) 90

b−a h= 2

(b − a )5 (4) f (ξ ) 2880

Simpson’s 3/8

3 5 (4) h f (ξ ) 80

b−a h= 3

(b − a )5 (4) f (ξ ) 6480

Method

Et

Simpson’s 3/8 rule is used when number of segments is odd.

Apply Simpson’s 1/3 and 3/8 Rules To handle multiple application with odd number of intervals f(x)

x 1/3 rule

3/8 rule

Example 9-5 Use Simpson’s 1/3 + 3/8 rule with n = 5 to numerically integrate f ( x ) = 0.2 + 25 x − 200 x 2 + 675 x 3 − 900 x 4 + 400 x 5 from a = 0 to b = 0.8. Note that the exact value of the integral can be determined analytically to be 1.640533. Solution: n = 5 (h = 0.16) : f(0) = 0.2 f(0.32) = 1.743 f(0.64) = 3.182

f(x)

f(0.16) = 1.297 f(0.48) = 3.186 f(0.8) = 0.232

For the first two segments use Simpson’s 1/3 : I =

0.16 (0.2 + 4(1.297) + 1.743) = 0.380 3

For the last three segments use Simpson’s 3/8 : I =

0.48 (1.743 + 3(3.186 + 3.182) + 0.232) 8

x 0

1/3 rule

= 1.265 Total integral : I = 0.380 + 1.265 = 1.645

0.16 0.32 0.48 0.64

εt =

0.8

3/8 rule

1.6405 − 1.645 × 100 % = 0.274 % 1.6405

MATLAB’s quad and quad8 Functions quad : low order method, quad8: high order method >> q = quad(’f',a,b) Approximates the integral of f(x) from a to b within a relative error of 1e-3 using an adaptive recursive Simpson's rule. ’f' is a string containing the name of the function. >> quad(‘sin’, 0, pi/2) Function f must return a vector of output values if given a vector of input values.

0.8

Example:

∫

0.2 + 25x − 200 x 2 + 675x3 − 900 x 4 + 400 x5 dx

0

poly5.m function p = poly5(x) p=0.2+25*x-200*x.^2+675*x.^3-900*x.^4+400*x.^5; >> quad(‘poly5’, 0, 0.8) ans = 1.6405

Example: Computing the Length of a Curve

x(t ) = sin(2t ),

y (t ) = cos(t ),

z (t ) = t

where t ∈ [ 0, 3π ] >> t = 0:0.1:3*pi; >> plot3(sin(2*t), cos(t), t) 10 8 6 4 2 0 1 0.5

1 0.5

0 0

-0.5

- 0.5 -1

-1

Length of the curve: Norm of derivative

3π

∫

4 cos(2t )2 + sin(t )2 + 1 dt

0

hcurve.m function f = hcurve(t) f = sqrt(4*cos(2*t).^2 + sin(t).^2 + 1); >> len = quad(‘hcurve’,0,3*pi) len = 17.2220

Example: Double Integration y max x max

∫ ∫

y min

f ( x, y ) dx dy

x min

For example: f (x, y) = y sin(x) + x cos(y) integrnd.m function out = integrnd(x,y) out = y*sin(x) + x*cos(y); >> xmin = pi; xmax = 2*pi; >> ymin = 0; ymax = pi; >> result = dblquad(‘integrnd’,xmin,xmax,ymin,ymax); result = -9.8698