lecture 4 numerical integration

8/4/2019 Lecture 4 Numerical Integration

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CLASS: 04

ME 262: Numerical Analysis Sessional

Department of Mechanical Engineering, BUET


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Numerical Integration

15 May 2011

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ME262 Numerical Analysis Sessional


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15 May 2011ME262 Numerical Analysis Sessional

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x0 x1 xnxn-1x

f(x)

Graphical Representation of Integral


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fn (x): linear fn (x): quadratic fn (x): higher-order

polynomial

Numerical Approximation of Integral

Newton-Cotes Closed Formulae -- Use both end points

Trapezoidal Rule : Linear

Simpsons 1/3-Rule : Quadratic

Simpsons 3/8-Rule : Cubic


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Approximate the function by a parabola

0 1 2( ) ( ) 4 ( ) ( )3

b

a

h f x dx f x f x f x

x0 x1x

f(x)

x2h h

L(x)

Simpsons 1/3-Rule


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n

abh

Piecewise Quadratic approximations

x0 x2x

f(x)

x4h h xn-2h xn

...

hx3x1 xn-1

Composite Simpsons Rule


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8

i = 0 1 2 3 4 5 6 . . n-1 n

Applicable only if

the number of segments is

even

0 1 2( ) 4 ( ) ( )3

hI f x f x f x

n-1 n- 2

0 i i ni =1,3,5 i = 2,4,6

h I = f(x ) + 4 f(x ) + 2 f(x ) + f(x )

3

Composite Simpsons 1/3 Rule


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Composite Simpsons 1/3 RuleAlgorithm for Uniform Spacing (Equal Segments)

n

abh

2

nm

13

m

k

2k - 2 2k -1 2kh

I = f x + 4f x f x

x0 x1 x2 x3 x4 x5 x6 . . . xn-1 xn


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Composite Simpsons 1/3 Rule

Evaluate the integral

n = 2

n = 4

dxxeI4

0

x2

%96.57411.82404)2(4032

)4()2(4)0(3

84

eefff

hI

%70.8975.56704)3(4)2(2)(40

3

1

)4()3(4)2(2)1(4)0(3

8642

eeeefffff

hI


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MATLAB Code

dxxeI4

0

x2function s = fvalue(x)s = x*exp(2*x);

a = 0; b = 4; n = 4;h = (b - a)/n;S = 0; m = n/2;

for k = 1 : m

S = S + fvalue(a+h*(2*k-2))+4*fvalue(a+h*(2*k-1))+fvalue(a+h*2*k);end

I = h*S/3;disp(I);

fvalue.m

simpsn3.m


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Approximate the function by a cubic polynomial

Simpsons 3/8-Rule

0 1 2 33

( ) ( ) 3 ( ) 3 ( ) ( )8

b

a

h f x dx f x f x f x f x

1

3m

k

3k -3 3k -2 3k -1 3k3h

I = f x + 3f x f x f x8

x0 x1x

f(x)

x2h h

L(x)

x3h

n must be a multiple of3

3

nm


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Example: Simpsons Rules

Evaluate the integral

Simpsons 1/3-Rule

Simpsons 3/8-Rule

dxxe4

0

x2

%.

.

..

.)(

)()()(

96579265216

41182409265216

4118240e4e2403

2

4f2f40f3

hdxxeI

84

4

0

x2

%71.30

926.5216

209.6819926.5216

209.6819832.11923)33933.552(3)18922.19(308

)4/3(3

)4(f)3

8

(f3)3

4

(f3)0(f8

h3

dxxeI

4

0

x2


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MATLAB Code

dxxeI4

0

x2function s = fvalue(x)s = x*exp(2*x);

a = 0; b = 4; n = 6;h = (b - a)/n;S = 0; m = n/3;

for k = 1 : m

S = S + fvalue(a+h*(3*k-3))+(3*fvalue(a+h*(3*k-2)))+(3*fvalue(a+h*(3*k-1)))+fvalue(a+h*3*k);end

I = (3*h*S)/8;disp(I);

fvalue.m

simpsn38.m


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MATLAB Code (Combined Rules)

x=[0.5 2 3 4 6 8 10 11 ]f=[320 268 233 226 226 212 142 107 ]n=8;h =x(2) - x(1);k = 1;

sum = 0;for j =2:n

if j~=nhf = x(j+1) - x(j);

elsehf = abs(h+1); % this is for last value

end

if abs(h-hf)


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MATLAB Code

if k ==1

sum = sum + (h/2)*(f(j-1)+f(j));

else

if k==2

sum = sum + (h/3)*(f(j-2)+(4*f(j-1))+f(j));

elsesum = sum + ((3*h)/8)*(f(j-3)+(3*f(j-2))+(3*f(j-1))+f(j));

end

k=1;

end

end

h = hf;

end

disp(sum);


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MATLAB Functions

Command Description

quad (function,a,b,tol) Uses an adaptive Simpsons rule to compute the integral ofthe functionfunctionwith a as the lower integration limitand b as the upper limit. The parameter tol is optional. tolindicates the specified error tolerance.

trapz (x,y) Uses trapezoidal integration to compute the integral ofywith respect to x, where the array y contains the functionvalues at the points contained in the array x.

(use of the trapz function)>> x = linspace(0, pi, 10);>> y = sin(x);>> trapz(x, y)

ans =

1.97965081121648

(use of the quad function)>> quad('sin(x)', 0, pi, 0.001)

ans =

1.99999349653496


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Curve Fitting

15 May 2011

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ME262 Numerical Analysis Sessional


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MATLAB Functions

Command Description

polyfit(x,y,n) Two vectors x and y fits a polynomial of order n through

the data points (xi,yi) and returns (n+1) coefficients of thepowers of x in the vector a. The coefficients are arrangedin the decreasing order of the powers of x, i.e.,

a = [an an-1 . a1 a0 ]polyval(a,x) A data vector x and the coefficients of a polynomial in a

row vector a, the command y = polyval(a,x) evaluates thepolynomial at the data points xi .


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Example

x = 5:5:50;y = [16 25 32 33 38 36 39 40 42 42];p = polyfit(x,y,2); % fit a line (2nd order polynomial)F_fit = polyval(p,x); % evaluate the polynomial at new pointsplot(x,y,'o',x,F_fit); % plot the data and fitted curve

xlabel('x');ylabel('y');title ('The variation of x with y');

x 5 10 15 20 25 30 35 40 50

y 16 25 32 33 38 36 39 49 42


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Example (Contd.)


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Example

% Power curve fit example y = a x^b ---- log(x) = log(a)+ b log(y)

x = 5:5:50;

y = [16 25 32 33 38 36 39 40 42 42];xbar = log10(x);ybar = log10(y);p = polyfit(xbar,ybar,1); % fit a line (1st order polynomial) p=[p1 p0]

a = 10^p(2); % since p(2) is p0ynew = a.*(x).^p(1);

plot(x,y,'o',x,ynew); % plot the data and fitted curvexlabel('x');ylabel('y');title ('The variation of x with y');disp([p]);disp([p(1) a]);

x 5 10 15 20 25 30 35 40 50

y 16 25 32 33 38 36 39 49 42


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THANK YOU

5/15/2011

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Its all about today!!

lecture 4 numerical integration

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