09 numerical integration
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This document discusses numerical integration techniques, including the trapezoidal rule and Simpson's rule. It begins by establishing the need for numerical integration when exact integrals cannot be calculated. It then derives the trapezoidal rule using geometric insight by approximating the area under a curve as trapezoids. The document explains how to apply the trapezoidal rule using equidistant points and presents an example. Finally, it introduces Simpson's rule, which uses quadratic interpolation between three points to better approximate the area under a curve compared to the trapezoidal rule. Students are assigned related homework problems.
![ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
After substitutions and
manipulation!
( ) [ ]210 4
3
2
0
yyy
h
dxxf
x
x
++≈∫](https://image.slidesharecdn.com/09numericalintegration-101108031213-phpapp01/85/09-numerical-integration-25-320.jpg)
![ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Working with three points!
( ) [ ]210 4
3
2
0
yyy
h
dxxf
x
x
++≈∫](https://image.slidesharecdn.com/09numericalintegration-101108031213-phpapp01/85/09-numerical-integration-26-320.jpg)
![ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
For 4-Intervals
( ) [ ]432210 44
3
4
0
yyyyyy
h
dxxf
x
x
+++++≈∫](https://image.slidesharecdn.com/09numericalintegration-101108031213-phpapp01/85/09-numerical-integration-27-320.jpg)
09 numerical integration
- 1. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Numerical Integration
- 2. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Objectives • The student should be able to – Understand the need for numerical integration – Derive the trapezoidal rule using geometric insight – Apply the trapezoidal rule – Apply Simpson’s rule
- 3. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Need for Numerical Integration! ( ) 6 11 01 2 1 3 1 23 1 1 0 231 0 2 =− ++= ++=++= ∫ x xx dxxxI ( ) 11 0 1 0 1 −−− −=−== ∫ eedxeI xx ∫ − = 1 0 2 dxeI x
- 4. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Area under the graph! • Definite integrations always result in the area under the graph (in x-y plane) • Are we capable of evaluating an approximate value for the area?
- 5. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Example • To perform the definite integration of the function between (x0 & x1), we may assume that the area is equal to that of the trapezium: ( ) ( )01 01 2 1 0 xx yy dxxf x x − + ≈∫
- 6. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Adding adjacent areas
- 7. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik The Trapezoidal Rule ( ) ( ) ( ) ( ) 2 2 12 12 01 01 yy xx yy xxI + −+ + −≈ Integrating from x0 to x2: ( ) ( ) ( ) ( ) 2 212112101001 yxxyxxyxxyxx I −+−+−+− ≈
- 8. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik The Trapezoidal Rule ( ) ( ) hxxxx =−=− 1201 If the points are equidistant 2 2110 hyhyhyhy I +++ ≈ ( )210 2 2 yyy h I ++≈
- 9. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Dividing the whole interval into “n” subintervals ++≈ ∑ − = n n i i yyy h I 1 1 0 2 2
- 10. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik The Algorithm • To integrate f(x) from a to b, determine the number of intervals “n” • Calculate the interval length h=(b-a)/n • Evaluate the function at the points yi=f(xi) where xi=x0+i*h • Evaluate the integral by performing the summation ++≈ ∑ − = n n i i yyy h I 1 1 0 2 2
- 11. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Note that X0=a Xn=b
- 12. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Example • Integrate • Using the trapezoidal rule • Use 2,3,&4 points and compare the results ∫= 1 0 2 dxxI
- 13. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Solution • Using 2 points (n=1), h=(1-0)/(1)=1 • Substituting: ( )21 2 1 yyI +≈ ( ) 5.010 2 1 =+≈I X Y 0 0 1 1 2 points, 1 interval
- 14. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Solution • Using 3 points (n=2), h=(1-0)/(2)=0.5 • Substituting: ( )321 2 2 5.0 yyyI ++≈ ( ) 375.0125.0*20 2 5.0 =++≈I X Y 0 0 0.5 0.25 1 1 3 points, 2 interval
- 15. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Solution • Using 4 points (n=3), h=(1-0)/(3)=0.333 • Substituting: ( )4321 22 2 333.0 yyyyI +++≈ ( ) 3519.01444.0*2111.0*20 2 333.0 =+++≈I X Y 0 0 0.33 0.111 0.667 0.444 1 1 4 points, 3 interval
- 16. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Let’s use Interpolation!
- 17. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Interpolation! • If we have a function that needs to be integrated between two points • We may use an approximate form of the function to integrate! • Polynomials are always integrable • Why don’t we use a polynomial to approximate the function, then evaluate the integral
- 18. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Example • To perform the definite integration of the function between (x0 & x1), we may interpolate the function between the two points as a line. ( ) ( )0 01 01 0 xx xx yy yxf − − − +≈
- 19. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Example • Performing the integration on the approximate function: ( ) ( )∫∫ − − − +≈= 1 0 1 0 0 01 01 0 x x x x dxxx xx yy ydxxfI 1 0 0 2 01 01 0 2 x x xx x xx yy xyI − − − +≈
- 20. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Example • Performing the integration on the approximate function: − − − +− − − − +≈ 00 2 0 01 01 0010 2 1 01 01 10 22 xx x xx yy xyxx x xx yy xyI ( ) ( ) 2 01 01 yy xxI + −≈ • Which is equivalent to the area of the trapezium!
- 21. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik The Trapezoidal Rule ( ) ( ) 2 01 01 yy xxI + −≈ ( ) ( ) ( ) ( ) 2 2 12 12 01 01 yy xx yy xxI + −+ + −≈ Integrating from x0 to x2:
- 22. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Simpson’s Rule Using a parabola to join three adjacent points!
- 23. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Quadratic Interpolation • If we get to interpolate a quadratic equation between every neighboring 3 points, we may use Newton’s interpolation formula: ( ) ( ) ( )( )103021 xxxxbxxbbxf −−+−+≈ ( ) ( ) ( )( )1010 2 3021 xxxxxxbxxbbxf ++−+−+≈
- 24. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Integrating ( ) ( ) ( )( )1010 2 3021 xxxxxxbxxbbxf ++−+−+≈ ( ) ( ) ( )( )∫∫ ++−+−+≈ 2 0 2 0 1010 2 3021 x x x x dxxxxxxxbxxbbdxxf ( ) ( ) 2 0 2 0 10 2 10 3 30 2 21 232 x x x x xxx x xx x bxx x bxbdxxf ++−+ −+≈∫
- 25. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik After substitutions and manipulation! ( ) [ ]210 4 3 2 0 yyy h dxxf x x ++≈∫
- 26. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Working with three points! ( ) [ ]210 4 3 2 0 yyy h dxxf x x ++≈∫
- 27. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik For 4-Intervals ( ) [ ]432210 44 3 4 0 yyyyyy h dxxf x x +++++≈∫
- 28. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik In General: Simpson’s Rule ( ) +++≈ ∑∑∫ − = − = n n i i n i i x x yyyy h dxxf n 2 ,..4,2 1 ,..3,1 0 24 30 NOTE: the number of intervals HAS TO BE even
- 29. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Example • Integrate • Using the Simpson rule • Use 3 points ∫= 1 0 2 dxxI
- 30. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Solution • Using 3 points (n=2), h=(1-0)/(2)=0.5 • Substituting: • Which is the exact solution! ( )210 4 3 5.0 yyyI ++≈ ( ) 3 1 125.0*40 3 5.0 =++≈I
- 31. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Homework #7 • Chapter 21, p. 610, numbers: 21.5, 21.6, 21.10, 21.11.