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# 08 numerical integration

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• 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 linear interpolation – Apply the trapezoidal rule – Derive Simpson’s rule using parabolic interpolation – 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 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
• 5. 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 − − − +≈
• 6. 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               − − − +≈
• 7. 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!
• 8. 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:
• 9. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik General Trapezoidal Rule • For all the points equally separated (xi+1-xi=h) • We may write the equation of the previous slide: ( ) ( ) ( ) ( ) ( )321 23 23 12 12 2 2 22 yyy h yy xx yy xxI ++= + −+ + −≈
• 10. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik In general       ++≈ ∑ − = n n i i yyy h I 1 1 0 2 2 Where n is the number if intervals and h=total interval/n
• 11. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Example • Integrate • Using the trapezoidal rule • Use 2 points and compare with the result using 3 points ∫= 1 0 2 dxxI
• 12. 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
• 13. 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
• 14. 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 ++−+−+≈
• 15. 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             ++−+      −+≈∫
• 16. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik After substitutions and manipulation! ( ) [ ]210 4 3 2 0 yyy h dxxf x x ++≈∫
• 17. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik For 4-Intervals ( ) [ ]23210 424 3 4 0 yyyyy h dxxf x x ++++≈∫
• 18. 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
• 19. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Example • Integrate • Using the Simpson rule • Use 3 points ∫= 1 0 2 dxxI
• 20. 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
• 21. ENEM602 Spring 2007 Dr. Eng. Mohammad Tawfik Homework #7 • Chapter 21, pp. 610-612, numbers: 21.1, 21.3, 21.5, 21.25, 21.28. • Due date: Week 8-12 May 2005