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Introductory Mathematical Analysis - Chapter 15 : Methods and Applications of Integration

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- 1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences ©2007 Pearson Education Asia Chapter 15Chapter 15 Methods and Applications of IntegrationMethods and Applications of Integration
- 2. ©2007 Pearson Education Asia INTRODUCTORY MATHEMATICAL ANALYSIS 0. Review of Algebra 1. Applications and More Algebra 2. Functions and Graphs 3. Lines, Parabolas, and Systems 4. Exponential and Logarithmic Functions 5. Mathematics of Finance 6. Matrix Algebra 7. Linear Programming 8. Introduction to Probability and Statistics
- 3. ©2007 Pearson Education Asia 9. Additional Topics in Probability 10. Limits and Continuity 11. Differentiation 12. Additional Differentiation Topics 13. Curve Sketching 14. Integration 15. Methods and Applications of Integration 16. Continuous Random Variables 17. Multivariable Calculus INTRODUCTORY MATHEMATICAL ANALYSIS
- 4. ©2007 Pearson Education Asia • To develop and apply the formula for integration by parts. • To show how to integrate a proper rational function. • To illustrate the use of the table of integrals. • To develop the concept of the average value of a function. • To solve a differential equation by using the method of separation of variables. • To develop the logistic function as a solution of a differential equation. • To define and evaluate improper integrals. Chapter 15: Methods and Applications of Integration Chapter ObjectivesChapter Objectives
- 5. ©2007 Pearson Education Asia Integration by Parts Integration by Partial Fractions Integration by Tables Average Value of a Function Differential Equations More Applications of Differential Equations Improper Integrals 15.1) 15.2) 15.3) Chapter 15: Methods and Applications of Integration Chapter OutlineChapter Outline 15.4) 15.5) 15.6) 15.7)
- 6. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.1 Integration by Parts15.1 Integration by Parts Example 1 – Integration by Parts Formula for Integration by Parts Find by integration by parts. Solution: Let and Thus, ∫∫ −= duvuvdvu dx x xln ∫ ( )( ) ( ) ( )[ ] Cxx dx x xxxdx x x +−= −= ∫∫ 2ln2 1 22ln ln 2/1 xu ln= dx x dv 1 = dx x du 1 = 2/12/1 2xdxxv == ∫ −
- 7. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.1 Integration by Parts Example 3 – Integration by Parts where u is the Entire Integrand Determine Solution: Let and Thus, .ln∫ dyy yv dydv = = ( )( ) [ ] Cyy Cyyy dy y yyydyy +−= +−= −= ∫∫ 1ln ln 1 lnln dy y du yu 1 ln = =
- 8. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.1 Integration by Parts Example 5 – Applying Integration by Parts Twice Determine Solution: Let and Thus, .122 ∫ + dxex x dxxdu xu 2 2 = = 2/12 12 + + = = x x ev dxedv dxxe ex dxx eex dxex x x xx x 2 )2( 22 12 122 12122 122 ∫ ∫∫ + + ++ + −= −= 1 1212 1212 12 42 22 C exe dx exe dxxe xx xx x +−= −= ++ ++ + ∫∫
- 9. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.1 Integration by Parts Example 5 – Applying Integration by Parts Twice Solution (cont’d): Cxx e C exeex dxex x xxx x + +−= ++−= + +++ + ∫ 2 1 2 422 2 12 1212122 122
- 10. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.2 Integration by Partial Fractions15.2 Integration by Partial Fractions Example 1 – Distinct Linear Factors • Express the integrand as partial fractions Determine by using partial fractions. Solution: Write the integral as Partial fractions: Thus, dx x x 273 12 2∫ − + . 9 12 3 1 2 dx x x ∫ − + ( )( ) ( ) ( ) 6 5 6 7 2 ,3ifand,3If 3333 12 9 12 =−=== − + + = −+ + = − + AxBx x B x A xx x x x Cxx x dx x dx dx x x + −++= − + + = − + ∫ ∫∫ 3ln 6 7 3ln 6 5 3 1 333 1 273 12 6 7 6 5 2
- 11. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.2 Integration by Partial Fractions Example 3 – An Integral with a Distinct Irreducible Quadratic Factor Determine by using partial fractions. Solution: Partial fractions: Equating coefficients of like powers of x, we have Thus, dx xxx x ∫ ++ −− 42 23 ( ) xCBxxxAx xx CBx x A xxx x )()1(42 11 42 2 22 ++++=−− ++ + += ++ −− 2,4,4 ==−= CBA ( ) C x xx Cxxx dx xx x x dx xx CBx x A + ++ = ++++−= ++ + + − = ++ + + ∫∫ 4 22 2 22 1 ln 1ln2ln4 1 244 1
- 12. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.2 Integration by Partial Fractions Example 5 – An Integral Not Requiring Partial Fractions Find Solution: This integral has the form Thus, . 13 32 2 dx xx x ∫ ++ + Cxxdx xx x +++= ++ + ∫ 13ln 13 32 2 2 . 1 du u∫
- 13. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.3 Integration by Tables15.3 Integration by Tables Example 1 – Integration by Tables • In the examples, the formula numbers refer to the Table of Selected Integrals given in Appendix B of the book. Find Solution: Formula 7 states Thus, ( ) . 32 2∫ + x dxx ( ) C bua a bua bbua duu + + ++= + ∫ ln 1 22 ( ) C x xdx x x + + ++= + ∫ 32 2 32ln 9 1 32 2
- 14. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.3 Integration by Tables Example 3 – Integration by Tables Find Solution: Formula 28 states Let u = 4x and a = √3, then du = 4 dx. . 316 2∫ +xx dx C u aau aauu du + −+ = + ∫ 22 22 ln 1 C x x xx dx + −+ = + ∫ 4 3316 ln 3 1 316 2 2
- 15. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.3 Integration by Tables Example 5 – Integration by Tables Find Solution: Formula 42 states If we let u = 4x, then du = 4 dx. Hence, ( ) .4ln7 2 dxxx∫ ( ) C n u n uu duuu nn n + + − + = ++ ∫ 2 11 11 ln ln ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) Cx x C xxx dxxxdxxx +−= + −= = ∫∫ 14ln3 9 7 9 4 3 4ln4 64 7 44ln4 4 7 4ln7 3 33 2 3 2
- 16. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.3 Integration by Tables Example 7 – Finding a Definite Integral by Using Tables Evaluate Solution: Formula 32 states Letting u = 2x and a2 = 2, we have du = 2 dx. Thus, ( ) . 24 4 1 2/32∫ +x dx ( ) C aua u au du + ± ± = ± ∫ 2222/322 ( ) C aua u au du + ± ± = ± ∫ 2222/322 ( ) ( ) 62 1 66 2 222 1 22 1 24 8 2 2 4 1 2/32 4 1 2/32 −= + = + = + ∫∫ u u u du x dx
- 17. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.4 Average Value of a Function15.4 Average Value of a Function Example 1 – Average Value of a Function • The average value of a function f (x) is given by Find the average value of the function f(x)=x2 over the interval [1, 2]. Solution: ( ) dxxf ab f b a 1 ∫− = ( ) 3 7 312 1 1 2 1 32 1 2 = = − = − = ∫ ∫ x dxx dxxf ab f b a
- 18. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.5 Differential Equations15.5 Differential Equations Example 1 – Separation of Variables • We will use separation of variables to solve differential equations. Solve Solution: Writing y’ as dy/dx, separating variables and integrating, .0,if' >−= yx x y y xCy dx x dy y x y dx dy lnln 11 1 −= −= −= ∫∫
- 19. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration Example 1 – Separation of Variables Solution (cont’d): 0, ln ln 1 1 >= = = − xC x C y e e y ey x C xC
- 20. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.5 Differential Equations Example 3 – Finding the Decay Constant and Half-Life If 60% of a radioactive substance remains after 50 days, find the decay constant and the half-life of the element. Solution: Let N be the size of the population at time t, tλ eNN − = 0 ( ) 01022.0 50 6.0ln 6.0 6.0and50When 50 00 0 ≈−= = == − λ eNN NNt λ days.82.67 2ln islifehalftheandThus, 01022.0 0 ≈≈ − λ eNN t
- 21. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.6 More Applications of Differential Equations15.6 More Applications of Differential Equations Logistic Function • The function is called the logistic function or the Verhulst– Pearl logistic function. Alternative Form of Logistic Function ct be M N − + = 1 t bC M N + = 1
- 22. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.6 More Applications of Differential Equations Example 1 – Logistic Growth of Club Membership Suppose the membership in a new country club is to be a maximum of 800 persons, due to limitations of the physical plant. One year ago the initial membership was 50 persons, and now there are 200. Provided that enrollment follows a logistic function, how many members will there be three years from now?
- 23. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.6 More Applications of Differential Equations Example 1 – Logistic Growth of Club Membership Solution: Let N be the number of members enrolled in t years, Thus, ( ) 15 11 800 50 1 ,0and800When =⇒ + =⇒ + = == b bbC M N tM t 5lnln 151 800 200 ,200and1When 5 1 =−=⇒ + = == − c e Nt c ( ) 781 151 800 4 5 1 ≈ + =N
- 24. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.6 More Applications of Differential Equations Example 3 – Time of Murder A wealthy industrialist was found murdered in his home. Police arrived on the scene at 11:00 P.M. The temperature of the body at that time was 31◦C, and one hour later it was 30◦C. The temperature of the room in which the body was found was 22◦C. Estimate the time at which the murder occurred. Solution: Let t = no. of hours after the body was discovered and T(t) = temperature of the body at time t. By Newton’s law of cooling, ( ) ( )22−=⇒−= Tk dt dT aTk dt dT
- 25. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.6 More Applications of Differential Equations Example 3 – Time of Murder Solution (cont’d): ( ) CktT dtk T dT +=− = − ∫∫ 22ln 22 ( ) ( ) 9ln02231ln ,0and31When =⇒+=− == CCk tT ( ) ( ) 9 8 ln9ln12230ln ,1and30When =⇒+=− == kk tT ( ) kt T InktT = − ⇒+=− 9 22 9ln22lnHence,
- 26. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.6 More Applications of Differential Equations Example 3 - Time of Murder Solution (cont’d): Accordingly, the murder occurred about 4.34 hours before the time of discovery of the body (11:00 P.M.). The industrialist was murdered at about 6:40 P.M. ( ) ( ) ( ) 34.4 9/8ln 9/15ln 9 8 ln2237ln ,37When −≈=⇒ =− = tt T
- 27. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.7 Improper Integrals15.7 Improper Integrals • The improper integral is defined as • The improper integral is defined as ( ) dxxf a ∫ ∞ ( ) ( ) dxxfdxxf r a r a lim ∫∫ ∞→ ∞ = ( ) ( ) ( ) dxxfdxxfdxxf 0 0 ∫∫∫ ∞ ∞− ∞ ∞− += ( )dxxf∫ ∞ ∞−
- 28. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.7 Improper Integrals Example 1 – Improper Integrals Determine whether the following improper integrals are convergent or divergent. For any convergent integral, determine its value. 2 1 2 1 0 2 limlim 1 a. 1 2 1 3 1 3 =+−= −== − ∞→ − ∞→ ∞ ∫∫ r r r r x dxxdx x [ ] 1limlimb. 0 00 === −∞→−∞→ ∞− ∫∫ r x r r x r x edxedxe [ ] ∞=== ∞→ − ∞→ ∞ ∫∫ r r r r xdxxdx x 1 2/1 1 2/1 1 2limlim 1 c.
- 29. ©2007 Pearson Education Asia Chapter 15: Methods and Applications of Integration 15.7 Improper Integrals Example 3 – Density Function In statistics, a function f is called a density function if f(x) ≥ 0 and . Suppose is a density function. Find k. Solution: ( ) 1=∫ ∞ ∞− dxxf ( ) ≥ = − elsewhere0 0for xke xf x ( ) ( ) [ ] 11lim1lim 101 0 0 00 0 =⇒=⇒= =+⇒=+ − ∞→ − ∞→ ∞ − ∞ ∞− ∫ ∫∫∫ kkedxke dxkedxxfdxxf rx r r x r x

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