Partial Fractions. dx dx x sec tan d 1 4 tan 2. 2 csc d. csc cot C. 2x 5. 2 ln. 2 x 2 5x 6 C. 2 ln. 2 ln x

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1 460_080.qd //04 :08 PM Page CHAPTER 8 Integration Techniques, L Hôpital s Rule, and Improper Integrals Section 8. Partial Fractions Understand the concept of a partial fraction decomposition. Use partial fraction decomposition with linear factors to integrate rational functions. Use partial fraction decomposition with quadratic factors to integrate rational functions. Partial Fractions θ sec Figure This section eamines a procedure for decomposing a rational function into simpler rational functions to which you can apply the basic integration formulas. This procedure is called the method of partial fractions. To see the benefit of the method of partial fractions, consider the integral 6 d. To evaluate this integral without partial fractions, you can complete the square and use trigonometric substitution (see Figure 8.) to obtain a, sec 6 d d sec tan d 4 tan d sec tan d csc d Mary Evans Picture Library JOHN BERNOULLI ( ) The method of partial fractions was introduced by John Bernoulli, a Swiss mathematician who was instrumental in the early development of calculus. John Bernoulli was a professor at the University of Basel and taught many outstanding students, the most famous of whom was Leonhard Euler. Now, suppose you had observed that 6. Then you could evaluate the integral easily, as follows. 6 d d Partial fraction decomposition This method is clearly preferable to trigonometric substitution. However, its use depends on the ability to factor the denominator, 6, and to find the partial fractions and ln ln csc cot C ln 6 6 C ln C ln C ln ln C. ln ln C. 6 C In this section, you will study techniques for finding partial fraction decompositions.

2 460_080.qd //04 :08 PM Page SECTION 8. Partial Fractions STUDY TIP In precalculus you learned how to combine functions such as. The method of partial fractions shows you how to reverse this process.?? Recall from algebra that every polynomial with real coefficients can be factored into linear and irreducible quadratic factors.* For instance, the polynomial 4 can be written as where is a linear factor, is a repeated linear factor, and is an irreducible quadratic factor. Using this factorization, you can write the partial fraction decomposition of the rational epression N 4 where N is a polynomial of degree less than, as follows. N A B C D E Decomposition of N/D into Partial Fractions. Divide if improper: If ND is an improper fraction (that is, if the degree of the numerator is greater than or equal to the degree of the denominator), divide the denominator into the numerator to obtain N D a polynomial N D where the degree of N is less than the degree of D. Then apply Steps,, and 4 to the proper rational epression N D.. Factor denominator: Completely factor the denominator into factors of the form p q m and a b c n where a b c is irreducible.. Linear factors: For each factor of the form p q m, the partial fraction decomposition must include the following sum of m fractions. A p q A p q... A m p q m 4. Quadratic factors: For each factor of the form a b c n, the partial fraction decomposition must include the following sum of n fractions. B C a b c B C a b c... B n C n a b c n * For a review of factorization techniques, see Precalculus, 6th edition, by Larson and Hostetler or Precalculus: A Graphing Approach, 4th edition, by Larson, Hostetler, and Edwards (Boston, Massachusetts: Houghton Mifflin, 004 and 00, respectively).

3 460_080.qd //04 :08 PM Page 4 4 CHAPTER 8 Integration Techniques, L Hôpital s Rule, and Improper Integrals Linear Factors Algebraic techniques for determining the constants in the numerators of a partial decomposition with linear or repeated linear factors are shown in Eamples and. EXAMPLE Distinct Linear Factors Write the partial fraction decomposition for Solution Because 6, you should include one partial fraction for each factor and write 6 A where A and B are to be determined. Multiplying this equation by the least common denominator ) yields the basic equation A B. B 6. Basic equation Because this equation is to be true for all, you can substitute any convenient values for to obtain equations in A and B. The most convenient values are the ones that make particular factors equal to 0. NOTE Note that the substitutions for in Eample are chosen for their convenience in determining values for A and B; is chosen to eliminate the term A, and is chosen to eliminate the term B. The goal is to make convenient substitutions whenever possible. To solve for A, let and obtain A B A B0 A. To solve for B, let and obtain A B A0 B B. So, the decomposition is 6 as shown at the beginning of this section. Let in basic equation. Let in basic equation. FOR FURTHER INFORMATION To learn a different method for finding the partial fraction decomposition, called the Heavyside Method, see the article Calculus to Algebra Connections in Partial Fraction Decomposition by Joseph Wiener and Will Watkins in The AMATYC Review. Be sure you see that the method of partial fractions is practical only for integrals of rational functions whose denominators factor nicely. For instance, if the denominator in Eample were changed to, its factorization as would be too cumbersome to use with partial fractions. In such cases, you should use completing the square or a computer algebra system to perform the integration. If you do this, you should obtain d ln ln C.

4 460_080.qd //04 :08 PM Page SECTION 8. Partial Fractions EXAMPLE Repeated Linear Factors Find 0 6 d. FOR FURTHER INFORMATION For an alternative approach to using partial fractions, see the article A Shortcut in Partial Fractions by Xun-Cheng Huang in The College Mathematics Journal. TECHNOLOGY Most computer algebra systems, such as Derive, Maple, Mathcad, Mathematica, and the TI-89, can be used to convert a rational function to its partial fraction decomposition. For instance, using Maple, you obtain the following. > convert 0 6, parfrac, 6 9 Solution Because ( you should include one fraction for each power of and and write 0 6 A B C. Multiplying by the least common denominator yields the basic equation 0 6 A B C. Basic equation To solve for A, let 0. This eliminates the B and C terms and yields 6 A 0 0 A 6. To solve for C, let. This eliminates the A and B terms and yields C The most convenient choices for have been used, so to find the value of B, you can use any other value of along with the calculated values of A and C. Using, A 6, and C 9 produces 0 6 A4 B C B So, it follows that C B 9 B. 0 6 d 6 9 d 6 ln ln 9 C ln 6 9 C. Try checking this result by differentiating. Include algebra in your check, simplifying the derivative until you have obtained the original integrand. NOTE It is necessary to make as many substitutions for as there are unknowns A, B, C,... to be determined. For instance, in Eample, three substitutions 0,, and were made to solve for A, B, and C.

5 460_080.qd //04 :08 PM Page 6 6 CHAPTER 8 Integration Techniques, L Hôpital s Rule, and Improper Integrals Quadratic Factors When using the method of partial fractions with linear factors, a convenient choice of immediately yields a value for one of the coefficients. With quadratic factors, a system of linear equations usually has to be solved, regardless of the choice of. EXAMPLE Find Solution Because Distinct Linear and Quadratic Factors you should include one partial fraction for each factor and write To solve for A, let 0 and obtain To solve for B, let and obtain 0 0 B 0 yields the basic At this point, C and D are yet to be determined. You can find these remaining constants by choosing two other values for and solving the resulting system of linear equations. If, then, using A and B, you can write 6 C D If, you have C D 8 C D. Solving the linear system by subtracting the first equation from the second C D yields C. Consequently, D 4, and it follows that d A B C D 4. Multiplying by the least common denominator 4 equation 4 8 A 4 B 4 C D. 8 A4 0 0 C D. C D d 4 A. B d ln ln ln 4 arctan C.

6 460_080.qd //04 :08 PM Page 7 SECTION 8. Partial Fractions 7 In Eamples,, and, the solution of the basic equation began with substituting values of that made the linear factors equal to 0. This method works well when the partial fraction decomposition involves linear factors. However, if the decomposition involves only quadratic factors, an alternative procedure is often more convenient. EXAMPLE 4 Repeated Quadratic Factors Find 8 d. Solution Include one partial fraction for each power of and write 8 A B C D. Multiplying by the least common denominator yields the basic equation 8 A B C D. Epanding the basic equation and collecting like terms produces 8 A A B B C D 8 A B A C B D. Now, you can equate the coefficients of like terms on opposite sides of the equation. 8 A 0 B D A B A C B D 0 B Using the known values A 8 and B 0, you can write A C 8 C 0 B D 0 D Finally, you can conclude that A C 8 d 8 4 ln C D 0. d C. TECHNOLOGY Use a computer algebra system to evaluate the integral in Eample 4 you might find that the form of the antiderivative is different. For instance, when you use a computer algebra system to work Eample 4, you obtain 8 d ln Is this result equivalent to that obtained in Eample 4? C.

7 460_080.qd //04 :08 PM Page 8 8 CHAPTER 8 Integration Techniques, L Hôpital s Rule, and Improper Integrals When integrating rational epressions, keep in mind that for improper rational epressions such as N D 7 7 you must first divide to obtain N D. The proper rational epression is then decomposed into its partial fractions by the usual methods. Here are some guidelines for solving the basic equation that is obtained in a partial fraction decomposition. Guidelines for Solving the Basic Equation Linear Factors. Substitute the roots of the distinct linear factors into the basic equation.. For repeated linear factors, use the coefficients determined in guideline to rewrite the basic equation. Then substitute other convenient values of and solve for the remaining coefficients. Quadratic Factors. Epand the basic equation.. Collect terms according to powers of.. Equate the coefficients of like powers to obtain a system of linear equations involving A, B, C, and so on. 4. Solve the system of linear equations. Before concluding this section, here are a few things you should remember. First, it is not necessary to use the partial fractions technique on all rational functions. For instance, the following integral is evaluated more easily by the Log Rule. 4 d ln 4 C Second, if the integrand is not in reduced form, reducing it may eliminate the need for partial fractions, as shown in the following integral. 4 d d 4 d d ln C Finally, partial fractions can be used with some quotients involving transcendental functions. For instance, the substitution u sin allows you to write cos u sin, du cos d sin sin d du uu.

8 460_080.qd //04 :08 PM Page 9 SECTION 8. Partial Fractions 9 Eercises for Section 8. In Eercises 6, write the form of the partial fraction decomposition of the rational epression. Do not solve for the constants In Eercises 7 8, use partial fractions to find the integral d d.. 4 d d d. 4. d 4 d. d 4 4 d d 8 4 d d d 4 4 d d d d 4 8 d d 4 d d d In Eercises 9, evaluate the definite integral. Use a graphing utility to verify your result d. d In Eercises 40, use a computer algebra system to determine the antiderivative that passes through the given point. Use the system to graph the resulting antiderivative , 0,. 0, 6. d, 6 9 d,, 4 4 d, d, d d In Eercises 4 46, use substitution to find the integral. sin sin cos cos d cos cos cos sec d sin tan tan d sin d e e 4 d In Eercises 47 0, use the method of partial fractions to verify the integration formula. 47. a b d a ln a b C 48. a d a ln a a C See for worked-out solutions to odd-numbered eercises. d, d, 4 d, d, e 6, 4, 0, 6, a b d a b a b ln a b C a b d a b a ln a b C Slope Fields In Eercises and, use a computer algebra system to graph the slope field for the differential equation and graph the solution through the given initial condition. dy dy.. d 4 d 6 4 y0 y0 Writing About Concepts e e e d. What is the first step when integrating Eplain. d? 4. Describe the decomposition of the proper rational function ND (a) if D p q m, and (b) if D a b c n, where a b c is irreducible. Eplain why you chose that method.. State the method you would use to evaluate each integral. Eplain why you chose that method. Do not integrate. (a) (b) (c) 8 d 8 d d

9 460_080.qd //04 :08 PM Page CHAPTER 8 Integration Techniques, L Hôpital s Rule, and Improper Integrals Writing About Concepts (continued) 6. Determine which value best approimates the area of the region between the -ais and the graph of f 0 over the interval,. Make your selection on the basis of a sketch of the region and not by performing any calculations. Eplain your reasoning. (a) 6 (b) 6 (c) (d) (e) 8 7. Area Find the area of the region bounded by the graphs of y 6, y 0, 0, and. 8. Area Find the area of the region bounded by the graphs of y 76 and y. 9. Modeling Data The predicted cost C (in hundreds of thousands of dollars) for a company to remove p% of a chemical from its waste water is shown in the table. p C A model for the data is given by C Use the model to find the average cost for removing between 7% and 80% of the chemical. 60. Logistic Growth In Chapter 6, the eponential growth equation was derived from the assumption that the rate of growth was proportional to the eisting quantity. In practice, there often eists some upper limit L past which growth cannot occur. In such cases, you assume the rate of growth to be proportional not only to the eisting quantity, but also to the difference between the eisting quantity y and the upper limit L. That is, dydt kyl y. In integral form, you can write this relationship as 4p 0 p00 p, dy yl y k dt. (a) A slope field for the differential equation dydt y y is shown. Draw a possible solution to the differential equation if y0, and another if y0. To print an enlarged copy of the graph, go to the website y 0 p < 00. (b) Where y0 is greater than, what is the sign of the slope of the solution? (c) For y > 0, find lim yt. t (d) Evaluate the two given integrals and solve for y as a function of t, where y 0 is the initial quantity. (e) Use the result of part (d) to find and graph the solutions in part (a). Use a graphing utility to graph the solutions and compare the results with the solutions in part (a). (f) The graph of the function y is a logistic curve. Show that the rate of growth is maimum at the point of inflection, and that this occurs when y L. 6. Volume and Centroid Consider the region bounded by the graphs of y, y 0, 0, and. Find the volume of the solid generated by revolving the region about the -ais. Find the centroid of the region. 6. Volume Consider the region bounded by the graph of y on the interval 0,. Find the volume of the solid generated by revolving this region about the -ais. 6. Epidemic Model A single infected individual enters a community of n susceptible individuals. Let be the number of newly infected individuals at time t. The common epidemic model assumes that the disease spreads at a rate proportional to the product of the total number infected and the number not yet infected. So, ddt k n and you obtain Solve for as a function of t. 64. Chemical Reactions In a chemical reaction, one unit of compound Y and one unit of compound Z are converted into a single unit of compound X. is the amount of compound X formed, and the rate of formation of X is proportional to the product of the amounts of unconverted compounds Y and Z. So, ddt ky 0 z 0, where y 0 and z 0 are the initial amounts of compounds Y and Z. From this equation you obtain (a) Perform the two integrations and solve for in terms of t. (b) Use the result of part (a) to find as t if () y 0 < z 0, () y 0 > z 0, and () y 0 z Evaluate 0 n d k dt. y 0 z 0 d k dt. 4 d in two different ways, one of which is partial fractions Prove Putnam Eam Challenge d. 4 t This problem was composed by the Committee on the Putnam Prize Competition. The Mathematical Association of America. All rights reserved.