1 / 89

TECHNIQUES OF INTEGRATION

Learn how to integrate rational functions by expressing them as a sum of simpler fractions using the technique of partial fractions.

ekillion
Download Presentation

TECHNIQUES OF INTEGRATION

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 7 TECHNIQUES OF INTEGRATION

  2. TECHNIQUES OF INTEGRATION 7.4Integration of Rational Functions by Partial Fractions • In this section, we will learn: • How to integrate rational functions • by reducing them to a sum of simpler fractions.

  3. PARTIAL FRACTIONS • We show how to integrate any rational function (a ratio of polynomials) by expressing it as a sum of simpler fractions, called partial fractions. • We already know how to integrate partial functions.

  4. INTEGRATION BY PARTIAL FRACTIONS • To illustrate the method, observe that, by taking the fractions 2/(x – 1) and 1/(x – 2) to a common denominator, we obtain:

  5. INTEGRATION BY PARTIAL FRACTIONS • If we now reverse the procedure, we see how to integrate the function on the right side of this equation:

  6. INTEGRATION BY PARTIAL FRACTIONS • To see how the method of partial fractions works in general, let’s consider a rational function • where P and Q are polynomials.

  7. PROPER FUNCTION • It’s possible to express f as a sum of simpler fractions if the degree of P is less than the degree of Q. • Such a rational function is called proper.

  8. DEGREE OF P • Recall that, if • where an≠ 0, then the degree of P is n and we write deg(P) = n.

  9. PARTIAL FRACTIONS • If f is improper, that is, deg(P) ≥ deg(Q), then we must take the preliminary step of dividing Q into P (by long division). • This is done until a remainder R(x) is obtained such that deg(R) < deg(Q).

  10. PARTIAL FRACTIONS Equation 1 • The division statement is • where S and R are also polynomials.

  11. PARTIAL FRACTIONS • As the following example illustrates, sometimes, this preliminary step is all that is required.

  12. PARTIAL FRACTIONS Example 1 • Find • The degree of the numerator is greater than that of the denominator. • So, we first perform the long division.

  13. PARTIAL FRACTIONS Example 1 • This enables us to write:

  14. PARTIAL FRACTIONS • The next step is to factor the denominator Q(x) as far as possible.

  15. FACTORISATION OF Q(x) • It can be shown that any polynomial Qcan be factored as a product of: • Linear factors (of the form ax +b) • Irreducible quadratic factors (of the form ax2 + bx +c, where b2 – 4ac < 0).

  16. FACTORISATION OF Q(x) • For instance, if Q(x) = x4 – 16, we could factor it as:

  17. FACTORISATION OF Q(x) • The third step is to express the proper rational function R(x)/Q(x) as a sum of partial fractionsof the form:

  18. FACTORISATION OF Q(x) • A theorem in algebra guarantees that it is always possible to do this. • We explain the details for the four cases that occur.

  19. CASE 1 • The denominator Q(x) is a product of distinct linear factors.

  20. CASE 1 • This means that we can write • Q(x) = (a1x + b1) (a2x + b2)…(akx + bk) • where no factor is repeated (and no factor is a constant multiple of another.

  21. CASE 1 Equation 2 • In this case, the partial fraction theorem states that there exist constants A1, A2, . . . , Ak such that:

  22. CASE 1 • These constants can be determined as in the following example.

  23. PARTIAL FRACTIONS Example 2 • Evaluate • The degree of the numerator is less than the degree of the denominator. • So, we don’t need to divide.

  24. PARTIAL FRACTIONS Example 2 • We factor the denominator as: 2x3 + 3x2 – 2x = x(2x2 + 3x – 2) = x(2x – 1)(x + 2) • It has three distinct linear factors.

  25. PARTIAL FRACTIONS E. g. 2—Equation 3 • So, the partial fraction decomposition of the integrand (Equation 2) has the form

  26. PARTIAL FRACTIONS E. g. 2—Equation 4 • To determine the values of A, B, and C, we multiply both sides of the equation by the product of the denominators, x(2x – 1)(x + 2), obtaining: • x2 + 2x + 1 = A(2x – 1)(x + 2) + Bx(x + 2) • + Cx(2x – 1)

  27. PARTIAL FRACTIONS E. g. 2—Equation 5 • Expanding the right side of Equation 4 and writing it in the standard form for polynomials, we get: • x2 + 2x + 1 = (2A + B + 2C)x2 • + (3A + 2B – C) – 2A

  28. PARTIAL FRACTIONS Example 2 • The polynomials in Equation 5 are identical. • So, their coefficients must be equal. • The coefficient of x2 on the right side, 2A +B + 2C, must equal that of x2 on the left side—namely, 1. • Likewise, the coefficients of x are equal and the constant terms are equal.

  29. PARTIAL FRACTIONS Example 2 • This gives the following system of equations for A, B, and C: • 2A + B + 2C = 1 • 3A + 2B – C = 2 • –2A = –1

  30. PARTIAL FRACTIONS Example 2 • Solving, we get: • A = ½ • B = 1/5 • C = –1/10

  31. PARTIAL FRACTIONS Example 2 • Hence,

  32. PARTIAL FRACTIONS Example 2 • In integrating the middle term, we have made the mental substitution u = 2x – 1, which gives du = 2 dx and dx =du/2.

  33. NOTE • We can use an alternative method to find the coefficients A, B, and Cin Example 2.

  34. NOTE • Equation 4 is an identity. • It is true for every value of x. • Let’s choose values of x that simplify the equation.

  35. NOTE • If we put x = 0 in Equation 4, the second and third terms on the right side vanish, and the equation becomes –2A = –1. • Hence, A = ½.

  36. NOTE • Likewise, x = ½ gives 5B/4 = 1/4 and x = –2 gives 10C = –1. • Hence, B = 1/5 and C = –1/10.

  37. NOTE • You may object that Equation 3 is not valid for x = 0, ½, or –2. • So, why should Equation 4 be valid for those values?

  38. NOTE • In fact, Equation 4 is true for all values of x, even x = 0, ½, and –2 . • See Exercise 69 for the reason.

  39. PARTIAL FRACTIONS Example 3 • Find , where a≠ 0. • The method of partial fractions gives: • Therefore,

  40. PARTIAL FRACTIONS Example 3 • We use the method of the preceding note. • We put x = a in the equation and get A(2a) = 1. So, A = 1/(2a). • If we put x = –a, we get B(–2a) = 1. So, B =–1/(2a).

  41. PARTIAL FRACTIONS Example 3 • Therefore,

  42. PARTIAL FRACTIONS E. g. 3—Formula 6 • Since ln x –ln y =ln(x/y), we can write the integral as: • See Exercises 55–56 for ways of using Formula 6.

  43. CASE 2 • Q(x) is a product of linear factors, some of which are repeated.

  44. CASE 2 • Suppose the first linear factor (a1x +b1) is repeated r times. • That is, (a1x +b1)r occurs in the factorization of Q(x).

  45. CASE 2 Equation 7 • Then, instead of the single term A1/(a1x +b1) in Equation 2, we would use:

  46. CASE 2 • By way of illustration, we could write: • However, we prefer to work out in detail a simpler example, as follows.

  47. PARTIAL FRACTIONS Example 4 • Find • The first step is to divide. • The result of long division is:

  48. PARTIAL FRACTIONS Example 4 • The second step is to factor the denominator Q(x) = x3 – x2 – x + 1. • Since Q(1) = 0, we know that x – 1 is a factor, and we obtain:

  49. PARTIAL FRACTIONS Example 4 • The linear factor x – 1 occurs twice. • So, the partial fraction decomposition is:

  50. PARTIAL FRACTIONS E. g. 4—Equation 8 • Multiplying by the least common denominator, (x – 1)2 (x + 1), we get:

More Related