Chapter 7 Techniques of Integration Solutions Stewart Eighth Edition
Chapter 7 Techniques of Integration Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
7. 4 Integration of Rational Functions by Partial Fractions Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (1 of 16) In this section we show to integrate any rational function (a ratio of polynomials) by expressing it as a sum of simpler fractions, called partial fractions, that we already know how to integrate. To illustrate the method, observe that by taking the fractions to a common denominator we obtain Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (2 of 16) If we now reverse the procedure, we see how to integrate the function on the right side of this equation: Stewart, Calculus Early Trancendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (3 of 16) To see how the method of partial fractions works in general, let's consider a rational function where P and Q are polynomials. It's possible to express f as a sum of simpler fractions provided that the degree of P is less than the degree of Q. Such a rational function is called proper. Stewart, Calculus Early Trancendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (4 of 16) We know that if where an ≠ 0, then the degree of P is n and we write deg(P) = n. 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) until a remainder R (x) is obtained such that deg(R) < deg(Q). Stewart, Calculus Early Trancendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (5 of 16) The division statement is where S and R are also polynomials. As the next example illustrates, sometimes this preliminary step is all that is required. Stewart, Calculus Early Trancendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 1 Find Solution: Since the degree of the numerator is greater than the degree of the denominator, we first perform the long division. This enables us to write Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (6 of 16) The next step is to factor the denominator Q(x) as far as possible. It can be shown that any polynomial Q can be factored as a product of linear factors (of the form ax + b) and irreducible quadratic factors (of the form For instance, if we could factor it as Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (7 of 16) The third step is to express the proper rational function as a sum of partial fractions of the form (from Equation 1) A theorem in algebra guarantees that it is always possible to do this. We explain the details for the four cases that occur. Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (8 of 16) Case I The denominator Q(x) is a product of distinct linear factors. This means that we can write where no factor is repeated (and no factor is a constant multiple of another). Stewart, Calculus Early Trancendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (9 of 16) In this case the partial fraction theorem states that there exist constants A 1, A 2, . . . , Ak such that These constants can be determined as in the following example. Stewart, Calculus Early Trancendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 2 Evaluate Solution: Since the degree of the numerator is less than the degree of the denominator, we don't need to divide. We factor the denominator as Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 2 – Solution (1 of 4) Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand (2) has the form To determine the values of A, B, and C, we multiply both sides of this equation by the product of the denominators, x(2 x − 1)(x + 2), obtaining Stewart, Calculus Early Trancendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 2 – Solution (2 of 4) Expanding the right side of Equation 4 and writing it in the standard form for polynomials, we get The polynomials in Equation 5 are identical, so their coefficients must be equal. The coefficient of on the right side, 2 A + B + 2 C, must equal the coefficient of on the left side—namely, 1. Likewise, the coefficients of x are equal and the constant terms are equal. Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 2 – Solution (3 of 4) This gives the following system of equations for A, B, and C: Solving, we get, and so Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 2 – Solution (4 of 4) In integrating the middle term we have made the mental substitution u = 2 x − 1, which gives du = 2 dx and Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (10 of 16) Note: We can use an alternative method to find the coefficients A, B, and C in Example 2. Equation 4 is an identity; it is true for every value of x. Let's choose values of x that simplify the equation. If we put x = 0 in Equation 4, then the second and third terms on the right side vanish and the equation then becomes − 2 A = − 1, or Likewise, gives 10 C = − 1, so Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (11 of 16) Case II Q(x) is a product of linear factors, some of which are repeated. Suppose the first linear factor (a 1 x + b 1) is repeated r times; that is, occurs in the factorization of Q(x). Then instead of the single term in Equation 2, we would use Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (12 of 16) By way of illustration, we could write but we prefer to work out in detail a simpler example. Stewart, Calculus Early Trancendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 4 Find Solution: The first step is to divide. The result of long division is Stewart, Calculus Early Trancendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 4 – Solution (1 of 4) The second step is to factor the denominator Since Q(1) = 0, we know that x − 1 is a factor and we obtain Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 4 – Solution (2 of 4) Since the linear factor x − 1 occurs twice, the partial fraction decomposition is Multiplying by the least common denominator, we get Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 4 – Solution (3 of 4) Now we equate coefficients: Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 4 – Solution (4 of 4) Solving, we obtain A = 1, B = 2, and C = − 1, so Stewart, Calculus Early Trancendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (13 of 16) Case III Q(x) contains irreducible quadratic factors, none of which is repeated. If Q(x) has the factor then, in addition to the partial fractions in Equations 2 and 7, the expression for will have a term of the form where A and B are constants to be determined. Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (14 of 16) For instance, the function given by has a partial fraction decomposition of the form The term given in (9) can be integrated by completing the square (if necessary) and using the formula Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 6 Evaluate Solution: Since the degree of the numerator is not less than the degree of the denominator, we first divide and obtain Stewart, Calculus Early Trancendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 6 – Solution (1 of 3) Notice that the quadratic is irreducible because its discriminant is This means it can't be factored, so we don't need to use the partial fraction technique. To integrate the given function we complete the square in the denominator: This suggests that we make the substitution u = 2 x − 1. Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 6 – Solution (2 of 3) Then du = 2 dx and Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 6 – Solution (3 of 3) Stewart, Calculus Early Trancendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (15 of 16) Note: Example 6 illustrates the general procedure for integrating a partial fraction of the form We complete the square in the denominator and then make a substitution that brings the integral into the form Then the first integral is a logarithm and the second is expressed in terms of Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration of Rational Functions by Partial Fractions (16 of 16) Case IV Q(x) contains a repeated irreducible quadratic factor. If Q(x) has the factor single partial fraction (9), the sum occurs in the partial fraction decomposition of then instead of the Each of the terms in (11) can be integrated by using a substitution or by first completing the square if necessary. Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 8 Evaluate Solution: The form of the partial fraction decomposition is Multiplying by we have Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 8 – Solution (1 of 2) If we equate coefficients, we get the system which has the solution A = 1, B = − 1, C = − 1, D = 1 and E = 0. Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 8 – Solution (2 of 2) Thus Stewart, Calculus Early Trancendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Rationalizing Substitutions Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Rationalizing Substitutions Some nonrational functions can be changed into rational functions by means of appropriate substitutions. In particular, when an integrand contains an expression of the form then the substitution may be effective. Other instances appear in the exercises. Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 9 Evaluate Solution: Let Then and dx = 2 u du. Therefore Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 9 – Solution We can evaluate this integral either by factoring and using partial fractions or by using Formula 6 with a = 2: Stewart, Calculus: Early Transcendentals, 8 th Edition. © 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 Techniques of Integration Solutions Stewart Eighth Edition
Source: https://slidetodoc.com/chapter-7-techniques-of-integration-stewart-calculus-early-3/