Math123 Lecture 1. Dr. Robert C. Busby. Lecturer: Office: Korman 266 Phone :

Transcription

1 Lecturer: Math1 Lecture 1 Dr. Robert C. Busby Office: Korman 66 Phone : rbusby@mcs.drexel.edu Course Web Site: (Links are case sensitive)

2 Integration of Rational Functions using Partial Fractions Example: We wish to evaluate the integral 5x 10 dx ( x 4)( x 1) As a first step, we need to break up the integrand into managable pieces. We will attempt to break the integrand into two pieces, each containing only one of the factors (x 4)(x 1) in its denominator. That is we want to write

3 5x 10 A B ( x 4)( x 1) ( x 4) ( x 1) for some numbers A and B. If the identity is true, then by putting the terms on the right in a form where we have a common denominator, we get: 5x 10 Ax ( 1) Bx ( 4) ( x 4)( x 1) ( x 4)( x 1) It then follows that 5x 10 Ax ( 1) Bx ( 4). This is not an equation to solve for x. It is an identity, meaning that it must be true for all possible values of x.

4 Method 1 for finding A and B. Collect like powers of x together on the right side of the equation 5x 10 Ax ( 1) Bx ( 4). To get 5x 10 ( A Bx ) ( A 4 B) It follows that ( A B) 5 and ( A 4 B) 10 We can solve these equations simultaneously. The first can be written A 5 B and substituted into the second to get (5 B) 4B 10 or 5B 15 It follows that 5B 15, or B. Thus A and so

5 5x 10 ( x 4)( x 1) ( x 4) ( x 1) Then 5x 10 dx dx dx ln x 4 ln x 1 C ( x 4)( x 1) ( x 4) ( x 1) Method for finding A and B. substitute values of x into the identity 5x 10 Ax ( 1) Bx ( 4). Any values will do, but the ones that produce the simplest results are x 4 and x 1. When substituted in, they produce the results 10 A(5) and 15 B( 5). Thus A and B, as before.

6 The terms on the right side of the equation 5x 10 ( x 4)( x 1) ( x 4) ( x 1) are called partial fractions. Any rational function px () qx () with p(x) having smaller degree than q(x), can be written as the sum of partial fractions, and then integrated. We must first explore what partial fractions to use.

7 Types of partial fractions Factoring a polynomial means breaking it up as the product of the smallest possible terms. Such terms may be linear, that is of the form (ax b) or quadratic, that is of the form ( ax bx c) Theorem: Every polynomial with real coefficients can be uniquely factored into linear factors and irreducible quadratic factors, that is quadratic factors that cannot be factored further without the use of complex numbers.

8 Linear and quadratic factors can occur in a polynomial more that once. If a linear factor (ax b) occurs m times then we will combine these factors together to produce ( ax b)m If the irreducible quadratic factor ( ax bx c) occurs m times, then these factors are combined to produce ( ax bx c)m Then we can restate the previous theorem as follows: Theorem: Every polynomial with real coefficients can be uniquely factored into distinct factors of the form ( ax b)m and ( ax bx c)m with a, b, and c real.

9 Suppose that we wish to integrate a rational function rx () px () qx () whose denominator has been factored as indicated in the previous theorem. It can be shown that the rational function can be written as the sum of partial fractions according to the following rules. Linear Factor Rule. For each factor of q having the form ( ax b)m the partial fraction decomposition contains a sum of the form A1 A A m ax b ( ax b ) ( ax b ) m

10 If the denominators have quadratic factors, we use the following rule. Quadratic Factor Rule. For each factor of q having the form ( ax bx c)m The partial fraction decomposition contains a sum of the form A1x B1 Ax B Amx Bm ( ax bx c) ( ax bx c) ( ax bx c) m

11 Example. Write out the form of the partial fraction expansion of the following rational functions, but do not solve for the unknown constants. (a) (b) (c) rx () x x 5 ( x 5)( x ) rx () x x 1 ( x )( x ) (x x 1) 4 rx () x x 5 ( x 6) ( x ) ( x 5)

12 Solution: x x 5 Ax B Cx D Ex F ( x 5)( x ) ( x 5) ( x ) ( x ) x x 1 A B C Dx E ( x )( x ) (x x 1) ( x ) ( x ) ( x ) (x x 1) x 4 x 5 ( x 6) ( x ) ( x 5) A B C D E F G ( x 6) ( x 6) ( x 6) ( x ) ( x ) ( x 5) ( x 5)

13 Example. Find the partial fraction decomposition of rx () x 7 ( x )( x 5) and integrate the resulting expression. Solution. By the linear factor rule, the partial fractions decomposition must contain factors of the form A and B ( x ) ( x 5) Thus x 7 A B ( x )( x 5) ( x ) ( x 5) We need to find the unknown coefficients A and B. To do this, put the right side over a common denominator, and equate numerators. We get

14 x 7 A B Ax ( 5) Bx ( ) ( x )( x 5) ( x ) ( x 5) ( x )( x 5) so x 7 Ax ( 5) Bx ( ) We can then find A and B by substituting two convenient numbers into this identity. Any two will do, but the simplest results come from using and 5. They are 10 A(8) (coming from x ) and B( 8) (coming from x 5) Thus A 5/4 and B ¼, therefore x ( x )( x 5) 4( x ) 4( x 5) x 7 5 dx 1 dx 5 so ln 1 dx x ln x 5 C ( x )( x 5) 4 ( x ) 4 ( x 5) 4 4

15 Example. Find the partial fraction decomposition of rx () x x 10 ( x 4)( x 1) and integrate the resulting expression. Solution. By the two factor rules, the partial fractions decomposition must contain factors of the form Thus A and Bx C ( x 4) ( x 1) x x 10 A Bx C ( x 4)( x 1) ( x 4) ( x 1)

16 x x 10 ( A Bx C Ax 1) ( Bx C)( x 4) ( x 4)( x 1) ( x 4) ( x 1) ( x 4)( x 1) so x x 10 Ax ( 1) ( Bx C)( x 4) There are three unknowns, so we choose three values of x. We pick x 4, x 0, and x 1. These yield respectively 4 A(17) 10 A 4C 8 A ( B C) Thus A ; 4C 1, so C ; B A C 8, so B 1. x x 10 x ( x 4)( x 1) ( x 4) ( x 1)

17 x x 10 dx x 1 ( 4)( 1) ( x 4) x x ( x 1) ( x 1) dx dx dx 1 ln x 4 ln x 1 tan 1 x C

18 Thus we have the following procedure. 1. Start with a proper fraction r(x) (order of numerator less than order of denominator) with denominator of order k and factored.. Set up the corresponding partial fraction expansion on the right, with k unknowns.. Put the right side over a common denominator, and equate numerators. Choose and plug in k values of x. Pick values that make one factor vanish, and if needed pick the other values small. 4. Solve for the unknown constants. 5. Plug the constants into the partial fraction expansion and integrate (by substitution).