Module 10 - Integration 3 (self Study)

SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Self Study Course

MODULE 10

INTEGRATION III

Module Topics 1. Integration of rational functions 2. Improper integrals

A:

Work Scheme based on JAMES (THIRD EDITION)

1. In the previous modules on integration you used simple substitutions to reduce integrals to one of the standard forms appearing on the Zformula sheet. For some integrals it is first necessary to complete the 1 square. For example, consider dx . The quadratic equation x2 + 2x + 5 = 0 does not x2 + 2x + 5 have real roots (since with the usual notation for the solution of a quadratic equation b2 − 4ac < 0) so the denominator in the integrand cannot be expressed as a product of linear factors with real coefficients. To proceed you first complete the square by writing x2 + 2x + 5 = x2 + 2x + 1 + 4 = (x + 1)2 + 4 = (x + 1)2 + 22 , and then reduce the integral to standard form by putting x + 1 = 2u (to make the denominator a multiple dx = 2, i.e. dx = 2 du and the integral can be evaluated as follows: of 1 + u2 ). With this substitution du   Z Z Z Z 1 dx dx x+1 2 du 2 du −1 = = tan + c. = = x2 + 2x + 5 (x + 1)2 + 22 (2u)2 + 22 22 (u2 + 1) 2 2

***Do Exercises 80(g), 82(j),(p) on p.550*** p(x) , where p(x) and q(x) are both polynomials, occur q(x) frequently in practice. You have evaluated integrals of this type when the numerator is the differential of the denominator by using a simple substitution. However, the general situation has not been considered until now. 2. Integrands of rational functions of the form

Partial fractions are often used in evaluating integrals of this type. You should remember that partial fractions were considered in some detail in section 14 of module 1, and you should remind yourself of the method by again working through that section. Partial fractions are also considered in J. and for this alternative account you can read pp.101-107. The summary of the method on pp.106 and 107 seems particularly useful. Note that in step 1 of this summary J. tells you to find, if necessary, the function r(x) by division whereas in module 1 it was suggested that you could find it as part of the partial fraction process. Both methods can be used - you should adopt the one that you find the most straightforward. Note also that in some situations J. uses the cover-up rule to find the constant coefficients which are associated with linear factors. This rule is a neat way of obtaining –1–

the constants but can only be used in some situations and it is not normally much quicker than using the general method outlined in module 1. After splitting up the integrand into partial fractions and then determining the coefficients, the resulting terms can be integrated using the methods that have been developed in previous modules, although one further type of integrand is looked at in more detail below. 3. You have seen inZ section 1 how completing the square is useful in reducing an integrand to standard 2x dx does not fall into this general category because of the appearance of form. The integral x2 + 2x + 5 x in the numerator. However, the numerator is close to the differential of the denominator, and the method of integration proceeds as follows: Z Z Z Z 2x (2x + 2) − 2 2x + 2 2 dx = dx = dx − dx . x2 + 2x + 5 x2 + 2x + 5 x2 + 2x + 5 x2 + 2x + 5 The two integrals in the final expression above are now considered in turn. The numerator in the first integral has been chosen to be the differential of the denominator, so can be integrated by using the substitution du u = x2 + 2x + 5, giving = 2x + 2 with the integral becoming dx Z Z du 2x + 2 dx = = ln |u| + c1 = ln |x2 + 2x + 5| + c1 . x2 + 2x + 5 u The second integral is a constant multiple of the one that was investigated in section 1, and so the answer can be written down. The full solution, therefore, is Z Z Z 2x + 2 2 2x dx = dx − dx x2 + 2x + 5 x2 + 2x + 5 x2 + 2x + 5     x+1 1 = ln |x2 + 2x + 5| + c1 − 2 tan−1 + c2 2 2   x+1 + c. = ln |x2 + 2x + 5| − tan−1 2

4. Turn to p.546 in J. and study Example 8.33. You will notice that all details of the determination of the partial fractions are omitted. You should be able to write down directly the solutions of the two integrals that appear in part (a), but the substitutions u = x + 2 and u = x − 4 respectively can be used if required. Similarly substitutions, if required, that are appropriate for the three integrals in part (b) are u = x − 1, u = x + 2 and u = x + 2 respectively. ***Do Exercises 79(a),(d),(f ),(i,harder) on p.550*** 5. To complete this module you need to study section 9.3 on improper integrals. Study first the introductory comments starting on p.595, including the definition of infinite discontinuity or singularity. J. uses the phrase infinite discontinuity in the subsequent sections but singularity is preferred by most authors. Study section 9.3.1 and work through Example 9.6. Continue by studying the comments at the bottom of p.597 and then work through Example 9.7. Finally, study section 9.3.2 on infinite integrals and Example 9.8. ***Do Exercises 14(e),(g) on p.599*** –2–

B:

Work Scheme based on STROUD (FIFTH EDITION)

Most of this module is covered in S. Completing the square is considered in Programme 16, starting with frame 4 on p.835. First READ through frames 1-18, which deal with two standard results you are not expected to remember in this course. The most relevant part of the programme begins at frames 20 and 21 which you should work through. Note that the result at the top of frame 21 is more general than the one you are given on the formula sheet and in the Data Book, and to obtain the answer after completing the square you will often need to introduce a further substitution. For instance, for Example 1 in frame 21 you need to proceed as follows: Z Z 1 1 dx = dx , 2 2 x + 16 x + 42 and it is then necessary to use the substitution x = 4u so that the denominator becomes a multiple of u2 + 1, which does appear in the list on the formula sheet. From the substitution x = 4u it follows that 1 du = , and hence the calculation can be completed: dx 4 Z Z Z Z   1 1 1 1 4 du 4 du −1 −1 x du du = tan tan + c. dx = du u + c = x2 + 42 (4u)2 + 42 42 (u2 + 1) 4(u2 + 1) 4 4 4 Work through frames 22-35 of Programme 16 with further examples on completing the square, noting that you would again need to use substitutions similar to the one discussed above before being able to use the standard results on the formula sheet. You have studied partial fractions in section 14 of module 1. Make sure you know that work or, alternatively, work through Programme F.7 in S. You can find how partial fractions are used to evaluate integrals by first studying frames 16-22 of Programme F.11 starting on p.343, although this section only covers denominators which are products of linear factors. For a fuller discussion turn to p.819 in Programme 15 and work through frames 31-42. S. does not contain any material on improper integrals, so for this short section you should study section 5 of one of the work schemes based on J.

Specimen Test 10 Z 1.

2.

1 dx. 4x2 + 4x + 2

Determine the indefinite integral

Use partial fractions to find the following indefinite integrals: Z Z 2x + 1 x+3 (i) dx, (ii) dx. (x + 2)(x + 1) (x + 2)2 Z

3.

Determine the indefinite integral

4.

Evaluate the improper integral

x2 dx. +1

x2 Z 1

∞

1 √ dx , x –3–

if it is defined.