1 Lesson 13: Methods of Integration

Transcription

1 Lesson 3: Methods of Integration Chapter 6 Material: pages in the textbook: Lesson 3 reviews integration by parts and presents integration via partial fraction decomposition as the third of the integration techniques used to compute antiderivatives of more complicated elementary functions. The substitution method reverses the chain rule and integration by parts rearranges the product rule of differentiation to aid in the determination of antiderivatives. It is important to review integration by parts in this lesson since many students are likely to struggle with this method. There a number of applications that require application of integration by parts. The computation of present and future value of an annuity or investment require the use of integration by parts. The things to cover are listed below: More on Integration by Parts: More examples of integration by parts should be covered in this lesson. Use this to review the method. The next part of the lesson will treat integration via partial fraction decomposition. The idea is to break apart a rational function of polynomials into simpler terms that can each be integrated. The presentation in the textbook develops integration through the definition of antiderivatives. In other presentations, the starting point is the determination of the area beneath a curve. The antiderivative is developed after the definite integral is defined and used in a number of examples. In this textbook, the presentation is in the opposite order. This really does not cause any problems. The definite integral needs to be illustrated along with other properties of the definite integral. One of the most important concepts in Chapter 6 is the presentation of the First Fundamental Theorem of Calculus. The essential part of the theorem is that antiderivatives are related to evaluation of definite integrals. It is very important in this lesson to cover the first fundamental theorem of calculus so students will understand the importance of developing antiderivative formulas. From a mathematical point of view, one can say that computing definite integrals is an application of antiderivatives to specific mathematical problems. Note that this means that all antiderivative formulas are available in any attempt to compute an antiderivative. Also, make sure that students understand that a definite integral is positive in special cases. Students should see examples where the evaluation of a definite integral produces positive and negative values and also cases where a definite integral returns a value of zero. In some ways, the case when a definite integral is zero is very important since this implies something about orthogonality. Even though orthogonality is not covered in this course it will likely appear in other mathematics courses students may need to take in their undergraduate programs. Prerequisite Content for Lesson 3 Students must be able to do the following: Students will need to use algebraic manipulations that are required in the integration by parts problems at the beginning of the lesson. Partial fraction decomposition is very important in this lesson. In order to compute antiderivatives of rational functions of polynomials the expression will typically need to be rewritten in a sum of simpler terms. Students should be able to compute integrals that end up producing logarithms. Students must understand how antiderivatives reverse differentiation formulas even in the case of substitution and integration by parts. It is important that students know how to compute antiderivatives. Students will need to review long division method for writing rational functions in a simpler sum of terms. Goals and Objectives for Lesson 3 Students who complete this lesson should be able to: effectively compute antiderivatives using integration by parts,

2 efficiently compute partial fraction decompositions including applying long division in appropriate cases, compute antiderivatives of rational functions of polynomials using partial fraction decomposition, compute definite integrals using the definition of the definite integral, and compute definite integrals using the first fundamental theorem of calculus relating the evaluation of antiderivatives to definite integrals. This lesson should provide students with an appropriate level of mathematical skills to take on most problems in a business curriculum. 2

3 . Lecture Notes for Lesson 3 - Day Lesson 3 should start with a brief review of integration by parts before continuing on to integration by partial fractions. In this lesson, it is important to cover in introduction for partial fraction decomposition so that progress can be made in the second day of lecture for this lesson. More on Integration by Parts If you presented the version of integration by parts that uses differentials, review the definition in class. That is, present the formula u dv = u v v du Use this form in a couple of examples. Problems like x ln(x) dx will provide simple enough examples to illustrate the notation. Examples 206 through 208 provide additional examples. Make sure students are able to apply integration by parts two or more times. That is, present an example like x 2 e x dx to illustrate that sometimes it is necessary to apply this method multiple times in the computation of a single antiderivative. Antiderivatives via Partial Fraction Decomposition In this part of the lesson, it is important for students to start with very simple examples and work towards more complicated examples. You should start with an indefinite integral of the form x dx Show students that the antiderivative can be obtained using the substitution u = x which implies dx = du. The antiderivative is dx = ln x x Repeat this type of example with integrands of the form and f(x) = 2 x f(x) = 7 x + 2 These should get students to see that this type of antiderivative is very simple. Simple Linear Factors The next step is to get students to understand how to take apart a rational function. For example, the sum of terms x + x + = x + (x )(x + ) + x (x )(x + ) = 2x x 2 3

4 This is easy to verify. Then ask students to compute the indefinite integral 2x x 2 dx using the algebraic relationship. The result should look like 2x x 2 dx = ( x + ) dx x + = x dx + = ln x + ln x + + B = ln (x )(x + ) + B = ln x 2 + B x + dx The integral can be evaluated using the decomposition into two terms and then properties of logarithms can be applied to come up with a final antiderivative. You should show students how to compute this antiderivative using substitution. In some cases substitution may be better (if it works). In this case, the substitution u = x 2 will work equally well in this problem. I suggest you compute the integral for students both ways to help them understand. More Linear Factors Problems As a next step up have students consider the computation of an antiderivative of the form x 2 5x + 6 dx Emphasize that the next step should be to factor the polynomials in both the numerator and denominator. Once this is done the partial fraction decomposition can proceed. In this case, x 2 5x + 6 dx = (x 3)(x 2) dx Once the polynomials are factored, a form for the decomposition is needed. In this case, the integrand can be written in the form (x 3)(x 2) = A x 3 + B x 2 There a couple of ways to compute values of A and B in this expression. One way is to clear fractions by multiplying by the polynomial in the denominator into both sides of the last equation. This results in = A (x 2) + B (x 3) Then setting x = 2 results in = A (2 2) + B (2 3) = A (0) + B ( ) So, B =. If we set x = 3, then = A (3 2) + B (3 3) = A + 0 = A So, A = and x 2 5x + 6 dx = So the indefinite integral is complete. = (x 3)(x 2) dx x 2 dx + ( ) (x 3) dx = ln x 2 ln x 3 + C = ln x 2 x 3 + C 4

5 A Couple of Remarks on Integration by Partial Fraction Decomposition From this point on the process is about the same for rational functions of polynomials. There will be some differences in the algebra needed to take apart the expression. However, make sure students understand the process and the fact that the resulting expression is much easier to integrate. All rational functions of polynomials can be integrated. It may take some algebraic knowledge. However, with some practice, these integrals can be computed. 5

6 .2 Lecture Notes for Lesson 3 - Day 2 Start by reviewing a simple example from the previous day. Something like 2 x 2 3x + 2 dx to refresh their memories about factoring this into linear factors. Then move on to more complicated rational functions. A Remark About the Form of the Decomposition Make sure students know that the form of the decomposition involves a polynomial of degree one less in the numerator than in the denominator. If the polynomial in the numerator has a higher degree than the polynomial in the denominator, something else needs to be done. Long Division and Other Algebra In this part of the Lesson, use an example like x x + 2 dx to illustrate the method. In this case, the polynomial in the numerator has the same degree as the polynomial in the denominator. Long division can be used to obtain ( x x + 2 x + 2 dx = x ) dx x + 2 ( = 2 ) dx x + 2 = x 2 ln x B Once an expression involving only a polynomial of degree one less than that in the denominator the integral can be computed. Higher Order Polynomials. Most students will not see polynomials in rational functions of polynomials of very high degree. Please run through as many examples as possible. Include examples where long division or other algebra is needed before the partial fraction decomposition is performed. The examples in the rest of Section 6.3 on pages 276 through 285 should provide illustrations of what students can do. Please work as many examples as you can for the students. A Remark About Quadratic Terms In order to handle all possible cases, differentiation and integration would need to be applied to trigonometric functions. To integrate all possibilities it is necessary to compute trigonometric integrals. Since this course does not cover calculus applied to trigonometric functions we cannot do all integrals in our list. 6

7 .3 Lecture Notes for Lesson 3 - Day 3 It won t hurt to review one or two examples of the integration methods from the previous 2 days of class. However, it is important to work through Section 6.4 where antiderivatives and definite integrals are related through the First Fundamental Theorem of Calculus. Definition of the Definite Integral Present the definition of a definite integral and the appropriate notation. Use the theorem to relate the ideas and then work as many problems as possible. I think the ideas in this section are simple and as long as you work many examples your students will be ok. The Mean Value Theorem Make sure that you cover the Mean Value Theorem for integrals in this section. There are many applications of this theorem to expected values and averages in statistics and many other quantitative courses students will see. 7

8 .4 Homework Another Integration by Parts Problem Problem : Compute the indefinite integral x sech 2 (x) dx by parts. Use the version in terms of differentials u dv = u v v du with u = x and dv = sech 2 (x) dx. 8

9 Integration of Rational Functions to Logarithms Problem 2: Compute the antiderivative in each of the following using a simple substitution. a. 2 8 x dx b. 7 x + 5 dx 9

10 Integration of Simple Linear Products Problem 3: Compute the antiderivative in each case using partial fraction decomposition. a. 2 x 2 7x + 2 dx b. 3 x 2 25 dx 0

11 Definite Integrals and Partial Fractions Problem 4: Compute the following definite integrals. a x 2 7x + 2 dx b. 2 0 x x 2 9 dx

12 Dividing First Problem 5: Compute the following antiderivatives. Reduce the polynomial in the numerator if needed. a. x + 2 x 9 dx b. x + 2 x 2 9 dx 2

13 c. x 2 x 2 9 dx 3

14 General Definite Integral Problems Problem 6: Use any integration methods in Chapter 6 to compute the following definite integrals. a. 0 x e x dx b. x e x2 dx 4

15 c. 3 2 x 2 dx 5

16 First Fundamental Theorem of Calculus Problem 7: Suppose that an antiderivative relationship of the form F (x) = f(x) dx and F (x) = 6 x ln(x + 3) + 4x 3 + B Compute the following definite integrals a. 4 3 f(x) dx b. 3 0 f(x) dx 6

17 Algebraic Properties of Definite Integrals Problem 8: Compute the following definite integrals using the algebraic properties of definite integrals. a. 3 3 f(x) dx b. b a f(x) dx if F (x) is an antiderivative of f(x) and F (a) F (b) = 3 7

18 c. b a f(x) dx if c a f(x) dx = 3 and b f(x) dx = 6. c 8

19 Multiple Integration by Parts Problem Problem 9: Compute the average value of the function on the given interval. Hint: You may use an antiderivative from a previous problem. a. f(x) = x 2 sinh(x), x [, 2] b. y(t) = (t 4) 3, t [, ] 9

20 Present Value Computation via Integration by Parts Problem 0: The present value of an object or annuity can be computed using an integral of the following form. P V = T 0 c(t) e r t dt Suppose that c(t) = 3t and r = Then an associated antiderivative can be obtained using integration by parts on the following. 3 t e 0.05 t dt Compute the antiderivative in this case and the present value, P V, when T = 0. 20