© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 1 of 58 Chapter 9 Techniques of Integration

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 2 of 58  Integration by Substitution  Integration by Parts  Evaluation of Definite Integrals  Approximation of Definite Integrals  Some Applications of the Integral  Improper Integrals Chapter Outline

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 3 of 58 § 9.1 Integration by Substitution

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 4 of 58  Differentiation and Integration Formulas  Integration by Substitution  Using Integration by Substitution Section Outline

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 5 of 58 Differentiation & Integration Formulas

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 6 of 58 Integration by Substitution If u = g(x), then

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 7 of 58 Using Integration by SubstitutionEXAMPLE SOLUTION Determine the integral by making an appropriate substitution. Let u = x 2 + 2x + 3, so that. That is, Therefore, And so we have Rewrite in terms of u. Bring the factor 1/2 outside. Integrate.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 8 of 58 Using Integration by SubstitutionCONTINUED Replace u with x 2 + 2x + 3.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 9 of 58 Using Integration by SubstitutionEXAMPLE SOLUTION Determine the integral by making an appropriate substitution. Let u = 1/x + 2, so that. Therefore, And so we have Rearrange factors. Rewrite in terms of u. Integrate. Rewrite u as 1/x + 2.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 10 of 58 Using Integration by SubstitutionEXAMPLE SOLUTION Determine the integral by making an appropriate substitution. Let u = cos2x, so that. Therefore, And so we have Rearrange factors. Rewrite in terms of u. Integrate. Rewrite u as cos2x.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 11 of 58 § 9.2 Integration by Parts

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 12 of 58  Integration by Parts  Using Integration by Parts Section Outline

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 13 of 58 Integration by Parts G(x) is an antiderivative of g(x).

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 14 of 58 Using Integration by PartsEXAMPLE SOLUTION Evaluate. Our calculations can be set up as follows: Then DifferentiateIntegrate

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 15 of 58 Using Integration by PartsCONTINUED

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 16 of 58 Using Integration by PartsEXAMPLE SOLUTION Evaluate. Our calculations can be set up as follows: Then Notice that the resultant integral cannot yet be solved using conventional methods. Therefore, we will attempt to use integration by parts again.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 17 of 58 Using Integration by Parts Our calculations can be set up as follows: Then Therefore, we have CONTINUED

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 18 of 58 Using Integration by PartsEXAMPLE SOLUTION Evaluate. Our calculations can be set up as follows: Then

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 19 of 58 Using Integration by PartsCONTINUED

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 20 of 58 § 9.3 Evaluation of Definite Integrals

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 21 of 58  The Definite Integral  Evaluating Definite Integrals  Change of Limits Rule  Finding the Area Under a Curve  Integration by Parts and Definite Integrals Section Outline

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 22 of 58 The Definite Integral where F΄(x) = f (x).

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 23 of 58 Evaluating Definite IntegralsEXAMPLE SOLUTION Evaluate. First let u = 1 + 2x and therefore du = 2dx. So, we have

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 24 of 58 Evaluating Definite Integrals Consequently, CONTINUED

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 25 of 58 Change of Limits Rule

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 26 of 58 Using the Change of Limits RuleEXAMPLE SOLUTION Evaluate using the Change of Limits Rule. First let u = 1 + 2x and therefore du = 2dx. When x = 0 we have u = 1 + 2(0) = 1. And when x = 1, u = 1 + 2(1) = 3. Thus

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 27 of 58 Finding the Area Under a CurveEXAMPLE SOLUTION Find the area of the shaded region. To find the area of the shaded region, we will integrate the given function. But we must know what our limits of integration will be. Therefore, we must determine the three x-intercepts of the function. This is the given function.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 28 of 58 Finding the Area Under a Curve Therefore, the left-most region (above the x-axis) starts at x = -3 and ends at x = 0. The right-most region (below the x-axis) starts at x = 0 and ends at x = 3. So, to find the area in the shaded regions, we will use the following. Replace y with 0 to find the x-intercepts. CONTINUED Set each factor equal to 0. Solve for x. Now let’s find an antiderivative for both integrals. We will use u = 9 – x 2 and du = -2xdx.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 29 of 58 Finding the Area Under a CurveCONTINUED Now we solve for the area.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 30 of 58 Finding the Area Under a CurveCONTINUED

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 31 of 58 Integration by Parts & Definite IntegralsEXAMPLE SOLUTION Evaluate. To solve this integral, we will need integration by parts. Our calculations can be set up as follows: Then

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 32 of 58 Integration by Parts & Definite Integrals Therefore, we have CONTINUED

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 33 of 58 § 9.4 Approximation of Definite Integrals

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 34 of 58  The Midpoint Rule  The Trapezoidal Rule  Simpson’s Rule  Error Analysis Section Outline

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 35 of 58 The Midpoint Rule

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 36 of 58 Using the Midpoint RuleEXAMPLE SOLUTION Approximate the following integral by the midpoint rule. We have Δx = (b – a)/n = (4 – 1)/3 = 1. The endpoints of the four subintervals begin at a = 1 and are spaced 1 unit apart. The first midpoint is at a + Δx/2 = 1.5. The midpoints are also spaced 1 unit apart. According to the midpoint rule, the integral is approximately equal to

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 37 of 58 The Trapezoidal Rule

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 38 of 58 Using the Trapezoidal RuleEXAMPLE SOLUTION Approximate the following integral by the trapezoidal rule. As in the last example, Δx = 1 and the endpoints of the subintervals are a 0 = 1, a 1 = 2, a 2 = 3, and a 3 = 4. The trapezoidal rule gives

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 39 of 58 Simpson’s Rule

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 40 of 58 Using Simpson’s RuleEXAMPLE SOLUTION Approximate the following integral by Simpson’s rule. As in the last example, Δx = 1 and the endpoints of the subintervals are a 0 = 1, a 1 = 2, a 2 = 3, and a 3 = 4. Simpson’s rule gives NOTE: Although this happens to be the exact answer, remember that Simpson’s Rule is still just an approximation and therefore it generally yields only an estimate of the exact answer, not the exact answer itself.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 41 of 58 Error Analysis

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 42 of 58 Using Error AnalysisEXAMPLE SOLUTION Obtain a bound on the error of using the midpoint rule with n = 3 to approximate Here a = 1, b = 4, and f (x) = (2x – 3) 3. Differentiating twice, we find that How large could | f ΄΄(x)| be if x satisfies 1 ≤ x ≤ 4? Since the function 48x – 72 is clearly increasing on the interval from 1 to 4 (in fact, it’s increasing everywhere), its greatest value occurs at x = 4. Therefore, its greatest value is

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 43 of 58 Using Error Analysis so we may take A = 120 in the preceding theorem. The error of approximation using the midpoint rule is at most CONTINUED NOTE: We have hitherto determined that the exact value of this integral is 78 and that the midpoint approximation for it (using n = 3) is 72. Therefore, this approximation was in error by 78 – 72 = 6. Our result in this exercise says that our midpoint approximation error should be no greater than 15. Since 6 is no greater than 15, this result suggests that our midpoint approximation was done correctly.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 44 of 58 § 9.5 Some Applications of the Integral

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 45 of 58  The Riemann Sum  Interest Compounded Continuously  Continuous Stream of Income  Population in a Ring Section Outline

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 46 of 58 The Riemann Sum Δx = (b – a)/n, t 1, t 2, …., t n are selected points from a partition [a, b].

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 47 of 58 Interest Compounded Continuously

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 48 of 58 Continuous Stream of Income

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 49 of 58 Continuous Stream of IncomeEXAMPLE SOLUTION Find the present value of a stream of earnings generated over the next 2 years at the rate of t thousand dollars per year at time t, assuming a 10% interest rate. Using the theorem above, we have K(t) = t, r = 0.06, T 1 = 0 and T 2 = 2. So, we have To evaluate, we will need to rewrite the integral as and then evaluate the first integral directly and the second using integration by parts. Upon doing this, we have

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 50 of 58 Continuous Stream of Income Now we evaluate the integral. CONTINUED So the present value of the described stream of earnings is $

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 51 of 58 Population in a Ring

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 52 of 58 Population in a RingEXAMPLE SOLUTION The population density of Philadelphia in 1940 was given by the function 60e -0.4t. Calculate the number of people who lived within 5 miles of the city center. Using the theorem above, we have D(t) = 60e -0.4t, a = 0 and b = 5. So, we have Using integration by parts, we have

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 53 of 58 § 9.6 Improper Integrals

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 54 of 58  Improper Integrals  Evaluating Improper Integrals Section Outline

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 55 of 58 Improper Integrals DefinitionExample Improper Integral: A tool used to calculate the area of a region that extends infinitely far to the right or left along the x-axis. Used when a limit of integration takes on a value of ∞ or -∞.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 56 of 58 Evaluating Improper IntegralsEXAMPLE SOLUTION Evaluate the improper integral if it is convergent. As b → ∞, approaches ∞, so the integral is divergent. Therefore, there is no solution.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 57 of 58 Evaluating Improper IntegralsEXAMPLE SOLUTION Evaluate the improper integral if it is convergent. As b → ∞, approaches 0, so the integral is convergent. Therefore,

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 58 of 58 Evaluating Improper IntegralsEXAMPLE SOLUTION Evaluate the improper integral if it is convergent. As b → -∞, e 4b approaches 0 so that approaches 1/4. Thus the improper integral converges and has value 1/4.