1. Consider: then: or It doesn’t matter whether the constant was 3 or -5, since when we take the derivative the constant disappears. However, when we try to reverse the operation: Given:find We don’t know what the constant is, so we put “C” in the answer to remind us that there might have been a constant. 6.1 Antiderivatives with Slope Fields

2. Given: and when, find the equation for. This is called an initial value problem. We need the initial values to find the constant. An equation containing a derivative is called a differential equation. It becomes an initial value problem when you are given the initial condition and asked to find the original equation. 6.1 Antiderivatives with Slope Fields

3. Initial value problems and differential equations can be illustrated with a slope field. 6.1 Antiderivatives with Slope Fields Definition Slope Field or Directional Field A slope field or a directional field for the first order differentiable equation is a plot of sort line segments with slopes f(x,y) for a lattice of points (x,y) in the plane.

4. 6.1 Antiderivatives with Slope Fields 0 0 0 0 1 0 0 2 0 0 3 0 1 0 2 1 1 2 2 0 4 -1 0 -2 -2 0 -4

5. If you know an initial condition, such as (1,-2), you can sketch the curve. By following the slope field, you get a rough picture of what the curve looks like. In this case, it is a parabola. 6.1 Antiderivatives with Slope Fields

6. Integrals such as are called definite integrals because we can find a definite value for the answer. The constant always cancels when finding a definite integral, so we leave it out! 6.1 Antiderivatives with Slope Fields

7. Integrals such as are called indefinite integrals because we can not find a definite value for the answer. When finding indefinite integrals, we always include the “plus C”. 6.1 Antiderivatives with Slope Fields

8. Definition Indefinite Integral The set of all antiderivatives of a function f(x) is the indefinite integral of f with respect to x and is denoted by

9. 6.1 Antiderivatives with Slope Fields

13. Find the position function for v(t) = t 3 - 2t 2 + t s(0) = 1 C = 1

14. 6.1 Antiderivatives with Slope Fields log cabin log cabin + C houseboat

15. The chain rule allows us to differentiate a wide variety of functions, but we are able to find antiderivatives for only a limited range of functions. We can sometimes use substitution to rewrite functions in a form that we can integrate. 6.2 Integration by Substitution

16. The variable of integration must match the variable in the expression. Don’t forget to substitute the value for u back into the problem! 6.2 Integration by Substitution

17. One of the clues that we look for is if we can find a function and its derivative in the integrand. The derivative of is.Note that this only worked because of the 2x in the original. Many integrals can not be done by substitution. 6.2 Integration by Substitution

18. Solve for dx. 6.2 Integration by Substitution

20. We solve for because we can find it in the integrand. 6.2 Integration by Substitution

22. The technique is a little different for definite integrals. We can find new limits, and then we don’t have to substitute back. new limit We could have substituted back and used the original limits. 6.2 Integration by Substitution

23. Wrong! The limits don’t match! Using the original limits: Leave the limits out until you substitute back. This is usually more work than finding new limits 6.2 Integration by Substitution

24. Don’t forget to use the new limits. 6.2 Integration by Substitution

27. Until then, remember that half the AP exam and half the nation’s college professors do not allow calculators. In another generation or so, we might be able to use the calculator to find all integrals. You must practice finding integrals by hand until you are good at it! 6.2 Integration by Substitution

28. Start with the product rule: This is the Integration by Parts formula. 6.3 Integrating by Parts

29. The Integration by Parts formula is a “product rule” for integration. u differentiates to zero (usually). dv is easy to integrate. Choose u in this order: LIPET Logs, Inverse trig, Polynomial, Exponential, Trig 6.3 Integrating by Parts

30. polynomial factor LIPET 6.3 Integrating by Parts

31. logarithmic factor LIPET 6.3 Integrating by Parts

32. This is still a product, so we need to use integration by parts again. LIPET 6.3 Integrating by Parts

33. A Shortcut: Tabular Integration Tabular integration works for integrals of the form: where: Differentiates to zero in several steps. Integrates repeatedly. 6.3 Integrating by Parts Also called tic-tac-toe method

34. Compare this with the same problem done the other way: 6.3 Integrating by Parts

36. LIPET This is the expression we started with! 6.3 Integrating by Parts

38. The goal of integrating by parts is to go from an integral that we don’t see how to integrate to an integral that we can evaluate.

39. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present (at least for awhile.) So does any population of living creatures. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interest-bearing account. If the rate of change is proportional to the amount present, the change can be modeled by: 6.4 Exponential Growth and Decay

40. Rate of change is proportional to the amount present. Divide both sides by y. Integrate both sides. 6.4 Exponential Growth and Decay

41. Exponentiate both sides. When multiplying like bases, add exponents. So added exponents can be written as multiplication. 6.4 Exponential Growth and Decay

42. Since is a constant, let. 6.4 Exponential Growth and Decay At,. This is the solution to our original initial value problem.

43. Exponential Change: If the constant k is positive then the equation represents growth. If k is negative then the equation represents decay. 6.4 Exponential Growth and Decay

44. Continuously Compounded Interest If money is invested in a fixed-interest account where the interest is added to the account k times per year, the amount present after t years is: If the money is added back more frequently, you will make a little more money. The best you can do is if the interest is added continuously. 6.4 Exponential Growth and Decay

45. Of course, the bank does not employ some clerk to continuously calculate your interest with an adding machine. We could calculate: Since the interest is proportional to the amount present, the equation becomes: Continuously Compounded Interest: You may also use: which is the same thing. 6.4 Exponential Growth and Decay

46. Radioactive Decay The equation for the amount of a radioactive element left after time t is: The half-life is the time required for half the material to decay. 6.4 Exponential Growth and Decay

47. Radioactive Decay 60 mg of radon, half-life of 1690 years. How much is left after 100 years? 6.4 Exponential Growth and Decay

48. 100 bacteria are present initially and double every 12 minutes. How long before there are 1,000,000 6.4 Exponential Growth and Decay

49. Half-life Half-life: 6.4 Exponential Growth and Decay

50. Espresso left in a cup will cool to the temperature of the surrounding air. The rate of cooling is proportional to the difference in temperature between the liquid and the air. If we solve the differential equation: we get: Newton’s Law of Cooling where is the temperature of the surrounding medium, which is a constant. 6.4 Exponential Growth and Decay

51. Years Bears 6.5 Population Growth

52. We have used the exponential growth equation to represent population growth. The exponential growth equation occurs when the rate of growth is proportional to the amount present. If we use P to represent the population, the differential equation becomes: The constant k is called the relative growth rate. 6.5 Population Growth

53. The population growth model becomes: However, real-life populations do not increase forever. There is some limiting factor such as food, living space or waste disposal. There is a maximum population, or carrying capacity, M. 6.5 Population Growth

54. A more realistic model is the logistic growth model where growth rate is proportional to both the amount present ( P ) and the fraction of the carrying capacity that remains: 6.5 Population Growth

55. The equation then becomes: Our book writes it this way: Logistics Differential Equation We can solve this differential equation to find the logistics growth model. 6.5 Population Growth

56. Partial Fractions Logistics Differential Equation 6.5 Population Growth

57. Logistics Differential Equation 6.5 Population Growth

58. Logistics Differential Equation 6.5 Population Growth

59. Logistics Growth Model 6.5 Population Growth

60. Logistic Growth Model Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears. Assuming a logistic growth model, when will the bear population reach 50? 75? 100? 6.5 Population Growth

61. Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears. Assuming a logistic growth model, when will the bear population reach 50? 75? 100? 6.5 Population Growth

62. At time zero, the population is 10. 6.5 Population Growth

63. After 10 years, the population is 23. 6.5 Population Growth

64. Years Bears We can graph this equation and use “trace” to find the solutions. y=50 at 22 yearsy=75 at 33 yearsy=100 at 75 years 6.5 Population Growth

65. Leonhard Euler 1707 - 1783 Leonhard Euler made a huge number of contributions to mathematics, almost half after he was totally blind. (When this portrait was made he had already lost most of the sight in his right eye.) 6.6 Euler’s Method

66. Leonhard Euler 1707 - 1783 It was Euler who originated the following notations: (base of natural log) (function notation) (pi) (summation) (finite change) 6.6 Euler’s Method

67. There are many differential equations that can not be solved. We can still find an approximate solution. We will practice with an easy one that can be solved. Initial value: 6.6 Euler’s Method

68. Recall the formula for local linearization 6.6 Euler’s Method where

69. 6.6 Euler’s Method (0,1) (.5,1) (1,1.5)

70. 6.6 Euler’s Method (1.5,2.5) (2,4)

71. Exact Solution: 6.6 Euler’s Method

72. It is more accurate if a smaller value is used for dx. This is called Euler’s Method. It gets less accurate as you move away from the initial value. 6.6 Euler’s Method

73. The book refers to an “Improved Euler’s Method”. We will not be using it, and you do not need to know it. The calculator also contains a similar but more complicated (and more accurate) formula called the Runge-Kutta method. You don’t need to know anything about it other than the fact that it is used more often in real life. This is the RK solution method on your calculator. 6.6 Euler’s Method