1. Riemann Sums and The Definite Integral

2. time velocity After 4 seconds, the object has gone 12 feet. Consider an object moving at a constant rate of 3 ft/sec. Since rate. time = distance: If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line. Estimating with Finite Sums

3. If the velocity is not constant, we might guess that the distance traveled is still equal to the area under the curve. Example: We could estimate the area under the curve by drawing rectangles touching at their left corners. This is called the Left- hand Rectangular Approximation Method (LRAM). Approximate area: Estimating with Finite Sums

4. We could also use a Right-hand Rectangular Approximation Method (RRAM). Approximate area: Estimating with Finite Sums

5. Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM). Approximate area: In this example there are four subintervals. As the number of subintervals increases, so does the accuracy. Estimating with Finite Sums

6. Approximate area: width of subinterval With 8 subintervals: The exact answer for this problem is. Estimating with Finite Sums

7. Circumscribed rectangles are all above the curve: Inscribed rectangles are all below the curve: Estimating with Finite Sums

8. We will be learning how to find the exact area under a curve if we have the equation for the curve. Rectangular approximation methods are still useful for finding the area under a curve if we do not have the equation. Estimating with Finite Sums

9. When we find the area under a curve by adding rectangles, the answer is called a Rieman sum. subinterval partition The width of a rectangle is called a subinterval. The entire interval is called the partition. Subintervals do not all have to be the same size. Definite Integrals

10. subinterval partition If the partition is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by. As gets smaller, the approximation for the area gets better. if P is a partition of the interval Definite Integrals

11. is called the definite integral of over. If we use subintervals of equal length, then the length of a subinterval is: The definite integral is then given by: Definite Integrals

12. Leibnitz introduced a simpler notation for the definite integral: Note: the very small change in x becomes dx. Definite Integrals

13. Integration Symbol lower limit of integration upper limit of integration integrand variable of integration (dummy variable) It is called a dummy variable because the answer does not depend on the variable chosen. Definite Integrals

14. We have the notation for integration, but we still need to learn how to evaluate the integral. Definite Integrals

15. Definition -- a Definite Integral If y = f(x) is non negative and integrable over a closed interval [a,b], then the DEFINITE INTEGRAL (area under the curve) y = f(x) from a to b is the integral of f(x) from a to b. A =

16. Area Where F(x) is the antiderivative of f(x)! Definite Integrals

18. Using integrals to find area works extremely well as long as we can find the antiderivative of the function. Sometimes, the function is too complicated to find the antiderivative. At other times, we don’t even have a function, but only measurements taken from a real-life object. What we need is an efficient method to estimate area when we can not find the antiderivative. Trapezoid Rule

19. Trapezoidal Rule: where [a,b] is partitioned into n subintervals of equal length h = (b – a)/n This gives us a better approximation than either left or right rectangles. Trapezoid Rule

21. Consider the function f whose graph is shown below. Use the Trapezoid Rule with n = 4 to estimate the value of A. 21 B. 22 C. 23 D. 24 E. 25 X X X X X

22. Trapezoidal Rule: Midpoint Rule

23. Rules for definite integrals

24. 5. Intervals can be added (or subtracted.) Definite Integrals and Antiderivatives

25. 6. Definite Integrals and Antiderivatives a b

26. Example

27. Example:

30. Definite Integrals and Antiderivatives Suppose that f and g are continuous functions and that Find 32*5=10 -5 – -3 = -2 -3 + 5 = 203*(-3)- 2 * 6 = -21

31. Example: Northeast Airlines determines that the marginal profit resulting from the sale of x seats on a jet traveling from Atlanta to Kansas City, in hundreds of dollars, is given by Find the total profit when 60 seats are sold.

32. Example (continued): We integrate to find P(60).

33. The average value of a function is the value that would give the same area if the function was a constant: Definite Integrals and Antiderivatives

34. Average Value of a Function Over an Interval

35. EXAMPLE SOLUTION Determine the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1. Using (2) with a = -1 and b = 1, the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1 is equal to An antiderivative of 1 – x is. Therefore, So, the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1 is 1.

36. Average Value of a Function Suppose f is integrable on [a,b]. Then the average value of f over [a,b] is Find the average value of over the interval [0,4]

37. For what value(s) in the interval does the function assume the average value? Definite Integrals and Antiderivatives

38. EXAMPLE SOLUTION (Average Temperature) During a certain 12-hour period the temperature at time t (measured in hours from the start of the period) was degrees. What was the average temperature during that period? The average temperature during the 12-hour period from t = 0 to t = 12 is

39. The mean value theorem for definite integrals says that for a continuous function, at some point on the interval the actual value will equal the average value. Mean Value Theorem (for definite integrals) If f is continuous on then at some point c in, Definite Integrals and Antiderivatives

40. If you were being sent to a desert island and could take only one equation with you, might well be your choice. Fundamental Theorem of Calculus

41. The Fundamental Theorem of Calculus, Part 1 If f is continuous on, then the function has a derivative at every point in, and Fundamental Theorem of Calculus

42. 2. Derivative matches upper limit of integration. First Fundamental Theorem: 1. Derivative of an integral. Fundamental Theorem of Calculus

43. 1. Derivative of an integral. 2. Derivative matches upper limit of integration. 3. Lower limit of integration is a constant. First Fundamental Theorem: Fundamental Theorem of Calculus

44. 1. Derivative of an integral. 2. Derivative matches upper limit of integration. 3. Lower limit of integration is a constant. New variable. First Fundamental Theorem: Fundamental Theorem of Calculus

45. 1. Derivative of an integral. 2. Derivative matches upper limit of integration. 3. Lower limit of integration is a constant. The long way: First Fundamental Theorem: Fundamental Theorem of Calculus

46. 1. Derivative of an integral. 2. Derivative matches upper limit of integration. 3. Lower limit of integration is a constant. Fundamental Theorem of Calculus

47. The upper limit of integration does not match the derivative, but we could use the chain rule. Fundamental Theorem of Calculus

48. The lower limit of integration is not a constant, but the upper limit is. We can change the sign of the integral and reverse the limits. Fundamental Theorem of Calculus

49. Neither limit of integration is a constant. It does not matter what constant we use! (Limits are reversed.) (Chain rule is used.) We split the integral into two parts. Fundamental Theorem of Calculus

50. The Fundamental Theorem of Calculus, Part 2 If f is continuous at every point of, and if F is any antiderivative of f on, then (Also called the Integral Evaluation Theorem) Fundamental Theorem of Calculus

51. Definite Integral Substitution

52. The substitution rule for definite integrals If g’ is continuous on [a,b], and f is continuous on the range of u=g(x) then

53. Find new limits new limit Exampl e

55. Don’t forget to use the new limits. Example

56. Integrals of Even and Odd Functions Theorem

57. Integrals of Even and Odd Functions Problem Solution An odd function is symmetric with respect to the origin. The definite integral from -a to a, in the case of the function shown in this picture, is the area of the blue domain minus the area of the red domain. By symmetry these areas are equal, hence the integral is 0. -a-a a