1. Announcements Topics: -sections 7.5 (additional techniques of integration) and 7.6 (applications of integration) * Read these sections and study solved examples in your textbook! Work On: -Practice problems from the textbook and assignments from the coursepack as assigned on the course web page (under the link “SCHEDULE + HOMEWORK”)

2. The Product Rule and Integration by Parts The product rule for derivatives leads to a technique of integration that breaks a complicated integral into simpler parts. Integration by Parts Formula: hopefully this is a simpler Integral to evaluate given integral that we cannot solve

3. The Product Rule and Integration by Parts Deriving the Formula Start by writing out the Product Rule: Solve for

4. The Product Rule and Integration by Parts Deriving the Formula Start by writing out the Product Rule: Solve for

5. The Product Rule and Integration by Parts Deriving the Formula Integrate both sides with respect to x:

6. The Product Rule and Integration by Parts Deriving the Formula Simplify:

7. Integration by Parts Template: Choose: Compute: easy to integrate part part which gets simpler after differentiation

8. Integration by Parts Example: Integrate each using integration by parts. (a)(b) (c)(d)

9. Strategy for Integration MethodApplies when… Basic antiderivative…the integrand is recognized as the reversal of a differentiation formula, such as Guess-and-check…the integrand differs from a basic antiderivative in that “x” is replaced by “ax+b”, for example Substitution…both a function and its derivative (up to a constant) appear in the integrand, such as Integration by parts…the integrand is the product of a power of x and one of sin x, cos x, and e x, such as …the integrand contains a single function whose derivative we know, such as

10. Strategy for Integration What if the integrand does not have a formula for its antiderivative? Example: impossible to integrate

11. Approximating Functions with Polynomials Recall: The quadratic approximation to around the base point x=0 is base point

12. Integration Using Taylor Polynomials We approximate the function with an appropriate Taylor polynomial and then integrate this Taylor polynomial instead! Example: easy to integrate impossible to integrate for x-values near 0

13. Integration Using Taylor Polynomials We can obtain a better approximation by using a higher degree Taylor polynomial to represent the integrand. Example:

14. The Definite Integral – Area Between Curves The area between the curves and and between and is Recall:

15. Area Between Curves Area between f and g on [a,b] =

16. The Definite Integral – Area Between Curves Examples: Sketch the region enclosed by the given curves and then find the area of the region. (a) (b)

17. estimate of the surface area of lake ontario

18. aerial distance hamilton - kingston approx. 290 km

19. aerial distance hamilton - kingston = 290km represents approximately 38km

20. aerial distance hamilton - kingston = 290km represents approximately 38km

21. aerial distance hamilton - kingston = 290km represents approximately 38km

22. 38 41.0 area = 38*41=1558 km^2

23. 38 55.3 area = 38*55.3=2101.4 km^2

24. 38 58.2 area = 38*58.2=2211.6 km^2

25. 38 63.0 area = 38*63.0=2394.0 km^2

26. 63.0 total area = 38*41.0+ 38*55.3+ 38*58.2+ 38*64.9+ 38*82.0+ 38*70.6+ 38*97.3+ 38*63.0= 20,227.4 km^2 38 41.0 55.3 58.2 64.982.0 70.6 97.3

27. represents approximately 17km

28. area = 17.2*17.2=295.8 km^2

29. area = 17.2*43.9=764.8 km^2

30. area = 17.2*55.3=951.7 km^2

31. area = 17.2*61.0=1050.1 km^2 total area = 19,233.5

32. average of the two estimates = (20,227.4 + 19,233.5)/2= 19,730.5 km^2

33. our estimate = 19,730.5 km^2 New York Times Almanac … 19,500 km^2 NOAA (National Oceanographic and Atmospheric Administration, U.S.) … 19,009 km^2 EPA (Environmental Protection Agency) … 18,960 km^2 Britannica Online … 19,011km^2

34. The Definite Integral - Average Value The average value of a function f on the interval from a to b is For a positive function, average height = area width

35. The Definite Integral - Average Value

36. Average Value Example: Find the average value of the function on the interval

37. Application Example: Several very skinny 2.0-m-long snakes are collected in the Amazon. Each snake has a density of where is measured in grams per centimeter and is measured in centimeters from the tip of the tail. Find the average density of the snake.

38. Application

39. (a) Find the total mass of each snake. (b) Find the average density of each snake.

40. estimate of the volume of a heart chamber from echocardiogram

41. echocardiogram (ECHO, cardiac ultrasound) is a sonogram of the heart ECHO uses standard ultrasound techniques to image two-dimensional slices of the heart latest ultrasound systems now employ 3D real- time imaging

42. ECHO uses standard ultrasound techniques to image two-dimensional slices of the heart

43. latest ultrasound systems now employ 3D real- time imaging

44. uses of ECHO (a) creating 2D picture of the cardiovascular system (shape of the heart) (b) assessment of quality of cardiac tissue (damage, thickening of walls within the heart) (c) estimate of the velocity of blood

45. uses of ECHO (d) investigate features of blood flow functioning of cardiac valves detection of abnormal communication between the left and right side of the heart leaking of blood through the valves strength at which blood is pumped out of heart (cardiac output)

47. 5.02 cm

48. 5.02/15 = 0.335 cm

49. 5.02 cm 5.02/15 = 0.335 cm 2.00 cm 2.50 cm

50. 5.02 cm 5.02/15 = 0.335 cm 2.00 cm 2.50 cm

51. 5.02 cm 5.02/15 = 0.335 cm 2.00 cm 2.50 cm

54. 0.335 cm 1.00 cm

55. 0.335 cm 1.25 cm

56. 0.335 cm 1.50 cm

57. 0.335 cm 1.61 cm

58. 0.335 cm 1.07 cm

60. all diameters, from bottom to top: all in cm 2.00, 2.50, 3.00, 3.21, 3.43, 3.57, 3.93, 4.07, 4.29, 4.29, 4.29, 4.14, 4.07, 3.43, 2.14 height: 5.02/15 cm

61. Approximating Volumes Riemann Sum

62. Integrals and Volumes Definition: Denote by A(x) the area of the cross-section of S by the plane perpendicular to the x-axis that passes through x. Assume that A(x) is continuous on [a,b]. Then the volume V of S is given by provided that the limit exists.

63. Volumes of Solids of Revolution Examples: Find the volume of the solid obtained by rotating the region R enclosed (bounded) by the given curves about the given axis. (a) (b)