1. A honey bee makes several trips from the hive to a flower garden. What is the total distance traveled by the bee? 200ft 100ft 700 feet 7.1 Integrals as Net Change

2. What is the displacement of the bee? 200ft -200ft 200ft -100ft 100 feet towards the hive 7.1 Integrals as Net Change

3. To find the displacement (position shift) from the velocity function, we just integrate the function. The negative areas below the x-axis subtract from the total displacement. 7.1 Integrals as Net Change

4. To find distance traveled we have to use absolute value. Find the roots of the velocity equation and integrate in pieces, just like when we found the area between a curve and the x-axis. (Take the absolute value of each integral.) Or you can use your calculator to integrate the absolute value of the velocity function. 7.1 Integrals as Net Change

5. velocity graph position graph Displacement: Distance Traveled: 7.1 Integrals as Net Change

6. In the linear motion equation: V(t) is a function of time. For a very small change in time, V(t) can be considered a constant. We add up all the small changes in S to get the total distance. 7.1 Integrals as Net Change

7. We add up all the small changes in S to get the total distance. As the number of subintervals becomes infinitely large (and the width becomes infinitely small), we have integration. 7.1 Integrals as Net Change

8. Let v(t) = 2t 3 – 14t 2 + 20t, Determine when the particle is moving to the right, left, and stopped. Find the particle’s distance and displacement after 3 seconds. v(t) = 2t 3 – 14t 2 + 20t v(t) = 2t(t 2 – 7t + 10) v(t) = 2t(t – 2)(t – 5) 0 2 5 0 ++0 ------- 0 +++++ Right (0,2) U (5,  ) Left (2,5) Stopped t = 0, 2s, 5s

9. 7.1 Integrals as Net Change Let v(t) = 2t 3 – 14t 2 + 20t, Determine when the particle is moving to the right, left, and stopped. Find the particle’s distance and displacement after 3 seconds.

10. This same technique is used in many different real- life problems. 7.1 Integrals as Net Change

11. National Potato Consumption The rate of potato consumption for a particular country was: where t is the number of years since 1970 and C is in millions of bushels per year. For a small, the rate of consumption is constant. The amount consumed during that short time is. 7.1 Integrals as Net Change

12. The amount consumed during that short time is. We add up all these small amounts to get the total consumption: From the beginning of 1972 to the end of 1973: million bushels 7.1 Integrals as Net Change

13. Work: Calculating the work is easy when the force and distance are constant. When the amount of force varies, we get to use calculus! 7.1 Integrals as Net Change

14. Hooke’s law for springs: x = distance that the spring is extended beyond its natural length k = spring constant 7.1 Integrals as Net Change

15. Hooke’s law for springs: It takes 10 Newtons to stretch a spring 2 meters beyond its natural length. F =10 N x =2 M How much work is done stretching the spring to 4 meters beyond its natural length? 7.1 Integrals as Net Change

16. F(x)F(x) x =4 M How much work is done stretching the spring to 4 meters beyond its natural length? For a very small change in x, the force is constant. newton-meters joules 7.1 Integrals as Net Change

17. How can we find the area between these two curves? We could and figure it out, but there is split the area into several sections, use subtraction an easier way. 7.2 Areas in the Plane

18. Consider a very thin vertical strip. The length of the strip is: or Since the width of the strip is a very small change in x, we could call it dx. 7.2 Areas in the Plane

19. Since the strip is a long thin rectangle, the area of the strip is: If we add all the strips, we get: 7.2 Areas in the Plane

21. The formula for the area between curves is: We will use this so much, that you won’t need to “memorize” the formula! 7.2 Areas in the Plane

22. If we try vertical strips, we have to integrate in two parts: We can find the same area using a horizontal strip. Since the width of the strip is dy, we find the length of the strip by solving for x in terms of y. 7.2 Areas in the Plane

23. We can find the same area using a horizontal strip. Since the width of the strip is dy, we find the length of the strip by solving for x in terms of y. length of strip width of strip 7.2 Areas in the Plane

24. 1 Decide on vertical or horizontal strips. (Pick whichever is easier to write formulas for the length of the strip, and/or whichever will let you integrate fewer times.) Sketch the curves. 2 3 Write an expression for the area of the strip.(If the width is dx, the length must be in terms of x.If the width is dy, the length must be in terms of y. 4 Find the limits of integration. (If using dx, the limits are x values; if using dy, the limits are y values.) 5 Integrate to find area. 7.2 Areas in the Plane

25. 3 3 3 Find the volume of the pyramid: Consider a horizontal slice through the pyramid. The volume of the slice is s 2 dh. If we put zero at the top of the pyramid and make down the positive direction, then s=h. 7.3 Volumes

26. s dh 0 3 h This correlates with the formula: 7.3 Volumes

27. Definition: Volume of a Solid The volume of a solid with known integrable cross section area A(x) from x = a to x = b is the integral of A from a to b,

28. Method of Slicing (p384): 1 Find a formula for V(x). (Note that V(x) is used instead of A(x).) Sketch the solid and a typical cross section. 2 3 Find the limits of integration. 4 Integrate V(x) to find volume. 7.3 Volumes

29. x y A 45 o wedge is cut from a cylinder of radius 3 as shown. Find the volume of the wedge. You could slice this wedge shape several ways, but the simplest cross section is a rectangle. 7.3 Volumes

30. x y If we let h equal the height of the slice then the volume of the slice is: Since the wedge is cut at a 45 o angle: x h 45 o Since 7.3 Volumes

31. x y Even though we started with a cylinder, p does not enter the calculation! 7.3 Volumes

32. Cavalieri’s Theorem: Two solids with equal altitudes and identical parallel cross sections have the same volume. Identical Cross Sections 7.3 Volumes

33. Suppose I start with this curve. My boss at the ACME Rocket Company has assigned me to build a nose cone in this shape. So I put a piece of wood in a lathe and turn it to a shape to match the curve. 7.3 Volumes

34. How could we find the volume of the cone? One way would be to cut it into a series of thin slices (flat cylinders) and add their volumes. The volume of each flat cylinder (disk) is: r= the y value of the function 7.3 Volumes

35. The volume of each flat cylinder (disk) is: If we add the volumes, we get: 7.3 Volumes

36. This application of the method of slicing is called the disk method. The shape of the slice is a disk, so we use the formula for the area of a circle to find the volume of the disk. 7.3 Volumes A shape rotated about the x-axis would be: A shape rotated about the y-axis would be:

37. The region between the curve, and the y-axis is revolved about the y-axis. Find the volume. y x We use a horizontal disk. The thickness is dy. The radius is the x value of the function. 7.3 Volumes

38. The region bounded by and is revolved about the y-axis. Find the volume. The “disk” now has a hole in it, making it a “washer”. If we use a horizontal slice: The volume of the washer is: 7.3 Volumes

39. The volume of the washer is: outer radius inner radius 7.3 Volumes

40. This application of the method of slicing is called the washer method. The shape of the slice is a circle with a hole in it, so we subtract the area of the inner circle from the area of the outer circle. The washer method formula is: 7.3 Volumes

41. If the same region is rotated about the line x = 2 : The outer radius is: R The inner radius is: r 7.3 Volumes

42. R r

43. Find the volume of the region bounded by,, and revolved about the y - axis. We can use the washer method if we split it into two parts: outer radius inner radius thickness of slice cylinder 7.3 Volumes

44. If we take a vertical sliceand revolve it about the y-axis we get a cylinder. cross section If we add all of the cylinders together, we can reconstruct the original object. Here is another way we could approach this problem: 7.3 Volumes

45. cross section The volume of a thin, hollow cylinder is given by: r is the x value of the function. h is the y value of the function. thickness is dx. 7.3 Volumes

46. If we add all the cylinders from the smallest to the largest: This is called the shell method because we use cylindrical shells. 7.3 Volumes

47. Definition: Volume of a Solid Shell Method The volume of a solid rotated about the y-axis is 7.3 Volumes

48. Definition: Volume of a Solid Shell Method The volume of a solid rotated about the x-axis is 7.3 Volumes

49. Find the volume generated when this shape is revolved about the y-axis. We can’t solve for x, so we can’t use a horizontal slice directly. 7.3 Volumes

50. Shell method: If we take a vertical slice and revolve it about the y-axis we get a cylinder. 7.3 Volumes

52. When the strip is parallel to the axis of rotation, use the shell method. When the strip is perpendicular to the axis of rotation, use the washer method. 7.3 Volumes

53. By the Pythagorean Theorem: We need to get dx out from under the radical. Length of Curve (Cartesian) 7.4 Lengths of Curves

54. if y is a smooth function on [a,b] if x is a smooth function on [c,d] Definition Arc Length: Length of a Smooth Curve If a smooth curve begins at (a,c) and ends at (b,d), a < c and b < d, then the length (arc length) of the curve is 7.4 Lengths of Curves

56. The curve should be a little longer than the straight line, so our answer seems reasonable. If we check the length of a straight line: 7.4 Lengths of Curves

58. If you have an equation that is easier to solve for x than for y, the length of the curve can be found the same way. Notice that x and y are reversed. 7.4 Lengths of Curves

59. Surface Area: r Consider a curve rotated about the x -axis: The surface area of this band is: The radius is the y -value of the function, so the whole area is given by: This is the same ds that we had in the “length of curve” formula, so the formula becomes: 7.4 Lengths of Curves

60. Surface Area: r Surface Area about x -axis (Cartesian): To rotate about the y -axis, just reverse x and y in the formula! 7.4 Lengths of Curves

61. Rotate about the y -axis. 7.4 Lengths of Curves

63. Rotate about the y -axis. From geometry: 7.4 Lengths of Curves

64. Example: rotated about x -axis. 7.4 Lengths of Curves

65. Example: Check: rotated about x -axis. 7.4 Lengths of Curves

66. A spring has a natural length of 1 m. A force of 24 N stretches the spring to 1.8 m. a Find k : b How much work would be needed to stretch the spring 3m beyond its natural length? 7.5 Applications

67. Over a very short distance, even a non-constant force doesn’t change much, so work becomes: If we add up all these small bits of work we get: 7.5 Applications

68. A leaky 5 lb bucket is raised 20 feet The rope weights 0.08 lb/ft. The bucket starts with 2 gal (16 lb) of water and is empty when it just reaches the top. Check: 7.5 Applications

69. Work: Bucket: Water:The force is proportional to remaining rope. Check: 7.5 Applications

70. Work: Bucket: Water: 7.5 Applications

71. Work: Bucket: Water: Rope: Check: Total: 7.5 Applications

72. 5 ft 10 ft 4 ft I want to pump the water out of this tank. How much work is done? The force is the weight of the water. The water at the bottom of the tank must be moved further than the water at the top. 7.5 Applications

73. 5 ft 10 ft 10 0 4 ft dx Consider the work to move one “slab” of water: 7.5 Applications

74. 5 ft 10 ft 10 0 4 ft dx forcedistance 7.5 Applications

75. 5 ft 10 ft 4 ft A 1 horsepower pump, rated at 550 ft-lb/sec, could empty the tank in just under 14 minutes! forcedistance 7.5 Applications

76. 10 ft 2 ft 10 ft A conical tank is filled to within 2 ft of the top with salad oil weighing 57 lb/ft 3. How much work is required to pump the oil to the rim? 7.5 Applications

77. Consider one slice (slab) first: 7.5 Applications

78. A conical tank if filled to within 2 ft of the top with salad oil weighing 57 lb/ft 3. How much work is required to pump the oil to the rim? 7.5 Applications

79. 10 ft 2 ft 10 ft A conical tank if filled to within 2 ft of the top with salad oil weighing 57 lb/ft 3. How much work is required to pump the oil to the rim? 7.5 Applications

80. What is the force on the bottom of the aquarium? 3 ft 2 ft 1 ft 7.5 Applications

81. If we had a 1 ft x 3 ft plate on the bottom of a 2 ft deep wading pool, the force on the plate is equal to the weight of the water above the plate. densitydepth pressure area All the other water in the pool doesn’t affect the answer! 7.5 Applications

82. What is the force on the front face of the aquarium? 3 ft 2 ft 1 ft Depth (and pressure) are not constant. If we consider a very thin horizontal strip, the depth doesn’t change much, and neither does the pressure. 7.5 Applications

83. 3 ft 2 ft 1 ft Depth (and pressure) are not constant. If we consider a very thin horizontal strip, the depth doesn’t change much, and neither does the pressure. 3 ft 2 ft 2 0 density depth area 7.5 Applications

84. 6 ft 3 ft 2 ft A flat plate is submerged vertically as shown. (It is a window in the shark pool at the city aquarium.) Find the force on one side of the plate. Depth of strip: Length of strip: Area of strip: We could have put the origin at the surface, but the math was easier this way. 7.5 Applications

85. 6 ft 3 ft 2 ft Depth of strip: Length of strip: Area of strip: densitydeptharea 7.5 Applications

86. 68% 95% 99.7% 34% 13.5% 2.35% Normal Distribution: For many real-life events, a frequency distribution plot appears in the shape of a “normal curve”. Examples: heights of 18 yr. old men standardized test scores lengths of pregnancies time for corn to pop The mean (or ) is in the middle of the curve. The shape of the curve is determined by the standard deviation. “68, 95, 99.7 rule” 7.5 Applications

87. 34% 13.5% 2.35% Normal Distribution: “68, 95, 99.7 rule” The area under the curve from a to b represents the probability of an event occurring within that range. Normal Probability Density Function: (Gaussian curve) 7.5 Applications