5.1 Area and Estimating with Finite Sums

Transcription

1 5.1 Area and Estimating with Finite Sums

2 Ideas for this section The ideas for this section are Left-Hand Sums

3 Ideas for this section The ideas for this section are Left-Hand Sums Right-Hand Sums

4 Ideas for this section The ideas for this section are Left-Hand Sums Right-Hand Sums Midpoint Rule

5 Ideas for this section The ideas for this section are Left-Hand Sums Right-Hand Sums Midpoint Rule Average value

6 Ideas for this section The ideas for this section are Left-Hand Sums Right-Hand Sums Midpoint Rule Average value Velocity problems

7 Ideas for this section The ideas for this section are Left-Hand Sums Right-Hand Sums Midpoint Rule Average value Velocity problems Distance v. Displacement

8 Measuring Distance Traveled Example A car is moving with increasing velocity as given in the table below. How far has the car traveled? time (sec) velocity (ft/sec)

9 Measuring Distance Traveled Example A car is moving with increasing velocity as given in the table below. How far has the car traveled? time (sec) velocity (ft/sec) Thoughts on how we can estimate this?

10 Measuring Distance Traveled Example A car is moving with increasing velocity as given in the table below. How far has the car traveled? time (sec) velocity (ft/sec) Thoughts on how we can estimate this? The best we can do is estimate by considering the velocity over 2 second intervals.

11 Measuring Distance Traveled Lower:

12 Measuring Distance Traveled Lower: 20(2) + 25(2) + 31(2) + 39(2) + 48(2) = 326 feet

13 Measuring Distance Traveled Lower: 20(2) + 25(2) + 31(2) + 39(2) + 48(2) = 326 feet Upper:

14 Measuring Distance Traveled Lower: 20(2) + 25(2) + 31(2) + 39(2) + 48(2) = 326 feet Upper: 25(2) + 31(2) + 39(2) + 48(2) + 60(2) = 406 feet

15 Measuring Distance Traveled Lower: 20(2) + 25(2) + 31(2) + 39(2) + 48(2) = 326 feet Upper: 25(2) + 31(2) + 39(2) + 48(2) + 60(2) = 406 feet So we know 326 distance 406 feet, which is a difference of 80 feet. We ll see the significance of this shortly.

16 Lower Estimate When we visually look at the underestimate, we see something like Figure: UnderEstimate

17 Upper Estimate When we look at the overestimate we see something like Figure: Over Estimate

18 Measuring More Often Example What if we were able to measure every second? time (sec) vel (ft/sec)

19 Measuring More Often Example What if we were able to measure every second? time (sec) vel (ft/sec) Lower:

20 Measuring More Often Example What if we were able to measure every second? time (sec) vel (ft/sec) Lower: 345 feet Upper:

21 Measuring More Often Example What if we were able to measure every second? time (sec) vel (ft/sec) Lower: 345 feet Upper: 385 feet

22 Measuring More Often Example What if we were able to measure every second? time (sec) vel (ft/sec) Lower: 345 feet Upper: 385 feet This gives us a difference of 40 feet.

23 Measuring More Often How wide will the interval be if we measured 1 2 second?

24 Measuring More Often How wide will the interval be if we measured 1 2 second? 20 feet

25 Measuring More Often How wide will the interval be if we measured 1 2 second? 20 feet 1 10 second?

26 Measuring More Often How wide will the interval be if we measured second? 20 feet second? 4 feet

27 Measuring More Often How wide will the interval be if we measured Conclusion: second? 20 feet second? 4 feet

28 Measuring More Often How wide will the interval be if we measured second? 20 feet second? 4 feet Conclusion: The smaller the time between estimates, the closer the upper and lower estimates. The difference between them is (right endpoint left endpoint) x

29 Measuring More Often How wide will the interval be if we measured second? 20 feet second? 4 feet Conclusion: The smaller the time between estimates, the closer the upper and lower estimates. The difference between them is (right endpoint left endpoint) x What is the number of observations used approaches? But first...

30 Left and Right Hand Rules Example Given the function f (x) = x 2, approximate the area under the curve on the interval [0, 4]for n = 4 using the Left-Hand Rule and the Right-Hand Rule.

31 Left and Right Hand Rules Example Given the function f (x) = x 2, approximate the area under the curve on the interval [0, 4]for n = 4 using the Left-Hand Rule and the Right-Hand Rule. We first need to figure out how large each of our subintervals needs to be.

32 Left and Right Hand Rules Figure: General Sum When we look at a general sum like this, we see that the left endpoint is x 0 = a and the right endpoint is x n = b. We use indices to make the number of endpoints general.

33 Left and Right Hand Rules Figure: General Sum In our case, we d have x 0 = 0, x 1 = 1 and x 2 = 2, x 3 = 3 and x 4 = 4.

34 Left and Right Hand Rules Figure: General Sum In our case, we d have x 0 = 0, x 1 = 1 and x 2 = 2, x 3 = 3 and x 4 = 4. Note that the distance between points is 4 units, and this comes from x = b a n

35 Left-Hand Rule In our case, x = = 1

36 Left-Hand Rule In our case, x = = 1 x 0 x 1 x 2 x 3 x 4

37 Left-Hand Rule In our case, x = = 1 0 x 0 x 1 x 2 x 3 x 4

38 Left-Hand Rule In our case, x = = x 0 x 1 x 2 x 3 x 4

39 Left-Hand Rule In our case, x = = x 0 x 1 x 2 x 3 x 4

40 Left-Hand Rule In our case, x = = x 0 x 1 x 2 x 3 x 4

41 Left-Hand Rule In our case, x = = x 0 x 1 x 2 x 3 x 4 Left-Hand Sum: 1( ) = 14

42 Right-Hand Rule x 0 x 1 x 2 x 3 x 4

43 Right-Hand Rule 1 x 0 x 1 x 2 x 3 x 4

44 Right-Hand Rule 1 4 x 0 x 1 x 2 x 3 x 4

45 Right-Hand Rule x 0 x 1 x 2 x 3 x 4

46 Right-Hand Rule x 0 x 1 x 2 x 3 x 4

47 Right-Hand Rule x 0 x 1 x 2 x 3 x 4 Right-Hand Sum: 1( ) = 30

48 Averages We could also find the average area under a curve by dividing an estimate by the length of the interval, which is to say average = 1 b a [estimate]

49 Averages We could also find the average area under a curve by dividing an estimate by the length of the interval, which is to say average = 1 b a [estimate] So with our last example, the average value of the underestimate would be

50 Averages We could also find the average area under a curve by dividing an estimate by the length of the interval, which is to say average = 1 b a [estimate] So with our last example, the average value of the underestimate would be 1 4 (14) = 7 2.

51 Larger n If we wanted to find an estimate for a large value of n, we would not want to do so by hand. There are utilities we can use to do so. Things like CAS or the TI-graphing calculator would do the trick. If you Google TI-calculator program and look for one with an address at the University of Arizona. There are two key programs I would look for here - All Sums and Rectangles.

52 Another Example Example Given the function f (x) = x 2 9, approximate the area under the curve on the interval [0, 3] for n = 4 using the Left-Hand Rule and the Right-Hand Rule.

53 Another Example Example Given the function f (x) = x 2 9, approximate the area under the curve on the interval [0, 3] for n = 4 using the Left-Hand Rule and the Right-Hand Rule. LHS:

54 Another Example Example Given the function f (x) = x 2 9, approximate the area under the curve on the interval [0, 3] for n = 4 using the Left-Hand Rule and the Right-Hand Rule. LHS: 3 4 ( ) 16 =

55 Another Example Example Given the function f (x) = x 2 9, approximate the area under the curve on the interval [0, 3] for n = 4 using the Left-Hand Rule and the Right-Hand Rule. LHS: 3 4 RHS: ( ) 16 =

56 Another Example Example Given the function f (x) = x 2 9, approximate the area under the curve on the interval [0, 3] for n = 4 using the Left-Hand Rule and the Right-Hand Rule. LHS: 3 4 RHS: 3 4 ( ) = ( ) =

57 Another Example Example Given the function f (x) = x 2 9, approximate the area under the curve on the interval [0, 3] for n = 4 using the Left-Hand Rule and the Right-Hand Rule. LHS: 3 4 RHS: 3 4 Average: ( ) = ( ) =

58 Another Example Example Given the function f (x) = x 2 9, approximate the area under the curve on the interval [0, 3] for n = 4 using the Left-Hand Rule and the Right-Hand Rule. LHS: 3 ( RHS: 3 ( Average: ( ) = ) = ) 14.3 =

59 Midpoint Rule We can also use another rule, called the Midpoint Rule to find estimates. This differs from both the Left-Hand Sum and the Right-Hand Sum in that we use no endpoints for our calculations, but rather the midpoint of the interval.

60 Midpoint Rule We can also use another rule, called the Midpoint Rule to find estimates. This differs from both the Left-Hand Sum and the Right-Hand Sum in that we use no endpoints for our calculations, but rather the midpoint of the interval. So if we look back at our example of f (x) = x 2 on [0, 4] for n = 4, we could apply the Midpoint Rule to approximate the area under the curve.

61 Midpoint Rule x 0 x 1 x 2 x 3 x 4

62 Midpoint Rule x 0 x 1 x 2 x 3 x 4

63 Midpoint Rule x 0 x 1 x 2 x 3 x 4

64 Midpoint Rule 1 4 x 0 x 1 x 2 x 3 x 4

65 Midpoint Rule 1 4 x 0 x 1 x 2 x 3 x 4

66 Midpoint Rule 1 4 x 0 x 1 x 2 x 3 x 4

67 Midpoint Rule x 0 x 1 x 2 x 3 x 4

68 Midpoint Rule x 0 x 1 x 2 x 3 x 4

69 Midpoint Rule x 0 x 1 x 2 x 3 x 4

70 Midpoint Rule x 0 x 1 x 2 x 3 x 4

71 Midpoint Rule x 0 x 1 x 2 x 3 x 4

72 Midpoint Rule x 0 x 1 x 2 x 3 x 4

73 Midpoint Rule x 0 x 1 x 2 x 3 x 4

74 Midpoint Rule x 0 x 1 x 2 x 3 x 4 MID(4) = 1 ( ) 4 = 21

75 Why not in the middle? Why does the midpoint rule not give the same answer as the average of LHS and RHS?

76 Why not in the middle? Why does the midpoint rule not give the same answer as the average of LHS and RHS? This is due to the concavity of the function - because it is concave up, the Midpoint Rule is closer to the Left-Hand Sum.

77 Another Example Example f (x) = x 2 + 2, [1, 3], n = 4. Use LHS, RHS, Midpoint Rule.

78 Another Example Example f (x) = x 2 + 2, [1, 3], n = 4. Use LHS, RHS, Midpoint Rule. LHS =

79 Another Example Example f (x) = x 2 + 2, [1, 3], n = 4. Use LHS, RHS, Midpoint Rule. LHS = 43 4

80 Another Example Example f (x) = x 2 + 2, [1, 3], n = 4. Use LHS, RHS, Midpoint Rule. LHS = 43 4 RHS =

81 Another Example Example f (x) = x 2 + 2, [1, 3], n = 4. Use LHS, RHS, Midpoint Rule. LHS = 43 4 RHS = 59 4

82 Another Example Example f (x) = x 2 + 2, [1, 3], n = 4. Use LHS, RHS, Midpoint Rule. LHS = 43 4 RHS = 59 4 Average =

83 Another Example Example f (x) = x 2 + 2, [1, 3], n = 4. Use LHS, RHS, Midpoint Rule. LHS = 43 4 RHS = 59 4 Average = 51 4

84 Another Example For the Midpoint Rule, we first need the midpoints.

85 Another Example For the Midpoint Rule, we first need the midpoints. [1, 1.5], midpoint is 1.25 = 5 4 [1.5, 2], midpoint is 1.75 = 7 4 [2, 2.5], midpoint is 2.25 = 9 4 [2.5, 3], midpoint is 2.75 = 11 4

86 Another Example For the Midpoint Rule, we first need the midpoints. ( 1 f 2 [1, 1.5], midpoint is 1.25 = 5 4 [1.5, 2], midpoint is 1.75 = 7 4 [2, 2.5], midpoint is 2.25 = 9 4 [2.5, 3], midpoint is 2.75 = 11 4 ( ) 5 + f 4 ( ) 7 + f 4 ( ) 9 + f 4 ( )) 11 4

87 Another Example For the Midpoint Rule, we first need the midpoints. ( 1 f 2 [1, 1.5], midpoint is 1.25 = 5 4 [1.5, 2], midpoint is 1.75 = 7 4 [2, 2.5], midpoint is 2.25 = 9 4 [2.5, 3], midpoint is 2.75 = 11 4 ( ) 5 + f 4 = 1 2 ( 7 4 ) ( ) ( )) f + f 4 4 ( ) 16

88 Another Example For the Midpoint Rule, we first need the midpoints. ( 1 f 2 [1, 1.5], midpoint is 1.25 = 5 4 [1.5, 2], midpoint is 1.75 = 7 4 [2, 2.5], midpoint is 2.25 = 9 4 [2.5, 3], midpoint is 2.75 = 11 4 ( ) 5 + f 4 = 1 2 ( 7 4 ) ( ) ( )) f + f 4 4 ( ) 16 = 101 8

89 Relationship Between Integration and Estimates Example f (x) = x on 0 x 2

90 Relationship Between Integration and Estimates Example f (x) = x on 0 x 2

91 Relationship Between Integration and Estimates Example f (x) = x on 0 x 2 4

92 Relationship Between Integration and Estimates Example f (x) = x on 0 x 2 4

93 Relationship Between Integration and Estimates Example f (x) = x on 0 x 2 4 3

94 Note Because this is a decreasing function, the LHS is the overestimate and the RHS is the underestimate. This is the opposite of what we saw for an increasing function.

95 Relationship Between Integration and Estimates Example f (x) = x on 0 x 2

96 Relationship Between Integration and Estimates Example f (x) = x on 0 x 2

97 Relationship Between Integration and Estimates Example f (x) = x on 0 x 2 2

98 Relationship Between Integration and Estimates Example f (x) = x on 0 x 2 2

99 Relationship Between Integration and Estimates Example f (x) = x on 0 x

100 Relationship Between Integration and Estimates Example f (x) = x on 0 x

101 Relationship Between Integration and Estimates Example f (x) = x on 0 x

102 Relationship Between Integration and Estimates Example f (x) = x on 0 x

103 Relationship Between Integration and Estimates Example f (x) = x on 0 x

104 Relationship Between Integration and Estimates Example f (x) = x on 0 x LHS:

105 Relationship Between Integration and Estimates Example f (x) = x on 0 x LHS: 1 2 ( ) ( 4 = 1 50 ) 2 4 = 25 4

106 If We Kept Going... If we could take an infinite number of subdivisions, then we would get the exact value of the area under the curve. This is the same as taking the integral of the function on the given interval. We will come back in the next section to see how to go from the sum to the integral. 2 0 ( x 2 + 4) dx = 16 3

107 If We Kept Going... If we could take an infinite number of subdivisions, then we would get the exact value of the area under the curve. This is the same as taking the integral of the function on the given interval. We will come back in the next section to see how to go from the sum to the integral. 2 0 ( x 2 + 4) dx = 16 3 Remember, the Fundamental Theorem of Calculus tells us that as long as f (x) is continuous on [a, b] and F (x) = f (x) then b a f (x) dx = F(b) F(a)

108 Distance v. Displacement Example Suppose you throw a ball from 6 feet off the ground straight up. The height is represented in the graph below. 50 ft 6 ft

109 Distance v. Displacement Example Suppose you throw a ball from 6 feet off the ground straight up. The height is represented in the graph below. 50 ft 6 ft What is the distance traveled?

110 Distance v. Displacement Example Suppose you throw a ball from 6 feet off the ground straight up. The height is represented in the graph below. 50 ft 6 ft What is the distance traveled? What is the displacement?

111 Applications Example Roger is training for a marathon. His friend rides behind him on a bicycle and clocks his speed every 15 minutes. Roger starts strong but after an hour and a half he is so exhausted that he has to stop. The data his friend recorded is time (min) speed (mph) Assuming Roger s speed is never increasing, give upper and lower estimates over the hour and a half.

112 Applications Since time is in mph, we need to have x in terms of hours too, so x = 1 4. Lower =

113 Applications Since time is in mph, we need to have x in terms of hours too, so x = 1 4. Lower = ( ) 1 miles Upper = 4 = 23 2 = 11.5

114 Applications Since time is in mph, we need to have x in terms of hours too, so x = 1 4. Lower = ( ) 1 4 = 23 2 = 11.5 miles Upper = ( ) 1 4 = 29 2 = 14.5 miles Average =

115 Applications Since time is in mph, we need to have x in terms of hours too, so x = 1 4. Lower = ( ) 1 4 = 23 2 = 11.5 miles Upper = ( ) 1 4 = 29 2 = 14.5 miles Average = 13 miles

116 Applications How often would we have to make measurements so that the difference between the upper and lower estimates are no more than.1?

117 Applications How often would we have to make measurements so that the difference between the upper and lower estimates are no more than.1? The difference is = 3, so in order to get to.1, we have to divide this difference by 30. Since this differences comes from measuring every hour, dividing by 30 gives 120 hours, or once every 30 seconds.

118 Applications If we wanted to find the average distance that Roger ran, we would probably want to base it on the average since that will be a better guess then the over or the underestimate. Since he ran for 90 minutes, in terms of hours, we d have 1.5 hours. So, the average distance would therefore be (13) = 26 3 miles hr.

119 Applications Example Coal gas is produced at a gasworks. Pollutants in the gas are removed by scrubbers, which become less and less efficient as time goes on. The following measurements, made at the start of each month, show the rate at which the pollutants are escaping (in tons/month) in the gas. time (months) rate (tons/month) Make an overestimate and underestimate of the total quantity of pollutants that escaped during the 6 month period.

120 Applications Which will be the underestimates - LHS or RHS?

121 Applications Which will be the underestimates - LHS or RHS? Because the amounts are increasing and we can assume they will never decrease within months, the LHS will be the underestimate.

122 Applications Which will be the underestimates - LHS or RHS? Because the amounts are increasing and we can assume they will never decrease within months, the LHS will be the underestimate. Under =

123 Applications Which will be the underestimates - LHS or RHS? Because the amounts are increasing and we can assume they will never decrease within months, the LHS will be the underestimate. Under = 1( ) = 59 tons

124 Applications Which will be the underestimates - LHS or RHS? Because the amounts are increasing and we can assume they will never decrease within months, the LHS will be the underestimate. Under = 1( ) = 59 tons Over =

125 Applications Which will be the underestimates - LHS or RHS? Because the amounts are increasing and we can assume they will never decrease within months, the LHS will be the underestimate. Under = 1( ) = 59 tons Over = 1( ) = 74 tons

126 Applications How often would measurements need to be made in order to find over and underestimates which differ by less than 1 ton from the exact quantity of pollutants that escaped during this 6 month period.

127 Applications How often would measurements need to be made in order to find over and underestimates which differ by less than 1 ton from the exact quantity of pollutants that escaped during this 6 month period. The difference between the lower and upper estimates is 15 tons and this results in monthly measurements. If we want to have the difference be no more than 1 ton, we would have to measure 15 times more often, so we would have to measure every 2 days (assuming that there are 30 days in a month).

128 Applications In general, if we want to find the number of observations we would need, we can simply look at the endpoints, the length of the interval and the error desired. So if we want the difference between the left and right hand sums to be less than D, we d have LHS RHS = D

129 Applications In general, if we want to find the number of observations we would need, we can simply look at the endpoints, the length of the interval and the error desired. So if we want the difference between the left and right hand sums to be less than D, we d have LHS RHS = D x (f (x 0 ) +... f (x n 1 ) f (x 1 )... f (x n ) = D

130 Applications In general, if we want to find the number of observations we would need, we can simply look at the endpoints, the length of the interval and the error desired. So if we want the difference between the left and right hand sums to be less than D, we d have LHS RHS = D x (f (x 0 ) +... f (x n 1 ) f (x 1 )... f (x n ) = D b a n (f (x 0) +... f (x n 1 ) f (x 1 )... f (x n ) = D

131 Applications In general, if we want to find the number of observations we would need, we can simply look at the endpoints, the length of the interval and the error desired. So if we want the difference between the left and right hand sums to be less than D, we d have LHS RHS = D x (f (x 0 ) +... f (x n 1 ) f (x 1 )... f (x n ) = D b a n (f (x 0) +... f (x n 1 ) f (x 1 )... f (x n ) = D n = (f (x 0) +... f (x n 1 ) f (x 1 )... f (x n ) (b a) D

132 Applications In general, if we want to find the number of observations we would need, we can simply look at the endpoints, the length of the interval and the error desired. So if we want the difference between the left and right hand sums to be less than D, we d have LHS RHS = D x (f (x 0 ) +... f (x n 1 ) f (x 1 )... f (x n ) = D b a n (f (x 0) +... f (x n 1 ) f (x 1 )... f (x n ) = D n = (f (x 0) +... f (x n 1 ) f (x 1 )... f (x n ) (b a) D That is, n = f (x 0) f (x n ) (b a) D

133 Applications Example We shoot an object straight up with an initial velocity of 400 ft/sec. If the only force is gravity (32 ft/sec 2 ), find a) the velocity at 10 seconds b) approximate distance travelled if we measure in 2 second intervals using i. a lower estimate ii. the Midpoint Rule

134 Applications a) To find the velocity, we need to figure out the velocity equation first.

135 Applications a) To find the velocity, we need to figure out the velocity equation first. We know we our velocity is slowing at a rate of 32 ft/sec 2, so

136 Applications a) To find the velocity, we need to figure out the velocity equation first. We know we our velocity is slowing at a rate of 32 ft/sec 2, so V(t) = t. When we evaluate, we get

137 Applications a) To find the velocity, we need to figure out the velocity equation first. We know we our velocity is slowing at a rate of 32 ft/sec 2, so V(t) = t. When we evaluate, we get V(10) = (10) = 80 ft/sec. b) i. lower estimate

138 Applications a) To find the velocity, we need to figure out the velocity equation first. We know we our velocity is slowing at a rate of 32 ft/sec 2, so V(t) = t. When we evaluate, we get V(10) = (10) = 80 ft/sec. b) i. lower estimate time velocity Since the function is decreasing, the lower estimate will be the RHS. D = 2( ) = 2080 ft ii. Midpoint Rule

139 Applications a) To find the velocity, we need to figure out the velocity equation first. We know we our velocity is slowing at a rate of 32 ft/sec 2, so V(t) = t. When we evaluate, we get V(10) = (10) = 80 ft/sec. b) i. lower estimate time velocity Since the function is decreasing, the lower estimate will be the RHS. D = 2( ) = 2080 ft ii. Midpoint Rule time velocity D = 2( ) = 2400 ft