#$&* course Mth 272 6/27 11:37pm 012. `query 12
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Given Solution: `a Dividing [1, 5] into four intervals each will have length ( 5 - 1 ) / 4 = 1. The four intervals are therefore [1, 2], [2, 3], [3, 4], [4,5]. The function values at the endpoints f(1) = 0, f(2) = .5, f(3) = .471, f(4) = .433 and f(5) = .4. The average altitudes of the trapezoids are therefore (0 + .5) / 2 = 0.25, (.5 + .471)/2 = 0.486, (.471 + .433) / 2 = 0.452 and (.433 + .4) / 2 = 0.417. Trapeoid areas are equal to ave height * width; since the width of each is 1 the areas will match the average heights 0.25, 0.486, 0.452, 0.417. The sum of these areas is the trapezoidal approximation 1.604. ** DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q5.6.32 (was 5.6.24 est pond area by trap and midpt (20 ft intervals, widths 50, 54, 82, 82, 73, 75, 80 ft). How can you tell from the shape of the point whether the trapezoidal or midpoint estimate will be greater? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Trapezoidal estimate: (0 + 50) / 2 = 25 (50 + 54) / 2 = 52 (54 + 82) / 2 = 68 (82 + 82) / 2 = 82 (82 + 73) / 2 = 77.5 (73 + 75) / 2 = 74 (75 + 80) / 2 = 77.5 (80 + 0) / 2 = 40 Multiply results by 20 to get 9920 ft^2 Midpoint: [20, 40], [40, 60], [60, 80], [80, 100], [100, 120], [120, 140], [140, 160] Midpoints are 30, 50, 70, 90, 110, 130, 150 Multiply each by 20 to get: 600, 1000, 1400, 1800, 2200, 2600, 3000 Add together to get 11,700 11,700 < 9920 so the midpoint estimate would exceed the trapezoidal estimate. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Since widths at 20-ft intervals are 50, 54, 82, 82, 73, 75, 80 ft the pond area can be approximated by a series of trapezoids with these altitudes. The average altitudes are therefore respectively (0 + 50 / 2) = 25 (50 + 54) / 2 = 52 (54 + 82) / 2 = 68 (82 + 82) / 2 = 82 etc., with corresponding areas 25 * 20 = 500 52 * 20 = 1040 etc., all areas in ft^2. The total area, according to the trapezoidal approximation, will therefore be 20 ft (25+52+68+82+77.5+74+77.5+40) ft = 9920 square feet. The midpoint widths would be calculated based on widths at positions 10, 30, 50, 70, ..., 150 ft. Due to the convex shape of the pond these estimates and will lie between the estimates made at 0, 20, 40, ..., 160 feet. The convex shape of the pond also ensures that the midpoint of each rectangle will 'hump above' the trapezoidal approximation, so the midpoint estimate will exceed the trapezoidal estimate. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Was that even the right way to do the midpoint estimate in this case? The math definitely corresponds to your solution. ------------------------------------------------ Self-critique Rating: