#$&* course MTH 272 10/16 about 12:25 pm 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:OK ********************************************* 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: A=[20*((50+0)/2 + (54+50)/2 + (82+54)/2 + (82+82)/2 + (73+82)/2 + (75+73)/2 + (80+75)/2 + (80+0)/2] = 9920 approx. A=10[(50+0)/2 + (80+0)/2] + 20[(54+50)/2 + (82+54)/2 + (82+82)/2 + (73+82)/2 + (75+73)/2 + (80+75)/2] = 650 + 8620 = 9270 approx. ???Am I wrong, in contradiction to your given solution, that my first problem is more similar to the midpoint rule and my second problem reflects the trapezoidal rule? The trapezoidal rule has the first and last intervals with a multiplier of 1 (10 in this case) and all intervals in between with a multiplier of 2 ( 20 in this case). Let me know if I am wrong in this thinking so I can understand better??? It appears to me that the midpoint rule would have a higher approximation while the trapezoidal rule would be more accurate. confidence rating #$&*:? ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. **