#$&* course 272 10/24 6:15pm If your solution to a stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a Dividing [1, 5] into four intervals each will have length ( 5 - 1 ) / 4 = 1. The four intervals are therefore [1, 2], [2, 3], [3, 4], [4,5]. The function values at the endpoints f(1) = 0, f(2) = .5, f(3) = .471, f(4) = .433 and f(5) = .4. The average altitudes of the trapezoids are therefore (0 + .5) / 2 = 0.25, (.5 + .471)/2 = 0.486, (.471 + .433) / 2 = 0.452 and (.433 + .4) / 2 = 0.417. Trapeoid areas are equal to ave height * width; since the width of each is 1 the areas will match the average heights 0.25, 0.486, 0.452, 0.417. The sum of these areas is the trapezoidal approximation 1.604. ** DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q5.6.32 (was 5.6.24 est pond area by trap and midpt (20 ft intervals, widths 50, 54, 82, 82, 73, 75, 80 ft). How can you tell from the shape of the point whether the trapezoidal or midpoint estimate will be greater? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Interval midpoints (0+50) /2 = 25 ft^2 (50+54) /2 = 52 ft^2 (54+82) /2 = 68 ft^2 (82+82) /2 = 82 ft^2 (82+73) /2 = 77.5 ft^2 (73+75) /2 = 74 ft^2 (75+80) /2 = 77.5 ft^2 (80+0) /2 = 40 ft^2 midpoints * 20 20 (25+52+68+82+77.5+74+77.5+40) = 9940 ft^2 area will approximately by 9940 ft^2 Since the first and last intervals go above the approximation, the midpoint estimation will definitely be larger than the trapezoidal estimate. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. Midpoint widths will be subject to some estimation errors. Each midpoint width will be multiplied by the 20 ft width of the interval. The total area obtained from midpoint estimates will exceed that of the trapezoidal approximation. The midpoint and trapezoidal rules won't differ much for the second through the seventh intervals, while on the first and the last the trapezoidal rule will seriously underestimate the area while the midpoint rule will seriously overestimate it. The pond itself isn't convex, but the first and last region are, and this is the source of the main difference between the two estimates. Put a little differently, the convex shape of these two regions ensures that the midpoint of the first and last regions will 'hump above' the trapezoidal approximation, so that for these two regions the midpoint estimate will very significantly exceed the trapezoidal estimate. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This section was very easy to visualize and follow. I am very comfortable applying the midpoint and trap rules for estimating purposes. ------------------------------------------------ Self-critique Rating: 3