course Mth 272
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RESPONSE --> delta x = 5-1/2(4) = 1/2 2f(1) = 0, 2f(2) = .5, 2f(3) = .4714, 2f(4) = .433, 2f(5) = .4 A = 1/2(0 + .5 + .4714 + .433 + .4) = 1.8044 confidence assessment: 3
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16:48:48 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
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RESPONSE --> self critique assessment: 3
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16:49:04 5.6.32 (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?
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RESPONSE --> trap. 8 spaces @ 20' = 160' length of the pond 160-0/2(8)=10 10(2(50)+2(54)+2(82)+2(82)+2(73)+2(75)+2(80)) = 9920 sq. ft. midpoint points are (10,25), (30,52), (50,68), (70,82), (90,77.5), (110,74), (130,77.5) (150,40) 160-0/8 = 20 20(25+52+68+82+77.5+74+77.5+40) = 9920 sq. ft. confidence assessment: 3
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16:49:07 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. **
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RESPONSE --> self critique assessment: 3
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16:49:09 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> confidence assessment: 3
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