course Mth 272
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RESPONSE --> delta x = 1-(-1)/4 = 1/2 midpoints are -.75, -.25, .25, .75 substitute midpoints into f(x) = 1-x^2 A = 1/2(.4375+.9375+.9375+.4375) = 1.375 int -1 to 1 (1-x^2)dx = x-1/3(x^3) 1-1/3(1^3) - (-1)-1/3(-1^3) = 1.3334 confidence assessment: 3
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20:00:00 Dividing [-1, 1] into four intervals each will have length ( 1 - (-1) ) / 4 = 1/2. The four intervals are therefore [-1, -.5], [-.5, 0], [0, 5], [.5,1]. The midpoints are -.75, .25, .25, .75. You have to evaluate 1 - x^2 at each midpoint. You get y values .4375, .9375, .9375 and .4375. These values will give you the altitudes of the rectangles used in the midpoint approximation. The width of each rectangle is the length 1/2 of the interval, so the areas of the rectangles will be 1/2 * .4375,1/2 * .9375, 1/2 * .9375 and 1/2 * .4375, or .21875, .46875, .46875, .21875. Adding these areas we get total area 1.375. The curve is concave down so the midpoints will give you values which are a little high. We confirm this by calculating the integral: The exact integral is integral(1 - x^2, x from 0 to 2). An antiderivative is x - 1/3 x^2; evaluating from -1 to 1 we find that the antiderivative changes from -2/3 to 2/3, a change of 4/3 = 1.333. So the accurate integral is 4/3 = 1.333 and our estimate 1.375 is indeed a little high. ** DER
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RESPONSE --> self critique assessment: 3
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20:00:14 5.6. 9 (was 5.6.12) (was 5.6.10 midpt rule n=4 for x^2-x^3 on [-1,0]
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RESPONSE --> delta x = 1/4 midpoints are -.125, -.375, -.625, -.875 1/4(.0176+.1934+.6348+1.4355) = .5703 int. -1 to 0 [x^2-x^3] dx = 1/3 x^3 - 1/4 x^4 A = .5833 confidence assessment: 3
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20:00:17 The four intervals are (-1, -3/4), (-3/4, -1/2), (-1/2, -1/4) and (-1/4, 0); in decimal form these are (-1, -.75), (-.75, -.5), (-.5, -.25) and (-.25, 0). The midpoints of these intervals are-7/8, -5/8, -3/8 and -1/8; in decimal form we get -.875, -.625, -.375, -.125. The values of the rectangle heights at the midpoints are found by evaluating x^2 - x^3 at the midpoints; we get respectively 735/512, 325/512, 99/512 and 9/512, or in decimal form 1.435546875; 0.634765625; 0.193359375; 0.017578125. The approximating rectangles each have width 1/4 or .25 so the areas arerespectively 735/2048 325/2048, 99/2048, 9/2048, or in decimal form 0.3588867187; 0.1586914062; 0.04833984375; 0.00439453125. The total area is (735 + 325 + 99 + 9) / 2048 = /2048 = 73/128, or in decimal form approximately .5703. An antiderivative of the function is x^3 / 3 - x^4 / 4; evaluating from -1 to 0 we obtain 1/3 + 1/4 = 7/12 = .5833... . So the midpoint approximation is low by about .013 units. ** DER
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RESPONSE --> self critique assessment: 3
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20:00:19
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RESPONSE --> self critique assessment:
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20:00:21 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> confidence assessment:
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