course Mth 272 €???K????????assignment #011011. `query 11
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15:20:09 5.5.4 (previous probme was 5.6.2 midpt rule n=4 for `sqrt(x) + 1 on [0,2]) 5.5.4 asks for an n = 4 midpoint-rule approximation to the integral of 1 - x^2 on the interval [-1, 1].
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RESPONSE --> first find the interval length 1-(-1)/ 4 = 1/2 the new intervals = [-1,-.5], [-.5,0], [0,.5], [.5,1] midpoints= -.75, -.25, .25, .75 then plug these in to 1-x^2 y= .4375, .9375, .9375, .4375 then use the midpoint formula to get the total area 1/2[.4375+ .9375+ .9375+ .4375] = 1.375 finding the exact interval take the antideriv. = x-x^3/3 then plug in 1 for 2/3 and -1 for -2/3 2/3-(-2/3) = 4/3 or 1.333 which is the accurate integral confidence assessment: 3
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15:20:15 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 --> got it self critique assessment: 3
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15:44:07 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 --> the new intervals = [-1,-3/4], [-3/4,-1/2], [-1/2,-1/4], [-1/4,0] midpoints= -7/8, -5/8, -3/8, -1/8 then plug these into the function = 735/512, 325/512, 99/512, 9/512 then use the midpoint formula 1/4 (735/512+ 325/512+ 99/512+ 9/512)= .5703 to find the accurate area, the antideriv is x^3/3- x^4/4 which by plugging in turns out to be 1/3 + 1/4 = .5833 confidence assessment: 3
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15:44:13 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 --> got it self critique assessment: 3
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15:45:47 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The assignment was very straight forward. I just followed the book and the midpoint rules and it ended up making sense to me pretty easily. confidence assessment: 3
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