#$&* course Mth 272 6/23 11 pm Question: `q5.5.4 (previous probme was 5.6.2 midpt rule n=4 for `sqrt(x) + 1 on [0,2])
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Given Solution: `a Dividing [-1, 1] into four intervals each will have length ( 1 - (-1) ) / 4 = 1/2. The four intervals are therefore [-1, -.5], [-.5, 0], [0, 5], [.5,1]. The midpoints are -.75, .25, .25, .75. You have to evaluate 1 - x^2 at each midpoint. You get y values .4375, .9375, .9375 and .4375. These values will give you the altitudes of the rectangles used in the midpoint approximation. The width of each rectangle is the length 1/2 of the interval, so the areas of the rectangles will be 1/2 * .4375,1/2 * .9375, 1/2 * .9375 and 1/2 * .4375, or .21875, .46875, .46875, .21875. Adding these areas we get total area 1.375. The curve is concave down so the midpoints will give you values which are a little high. We confirm this by calculating the integral: The exact integral is integral(1 - x^2, x from -1 to 1). An antiderivative is x - 1/3 x^2; evaluating from -1 to 1 we find that the antiderivative changes from -2/3 to 2/3, a change of 4/3 = 1.333. So the accurate integral is 4/3 = 1.333 and our estimate 1.375 is indeed a little high. ** DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: Ok ********************************************* Question: `q 5.6. 9 (was 5.6.12) (was 5.6.10 midpt rule n=4 for x^2-x^3 on [-1,0] YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (-1 + 0)/4= -1/4 [-1, -.75] [-.75, -.5] [-.5, -.25] [-.25, 0] midpoint for intervals: -.875, -.625, -.375, -.125 integral x^2 - x^3 = 1/4 ( 1.43 + .634 + .193 + .017)= .5703 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: Ok ********************************************* Question: `q Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Found this to be easy. ------------------------------------------------ Self-critique Rating: Ok "