#$&* course Mth 272 October 15 around 2:55 011. `query 11
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Given Solution: `a Dividing [-1, 1] into four intervals each will have length ( 1 - (-1) ) / 4 = 1/2. The four intervals are therefore [-1, -.5], [-.5, 0], [0, 5], [.5,1]. The midpoints are -.75, .25, .25, .75. You have to evaluate 1 - x^2 at each midpoint. You get y values .4375, .9375, .9375 and .4375. These values will give you the altitudes of the rectangles used in the midpoint approximation. The width of each rectangle is the length 1/2 of the interval, so the areas of the rectangles will be 1/2 * .4375,1/2 * .9375, 1/2 * .9375 and 1/2 * .4375, or .21875, .46875, .46875, .21875. Adding these areas we get total area 1.375. The curve is concave down so the midpoints will give you values which are a little high. We confirm this by calculating the integral: The exact integral is integral(1 - x^2, x from -1 to 1). An antiderivative is x - 1/3 x^2; evaluating from -1 to 1 we find that the antiderivative changes from -2/3 to 2/3, a change of 4/3 = 1.333. So the accurate integral is 4/3 = 1.333 and our estimate 1.375 is indeed a little high. ** DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t confirm by doing the integral. ------------------------------------------------ self-critique rating #$&*: 3 ********************************************* Question: `q 5.6. 9 midpt rule n=4 for x^2-x^3 on [-1,0] YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Four intervals are [-1,-0.75], [-0.75,-0.5], [-0.5,-0.25], [-0.25,0] The midpoints are then -0.875, -0.625, -0.375, -0.125 x^2-x^3 (-0.875)^2-(-0.875)^3=1.44 (-0.625)^2-(-0.625)^3=0.635 (-0.375)^2-(-0.375)^3=0.193 (-0.125)^2-(-0.125)^3=0.018 The area of rectangles are 0.25*1.44=0.359 0.25*0.635=0.159 0.25*0.193=0.048 0.25*0.018=0.004 The total area is 0.359+0.159+0.048+0.004=0.57 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The four intervals are (-1, -3/4), (-3/4, -1/2), (-1/2, -1/4) and (-1/4, 0); in decimal form these are (-1, -.75), (-.75, -.5), (-.5, -.25) and (-.25, 0). The midpoints of these intervals are-7/8, -5/8, -3/8 and -1/8; in decimal form we get -.875, -.625, -.375, -.125. The values of the rectangle heights at the midpoints are found by evaluating x^2 - x^3 at the midpoints; we get respectively 735/512, 325/512, 99/512 and 9/512, or in decimal form 1.435546875; 0.634765625; 0.193359375; 0.017578125. The approximating rectangles each have width 1/4 or .25 so the areas arerespectively 735/2048 325/2048, 99/2048, 9/2048, or in decimal form 0.3588867187; 0.1586914062; 0.04833984375; 0.00439453125. The total area is (735 + 325 + 99 + 9) / 2048 = /2048 = 73/128, or in decimal form approximately .5703. An antiderivative of the function is x^3 / 3 - x^4 / 4; evaluating from -1 to 0 we obtain 1/3 + 1/4 = 7/12 = .5833... . So the midpoint approximation is low by about .013 units. ** DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t do the antiderivative to check how close the approximation is. ------------------------------------------------ self-critique rating #$&*: 3 ********************************************* Question: `q Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I keep forgetting to take the antiderivative to check how close the approximation really is. ------------------------------------------------ self-critique rating #$&*: 3 "