course Mth 272 011. `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 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* 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: To use the midpoint rule, we must first break our interval into n=4 rectangles. Our new intervals become [-1,-3/4][-3/4, -1/2][-1/2, -1/4][-1/4,0] and their midpoints are -7/8, -5/8, -3/8, and -1/8. Now we can use our midpoint rule to estimate the area. So Area = ½( f(-7/8) + f(-5/8) + f(-3/8) + f(-1/8) = ½(735/2048 + 325/2048 + 99/2048 + 9/2048) = .5703. Now we need to find the actual integral and compare our midpoint rule answer to it. Using the rules of integration, we find that the integral of our original equation is 1/3x^3 – 1/4x^4. Now we can use the Fundamental Theorem of Calculus to find the actual area. On the interval (-1,0), we find that the area equals .5833. So our midpoint rule estimate was lower than the actual area. 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): Self-critique Rating: ********************************************* Question: `q Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I misunderstood the midpoint rule at first, I thought I had to take the integral of the equation when it was given to me in the first few homework problems. But now I understand it better and hopefully will not make the same mistake. "