If your solution to a stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
011. `query 11
Question: `q5.5.4 (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].
Your solution:
Confidence Assessment:
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
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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]
Your solution:
Confidence Assessment:
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):
Self-critique Rating: