course mth 272 If your solution to a stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a The graphs cross at (0, -1), (1,0), and (2, 1) as we easily find by solving the equation (x-1)^3 = (x - 1). Since (1, 0) lies between the endpoints of our interval we have to be careful about which function lies above which, and we'll have to split the calculation into two separate intervals. • The x-1 graph lies above the (x-1)^3 graph from 0 to 1, so the area on this interval will be the integral of (x-1) - (x-1)^3 between these limits. • The (x-1)^3 graph lies above the (x-1) graph from 1 to 2, so the area on this interval will be the integral of (x-1)^3 - (x-1) between these limits. Each integral is equal to .25, so the total area is the sum .25 + .25 = .5 of these areas. ** DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Well I got negative but then remembered from my last calculus class that area can’t be negative so I turned it postitive. ------------------------------------------------ Self-critique rating #$&*:3 ********************************************* Question: `q 5.5.2 (previously 5.5.10 (was 5.5.10) ) describe the region for integral of (1-x^2) - (x^2-1) from -1 to 1 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (1-x^2) - (x^2-1) = 2 - 2 x^2 = 2x – 2/3 x^3 Plug in 1, and -1 and you get 8/3 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a*& The graph of 1 - x^2 is an upside down parabola with vertex at (0, 1), intercepting the x axis at x = -1 and x = 1. The graph of x^2 - 1 is a rightside up parabols with vertex at (0, -1), intercepting the x axis at x = -1 and x = 1. The region between the graphs is close to a circle passing thru (-1,0), (0,1), (1,0) and (0, -1), but the region is not exactly circular since it is formed by two parabolas. The graphs aren't vertical at (1,0) and (-1,0), for example, and a circle would be. The parabolas curve in such a way as to stay inside the circular region, so the region between the parabolas will have a bit less area than the circle. The integrand (1-x^2) - (x^2-1) can be simplified to 2 - 2 x^2. An antiderivative would be 2x - 2/3 x^3. Evaluating this at -1 and 1 we obtain integral 8/3. The area of the region is 8/3 = 2.67 approx.. Note that the area of the circle described above would be pi = 3.14, approx., a bit bigger than the area of the region between the parabolas. *&*& STUDENT ERROR: The graph is a circular region centered on the point (0,0) INSTRUCTOR COMMENT: The region is not exactly circular, (for example the graphs aren't vertical at (1,0) and (-1,0), for example), but it's fairly close to the circle. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*:3