using the information given above, answer the following questions, and attach a
ID: 3118542 • Letter: U
Question
using the information given above, answer the following questions, and attach a maple worksheet to your responses. You may insert comments right into the maple file (preferable, but not required ) or include them separately. Obviously, the answer to the problem out lined above is not -5/6. Showing all your steps and including an explanation,tell me the correct solution to: What is the area under the curve for the function f(x) = x2 - x - 2 for x [-2,3]? in other words solve: Given the same f(x), What is the solution for the area under the curve if the interval is: x [-5,-2]? What is the solution for the area under the curve if the interval is: x [0,2]? What is the solution for the area under the curve if the interval Is: x [1,3]? This graph consists of segments and a semicircle, as shown in the figure below. Evaluate the following definite integrals. Again, please attach a Maple file (this may be combined with the file for problem Define the function over the four intervals, which will allow you to find the area under the curve from x [-4,6] In order to do this, you will need to review the techniques of finding the equations of lines and circles, given the data from the image. Show all of your steps. Integrate over each integral so that you arrive at: Integrate over each integral so that you arrive at: I want to see the solution as a decimal (hint: you may have to use the fsolve command, rather than just plain solve). Does your answer make sense, given the four separate integrals? What other comments do you have about this problem?Explanation / Answer
1) a)f(x) = x^2-x-2 = x^2-2x+x-2 = x(x-2)+1(x-2) = (x+1)(x-2) f(x) = 0 =>x = -1,2 Area under the curve in the interval [-2,3] = |Area under the curve in the interval [-2,-1]| + |Area under the curve in the interval [-1,2]| + |Area under the curve in the interval [2,3]| = 1.833 + 4.500 + 1.833 = 8.167 b)Area under the curve in the interval [-5,-2] = 43.500 c)Area under the curve in the interval [0,2] = 3.333 d)Area under the curve in the interval [1,3] = |Area under the curve in the interval [1,2]| + |Area under the curve in the interval [2,3]| = 1.167 + 1.833 = 3.000 2) a)for [-4,-2], y+1 = 0.5(x+4) => y = 0.5x + 1 for[-2,2], x^2+y^2 = 4 for[2,4], y-2 = x-4 => y = x-2 for[4,6], y-2 = -1(x-4) =>y = -x+6 b)Area under the curve = (0.5*2*1) + ((11/7)*2^2) + (0.5*2*2) + (0.5*2*2) = 11.2857
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