Solução_Calculo_Stewart_6e

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F. TX.10 SECTION 5.1 AREAS AND DISTANCES ¤ 235 7. Here is one possible algorithm (ordered sequence of operations) for calculating the sums: 1 Let SUM = 0, X_MIN = 0, X_MAX = 1, N = 10 (depending on which sum we are calculating), DELTA_X = (X_MAX - X_MIN)/N, and RIGHT_ENDPOINT = X_MIN + DELTA_X. 2 Repeat steps 2a, 2b in sequence until RIGHT_ENDPOINT > X_MAX. 2a Add (RIGHT_ENDPOINT)^4 to SUM. 2b Add DELTA_X to RIGHT_ENDPOINT. At the end of this procedure, (DELTA_X)·(SUM) is equal to the answer we are looking for. We find that R 10 = 1 10 10 i=1 R 100 = 1 100 i 10 4 ≈ 0.2533, R 30 = 1 30 100 i=1 30 i=1 i 30 4 ≈ 0.2170, R 50 = 1 50 4 i ≈ 0.2050. It appears that the exact area is 0.2. 100 50 i=1 4 i ≈ 0.2101, and 50 The following display shows the program SUMRIGHT and its output from a TI-83 Plus calculator. To generalize the program, we have input (rather than assign) values for Xmin, Xmax, and N. Also, the function, x 4 , is assigned to Y 1, enabling us to evaluate any right sum merely by changing Y 1 and running the program. 9. In Maple, we have to perform a number of steps before getting a numerical answer. After loading the student package [command: with(student);] weusethecommand left_sum:=leftsum(1/(xˆ2+1),x=0..1,10 [or 30, or50]); which gives us the expression in summation notation. To get a numerical approximation to the sum, we use evalf(left_sum);. Mathematica does not have a special command for these sums, so we must type them in manually. For example, the firstleftsumisgivenby (1/10)*Sum[1/(((i-1)/10)ˆ2+1)],{i,1,10}], and we use the N command on the resulting output to get a numerical approximation. In Derive, we use the LEFT_RIEMANN command to get the left sums, but must define the right sums ourselves. (We can define a new function using LEFT_RIEMANN with k ranging from 1 to n instead of from 0 to n − 1.) (a) With f(x) = 1 x 2 +1 , 0 ≤ x ≤ 1, the left sums are of the form Ln = 1 n L 30 ≈ 0.7937,andL 50 ≈ 0.7904. The right sums are of the form R n = 1 n R 30 ≈ 0.7770,andR 50 ≈ 0.7804. n i=1 i−1 n n i=1 1 2 . Specifically, L10 ≈ 0.8100, +1 1 i 2 .Specifically, R10 ≈ 0.7600, n +1
F. 236 ¤ CHAPTER 5 INTEGRALS TX.10 (b) In Maple, we use the leftbox (with the same arguments as left_sum)andrightbox commands to generate the graphs. left endpoints, n =10 left endpoints, n =30 left endpoints, n =50 right endpoints, n =10 right endpoints, n =30 right endpoints, n =50 (c) We know that since y =1/(x 2 +1)is a decreasing function on (0, 1), all of the left sums are larger than the actual area, and all of the right sums are smaller than the actual area. Since the left sum with n =50is about 0.7904 < 0.791 and the right sum with n =50is about 0.7804 > 0.780, we conclude that 0.780
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Solução_Calculo_Stewart_6e

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