Calculus with Analytic Geometry by edwards & Penney soln ch14

Calculus with Analytic Geometry

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Unformatted text preview: Section 14.1 C14S01.001: Part (a): f (1 , − 2) · 1 + f (2 , − 2) · 1 + f (1 , − 1) · 1 + f (2 , − 1) · 1 + f (1 , 0) · 1 + f (2 , 0) · 1 = 198 . Part (b): f (2 , − 1) · 1 + f (3 , − 1) · 1 + f (2 , 0) · 1 + f (3 , 0) · 1 + f (2 , 1) · 1 + f (3 , 1) · 1 = 480 . The average of the two answers is 339, fairly close to the exact value 312 of the integral. C14S01.002: Part (a): f (1 , − 1) · 1 + f (2 , − 1) · 1 + f (1 , 0) · 1 + f (2 , 0) · 1 + f (1 , 1) · 1 + f (2 , 1) · 1 = 144 . Part (b): f (2 , − 2) · 1 + f (3 , − 2) · 1 + f (2 , − 1) · 1 + f (3 , − 1) · 1 + f (2 , 0) · 1 + f (3 , 0) · 1 = 570 . The average of the two answers is 357, fairly close to the exactly value 312 of the integral. The computations shown here can be automated in computer algebra systems. For example, in Mathematica 3 . 0, after defining f ( x, y ) = 4 x 3 + 6 xy 2 , you could proceed as follows. x[i ] := i + 1; y[j ] := j - 2; deltax = x[1] - x[0]; deltay = y[1] - y[0]; ( ∗ Part (a): ∗ ) xstar[i ] := x[i-1]; ystar[j ] := y[j] Sum[ Sum[ f[xstar[i],ystar[j]] ∗ deltax ∗ deltay, { j, 1, 3 } , { i, 1, 2 } ] 144 ( ∗ Part (b): ∗ ) xstar[i ] := x[i]; ystar[j ] := y[j-1] Sum[ Sum[ f[xstar[i],ystar[j]] ∗ deltax ∗ deltay, { j, 1, 3 } , { i, 1, 2 } ] 570 The idea is that to work another such problem, all you need to do is redefine f , xstar and ystar , and the limits on i and j . C14S01.003: We omit ∆ x and ∆ y from the computation because each is equal to 1. f ( 1 2 , 1 2 ) + f ( 3 2 , 1 2 ) + f ( 1 2 , 3 2 ) + f ( 3 2 , 3 2 ) = 8 . This is also the exact value of the iterated integral. C14S01.004: We omit ∆ x and ∆ y from the computation because each is equal to 1. f ( 1 2 , 1 2 ) + f ( 3 2 , 1 2 ) + f ( 1 2 , 3 2 ) + f ( 3 2 , 3 2 ) = 4 . This is also the exact value of the iterated integral. In a Mathematica 3 . 0 solution similar to the one in Problem 2, we would use xstar[i ] := (x[i] + x[i-1])/2; ystar[j ] := (y[j] + y[j-1])/2 1 C14S01.005: The Riemann sum is f (2 , − 1) · 2 + f (4 , − 1) · 2 + f (2 , 0) · 2 + f (4 , 0) · 2 = 88 . The true value of the integral is 416 3 ≈ 138 . 666666666667. C14S01.006: We omit ∆ x = 1 and ∆ y = 1 from the computation. f (1 , 1) + f (2 , 1) + f (1 , 2) + f (2 , 2) + f (1 , 3) + f (2 , 3) = 43 . The true value of the integral is 26. The midpoint approximation gives the very close Riemann sum 25. C14S01.007: We factor out of each term in the sum the product ∆ x · ∆ y = 1 4 π 2 . The Riemann sum then takes the form 1 4 π 2 · £ f ( 1 4 π, 1 4 π ) + f ( 3 4 π, 1 4 π ) + f ( 1 4 π, 3 4 π ) + f ( 3 4 π, 3 4 π )¤ = 1 2 π 2 ≈ 4 . 935 . The true value of the integral is 4....
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Calculus with Analytic Geometry by edwards & Penney soln ch14

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