57326-0136147054_13

# 57326-0136147054_13 - Section 13.1 13.1.1: Part (a): f (1,...

This preview shows pages 1–4. Sign up to view the full content.

Section 13.1 13.1.1: Part (a): f (1 , 2) · 1+ f (2 , 2) · f (1 , 1) · f (2 , 1) · f (1 , 0) · f (2 , 0) · 1 = 198 . Part (b): f (2 , 1) · f (3 , 1) · f (2 , 0) · f (3 , 0) · f (2 , 1) · f (3 , 1) · 1 = 480 . The average of the two answers is 335, fairly close to the exact value 312 of the integral. 13.1.2: Part (a): f (1 , 1) · f (2 , 1) · f (1 , 0) · f (2 , 0) · f (1 , 1) · f (2 , 1) · 1 = 144 . Part (b): f (2 , 2) · f (3 , 2) · f (2 , 1) · f (3 , 1) · f (2 , 0) · 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 , after de±ning 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 rede±ne f , xstar and ystar ,andth e limits on i and j . 13.1.3: 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 . 1654

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is also the exact value of the iterated integral. 13.1.4: 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 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 13.1.5: The Riemann sum is f (2 , 1) · 2+ f (4 , 1) · f (2 , 0) · f (4 , 0) · 2=88 . The true value of the integral is 416 3 138 . 666666666667. 13.1.6: x =1and∆ 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 fairly close Riemann sum 25. 13.1.7: 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 1 4 π ) + f ( 1 4 3 4 π ) + f ( 3 4 3 4 π = 1 2 π 2 4 . 935 . The true value of the integral is 4. 13.1.8: We factor out of each term in the sum the product ∆ x · y = 1 6 π . The Riemann sum then takes the form 1 6 π · ± f ( 1 4 , 1 6 π ) + f ( 3 4 , 1 6 π ) + f ( 1 4 , 1 2 π ) + f ( 3 4 , 1 2 π ) + f ( 1 4 , 5 6 π ) + f ( 3 4 , 5 6 π = 1 2 π 1 . 571 . Mathematica reports that the true value of the integral is - 1 4 CosIntegral[4 π ]+ 1 4 (EulerGamma + Log[4 π ]) and when we asked for a numerical value with the command N[ % ] , it returned the approximation 0 . 77858913775068568 1655
13.1.9: Because f ( x, y )= x 2 y 2 is increasing in both the positive x -direction and the positive y -direction on [1 , 3] × [2 , 5], L 5 M 5 U . 13.1.10: Because f ( x, y p 100 x 2 y 2 is decreasing in both the positive x -direction and the positive y -direction on [1 , 4] × , U 5 M 5 L .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This document was uploaded on 11/28/2010.

### Page1 / 174

57326-0136147054_13 - Section 13.1 13.1.1: Part (a): f (1,...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online