128ahw7sum10 - MATH 128A, SUMMER 2010, HOMEWORK 7 SOLUTION...

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MATH 128A, SUMMER 2010, HOMEWORK 7 SOLUTION BENJAMIN JOHNSON Homework 7: Due Monday, July 19 4.4; 1d, 3d 4.5; 2a 4.6; 1d, 9 4.7; 1f, 8 section 4.4 1. Use the composite Trapezoid rule with the indicated values of n to approximate the following integrals. d. R x 0 x 2 cos xdx , n = 6 solution: Dividing the interval [0 ] into six pieces, the points are x 0 = 0, x 1 = π 6 , x 2 = π 3 , x 3 = π 2 , x 4 = 2 π 3 , x 5 = 5 π 6 , and x 6 = π . We use the formula h 2 h f ( x 0 ) + 2 n - 1 j =1 f ( x j ) + f ( x n ) i , obtaining π 12 h 0 + 2 ± π 2 36 · 3 2 + π 2 9 · 1 2 + π 2 4 · 0 + 4 π 2 9 · - 1 2 + 25 π 2 36 · - 3 2 ² - π 2 i ≈ - 6 . 42872. 3. Use the composite Simpson’s rule to approximate the integrals in Exercise 1. d. R x 0 x 2 cos xdx , n = 6 solution: Same coordinates and parameters as in the solution to Exercise 1, except we now use the formula: h 3 f ( x 0 ) + 2 n/ 2 - 1 X j =1 f ( x 2 j ) + 4 n/ 2 X j =1 f ( x 2 j - 1 ) + f ( x n ) . This gives π 18 h 0 + 2 ± π 2 18 - 4 π 2 18 ² + 4 ± 3 π 2 72 + 0 - 25 3 π 2 72 ² - π 2 i ≈ - 6 . 27487. Note: The exact value for this integral is - 2 π ≈ - 6 . 28319. section 4.5 2. Use Romberg integration to compute R 3 , 3 for the following integrals. a. R 1 - 1 (cos x ) 2 dx solution: We have R 1 , 1 = 1 - ( - 1) 2 ( cos 2 ( - 1) + cos 2 (1)) 0 . 583853. For R 2 , 1 , we get 1 2 (cos 2 ( - 1) + 2 cos 2 0 + cos 2 1) 1 . 29193 Then R 2 , 2 = 4 R 2 , 1 - R 1 , 1 3 = 4 · 1 . 29193 - 0 . 583853 3 1 . 52795. Using the recursive formula,
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128ahw7sum10 - MATH 128A, SUMMER 2010, HOMEWORK 7 SOLUTION...

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