128ahw8sum10 - MATH 128A, SUMMER 2010, HOMEWORK 8 SOLUTION...

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Unformatted text preview: MATH 128A, SUMMER 2010, HOMEWORK 8 SOLUTION BENJAMIN JOHNSON Homework 8: Due Wednesday, July 21 4.8; 1d, 5d 4.9; 1a, 8 5.1; 4b, 8a 5.2; 2b section 4.8 1 Use Algorithm 4.4 with n = m = 4 to approximate the following double integrals, and compare the results to the exact answers. d. R 1 . 5 1 R x ( x 2 + y ) dy dx solution: Algorithm 4.4 is just the composite Simpsons rule applied to both integrals. With n = 4, the rule is R b a f ( x ) dx b- a 12 f ( a ) + f ( b ) + 2 f ( a + b 2 ) + 4( f ( 3 a + b 4 ) + f ( a +3 b 4 )) . Using this, we get Z 1 . 5 1 Z x ( x 2 + y ) dy dx Z 1 . 5 1 x 12 " 2 x 2 + x + 2 2 x 2 + x 2 ! + 4 x 2 + 3 x 2 + 2 x 2 !# dx 1 . 4766841026926054 The exact value of the integral is 1 . 48381, giving an approximation error of 0.00712126. 5 Use Algorithm 4.5 with n = m = 2 to approximate the integrals in Exercise 1, and compare the results to those obtained in Exercise 1. d. R 1 . 5 1 R x ( x 2 + y ) dy dx solution: Algorithm 4.5 is just the Gaussian quadrature approximation applied to a double integral. With n = 2 the rule says R 1- 1 f ( x ) dx f ( r 2 , 1 ) + f ( r 2 , 2 ) where r 2 , 1 = . 5773502692 and r 2 , 2 =- . 5773502692. For integrals over [ a,b ] 6 = [- 1 , 1], we need to do a variable substitution and then the rule can be restated as...
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128ahw8sum10 - MATH 128A, SUMMER 2010, HOMEWORK 8 SOLUTION...

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