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Unformatted text preview: . 5% ± d = 2082 + 1765 . 21765 . 2  × 100% = 18 . 0% ± e = 1785 + 1765 . 21765 . 2  × 100% = 1 . 1% 2. [3pt] Use Romberg integration to evaluate Z 2 e 2 x sin( x ) 1 + x 2 dx to an accuracy less than 0.5%. Use an estimate of the percent relative error deﬁned by ± a = ± ± ± ± I 1 ,kI 2 ,k1 I 1 ,k ± ± ± ± × 100% . Q2: I 1 , 1 ( h = 2) = 2 2 ( f (0) + f (2)) = 9 . 9292 I 2 , 1 ( h = 1) = 1 2 ( f (0) + 2 f (1) + f (2)) = 8 . 0734 I 3 , 1 ( h = 0 . 5) = . 5 2 ( f (0) + 2( f (0 . 5) + f (1) + f (1 . 5)) + f (2)) = 7 . 6403 I 1 , 2 = 4 I 2 , 1I 1 , 1 41 = 7 . 4548 I 2 , 2 = 4 I 3 , 1I 2 , 1 41 = 7 . 4959 I 1 , 3 = 4 2 I 2 , 2I 1 , 2 4 21 = 7 . 4986 ± =  I 1 , 3I 2 , 2 I 1 , 3  =  7 . 49867 . 4959 7 . 4986  = 0 . 036% < . 5% so we have Z 2 e 2 x sin( x ) 1 + x 2 dx ≈ 7 . 4986 2...
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This note was uploaded on 11/14/2011 for the course EAME 250 taught by Professor Lee during the Spring '11 term at Case Western.
 Spring '11
 LEE

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