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Unformatted text preview: roots, in this case x = 3 and x =2 . We get 9 = 25 A, so A = 9 25 from x = 3 and 4 =5 C, so C =4 5 from x =2 . Now we apply the second method for nding coecients, and equate coecients of x 2 on both sides above, 1 = A + B, so using our solution for A, we get B = 16 25 . Then Z x 2 ( x3)( x + 2) 2 dx = Z 9 25 x3 + Z 16 25 x + 2Z 4 5 ( x + 2) 2 = 9 25 ln ( x3) + 16 25 ln ( x + 2) + 4 5 ( x +2) + C. Problem 7.7.8. Find the approximations to Z 1 2 sin 2 ( x ) dx given by the Midpoint Rule M n , the Trapezoid Rule T n and Simpsons Rule S n when n = 4 . 4 Solution. The partition is 0 = x < 1 8 < 1 4 < 3 8 < 1 2 = x 4 , where x 1 = 1 8 , x 2 = 1 4 , and x 3 = 3 8 . So the Trapezoid rule gives T 4 = 1 8 [sin 2 0 + 2 sin 2 1 8 + 2 sin 2 1 4 + 2 sin 2 3 8 sin 2 1 2 ] . (Stop!)...
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This note was uploaded on 02/29/2008 for the course MATH 22 taught by Professor Dodson during the Summer '05 term at Lehigh University .
 Summer '05
 Dodson
 Fractions, Improper Integrals, Integrals

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