sum7.4.20 - 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

Week 3 Homework: Wed, Thurs, Mon 7.3 trig substitution – text ex 3, #27 7.4 partial fractions – #21 below, #25 7.7 approximate integration — answers as sums 7.8 improper integrals (first case) 8.2 surface areas (rotation about x -axis)

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2 Homework notes: In 7.3, #27 done in-class. This week’s problems are #3, 12, 24 (added), 27. For 7.4, just the problems from the Fall 2004 list between #9 and #38. Problems listed from 7.5 remain on this week’s syllabus. For 7.7, see example below; decimal approximations will not be on exam, answers as explicit sums will be. 7.8 - 5,8,9 and 8.2 - 5, 8, 10, 11 as covered in class toda Problem 7.4.20: Evaluate Z x 2 ( x - 3)( x + 2) 2 dx. Solution: The method of partial fractions says there are A, B, C so x 2 ( x - 3)( x + 2) 2 = A x - 3 + B x + 2 + C ( x + 2) 2 . As in previous cases, we first clear denominators, multiplying both sides by ( x - 3)( x + 2) 2 , so x 2 = A ( x + 2) 2 + B ( x - 3)( x + 2) + C ( x - 3) .

Again as in previous cases, we plug in the distinct real

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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 ﬁnding coeﬃcients, and equate coeﬃcients 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 ( x-3)( x + 2) 2 dx = Z 9 25 x-3 + Z 16 25 x + 2-Z 4 5 ( x + 2) 2 = 9 25 ln ( x-3) + 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 Simpson’s 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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