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Fall 2007 - Hohnhold's Class - Exam 2 (Version B)

Fall 2007 - Hohnhold's Class - Exam 2 (Version B) - Midterm...

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Midterm 2 for Math 20B. Fall Quarter 2007, UCSD Henning Hohnhold Name: Section: #1: #2: #3: #4: Total: (1) Sequences and Series. (a) (5 points) Explain briefly which of the following two sequences are convergent. If a sequence converges, find its limit. ( i ) a n = 7( n 3 + 5) (4 n 2 + 1)( n + 1) ( ii ) a n = ( - 1) n 7( n 3 + 5) (4 n 2 + 1)( n + 1) We have lim n →∞ 7( n 3 + 5) (4 n 2 + 1)( n + 1) = lim n →∞ 7 n 3 + 35 4 n 3 + 4 n 2 + n + 1 = lim n →∞ n 3 (7 - 35 /n 3 ) n 3 (4 + 4 /n + 1 /n 2 + 1 /n 3 ) = 7 4 The sequence in (ii) diverges, since for large n its values jump back and forth between numbers close to 7 4 and close to - 7 4 . (b) (5 points) Determine whether the following series are convergent. If so, find their value. ( i ) 7 + 7 4 + 7 16 + 7 64 + 7 256 + ... ( ii ) 1 2 + 3 4 + 9 8 + 27 16 + 81 32 + ... The second series is the geometric series 1 2 n =0 ( 3 2 ) n ; it diverges because 3 2 > 1. The first series is a convergent geometric series: 7 + 7 4 + 7 16 + 7 64 + 7 256 + ... = 7 X n =0 1 4 n = 7 1 1 - 1 4 = 28 3
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(2) Partial fraction decompositions. (a) (9 points) Use the method of partial fractions to find the indefinite integral of the rational function g ( x ) = - x 2 - 3 x + 18 x 3 + x 2 + 9 x + 9 . Hint: x 3 + x 2 + 9 x + 9 = ( x + 1)( x 2 + 9) . Using the hint, we set up the PFD as - x 2 - 3 x + 18 x 3 + x 2 + 9 x + 9 = A x + 1 + Bx + C x 2 + 9 . This is equivalent to - x 2 - 3 x + 18 = A ( x 2 + 9) + ( Bx + C )( x + 1). Choosing x = - 1 this gives - 1 + 3 + 18 = 10 A , so A = 2. Multiplying out, we see - x 2 - 3 x + 18 = ( A + B ) x 2 + ( C + B ) x + 9 A + C so that B = - 1 - A = - 3 and C = - 9 A + 18 = 0. Finally, we integrate: Z g ( x ) dx = Z 2 x + 1 - 3 x x 2 + 9 dx = 2 ln( x + 1) - 3 2 ln( x 2 + 9) + C, where we used the substitution rule to integrate the second term. (b) (4 points) Find the indefinite integral of f ( x ) = - 2 x 3 - 3 x 2 - 21 x x 3 + x 2 + 9 x + 9 . Hint: Use long division and your result from part (a). Long division gives (after only one step): f ( x ) = - 2 x 3 - 3 x 2 - 21 x x 3 + x 2 + 9 x + 9 = - 2 + - x 2 - 3 x + 18 x 3 + x 2 + 9 x + 9 = 1 + g ( x ) So we can use (a) to compute the indefinite integral of f ( x ): Z f ( x ) dx = Z 1 + g ( x ) dx = - 2 x + 2 ln( x + 1) - 3 2 ln( x 2 + 9) + C.
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