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

# Fall 2007 - Hohnhold's Class - Exam 2 (Version A) - 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 = (10 n + 1)( n - 2) 3 n 2 + 3 ( ii ) a n = ( - 1) n (10 n + 1)( n - 2) 3 n 2 + 3 We have lim n →∞ (10 n + 1)( n - 2) 3 n 2 + 3 = lim n →∞ 10 n 2 - 19 n - 2 3 n 2 + 3 = lim n →∞ n 2 (10 - 19 /n - 2 /n 2 ) n 2 (3 + 3 /n 2 ) = 10 3 The sequence in (ii) diverges, since for large n its values jump back and forth between numbers close to 10 3 and close to - 10 3 . (b) (5 points) Determine whether the following series are convergent. If so, find their value. ( i ) 3 + 3 5 + 3 25 + 3 125 + 3 625 + ... ( ii ) 1 3 + 5 9 + 25 27 + 125 81 + 625 243 + ... The second series is the geometric series 1 3 n =0 ( 5 3 ) n ; it diverges because 5 3 > 1. The first series is a convergent geometric series: 3 + 3 5 + 3 25 + 3 125 + 3 625 + ... = 3 X n =0 1 5 n = 3 1 1 - 1 5 = 15 4

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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 ) = 4 x 2 - 3 x + 4 x 3 - x 2 + 4 x - 4 . Hint: x 3 - x 2 + 4 x - 4 = ( x - 1)( x 2 + 4) . Using the hint, we set up the PFD as 4 x 2 - 3 x + 4 x 3 - x 2 + 4 x - 4 = A x - 1 + Bx + C x 2 + 4 . This is equivalent to 4 x 2 - 3 x + 4 = A ( x 2 + 4) + ( Bx + C )( x - 1). Choosing x = 1 this gives 4 - 3 + 4 = 5 A , so A = 1. Multiplying out, we see 4 x 2 - 3 x + 4 = ( A + B ) x 2 + ( C - B ) x + 4 A - C so that B = 4 - A = 3 and C = 4 A - 4 = 0. Finally, we integrate: Z g ( x ) dx = Z 1 x - 1 + 3 x x 2 + 4 dx = ln( x - 1) + 3 2 ln( x 2 + 4) + C, where we used the substitution rule to integrate the second term. (b) (4 points) Find the indefinite integral of f ( x ) = x 3 + 3 x 2 + x x 3 - x 2 + 4 x - 4 . Hint: Use long division and your result from part (a). Long division gives (after only one step): f ( x ) = x 3 + 3 x 2 + x x 3 - x 2 + 4 x - 4 = 1 + 4 x 2 - 3 x + 4 x 3 - x 2 + 4 x - 4 = 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 = x + ln( x - 1) + 3 2 ln( x 2 + 4) + C.
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