152review1Sp11Ans - 3 5 x 1 4 You are not asked to find a...

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Math 152, Spring 2011, Some Answers to Review for Midterm 1, sections 1–3, 7–9 Please refer to http://www.math.rutgers.edu/courses/152/152Review/152notes1Sp10.pdf for the solutions to the first few problems. Here is the rest: (10) Evaluate the following integrals. Careful: these are improper integrals. ( c ) Z 2 0 dx 2 x - x 2 = π ( f ) Z -∞ dx x 2 + 4 x + 9 = π/ 5 (11) Evaluate Z sin(ln x ) d x using two integrations by parts. Would another method work? Does Z 1 0 sin(ln x ) d x converge? Provide a detailed explana- tion. Z 1 0 sin(ln x ) d x = lim a 0 + Z 1 a sin(ln x ) d x = lim a 0 + ( x/ 2)(sin(ln( x ) - cos(ln( x )) | 1 a = lim a 0 + ( - 1 / 2) - ( a/ 2)(sin(ln( a ) - cos(ln( a )) = - 1 / 2 (here we use the fact that both sine and cosine are bounded functions, and the squeeze theorem). The integral converges to - 1 / 2. (12) The integral R 2 1 4 x 1 / 3 +5 x 1 / 4 dx diverges to . Find some A > 0 so that R A 2 1 4 x 1 / 3 +5 x 1 / 4 dx > 10 10 . Note: You are not asked to find an explicit antiderivative of
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Unformatted text preview: / 3 +5 x 1 / 4 . You are not asked to find a ”best possible” A . You are asked to find a valid A and to support your answer with some reasoning. Answers may vary. When using the underestimating function 1 / (9 x 1 / 3 ) we naturally get A > (6 · 10 10 + 2 2 / 3 ) 3 / 2 (13) 1. Suppose A is a positive real number and let m A be the average of (sin( Ax )) 3 on the interval [0,2]. Compute m A . Note: The answer will have several terms and will not be simple. m A = (1 / (2 A ))(cos 3 (2 A ) / 3-cos(2 A ) + 2 / 3) 2. What is lim A →∞ m A ? Note: This answer should be simple. Explain briefly why it is correct. lim A →∞ m A = 0. (here we use the fact that cosine is a bounded function, and the squeeze theorem). p. 1...
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