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Unformatted text preview: Final Exam Key, 2 hours and 45 minutes April 24, 2003 The average was about 60/100 with the lowest scores on the recent material power series and polar coordinates. Warning  this key was written days after the exam was graded, so I don’t expect too much interest in it, and I haven’t checked it carefully. The basic methods given are all correct. 1) 1 2 R x 4 dx = 32 / 10 = 16 / 5. 2A) From memory, ln(1 + x ) = x x 2 / 2 + x 3 / 3 . . . . So, ln(2) ≈ 1 1 / 2 + 1 / 3 = 5 / 6. Since this is a decreasing alternating series, this answer is too big, but the error is less than the next term, 1/4. 2B) ln(2) = R 2 1 dt t ≈ 2 1 2 · 3 (1 + 4 · 2 3 + 1 / 2) = 25 / 36. Here, for example, y 1 = f ( x 1 ) = f (1 . 5) = 1 / 1 . 5 = 2 / 3. 3a) Set u = tan x and get (tan 3 x ) / 3 + C . 3b) Do IBP twice to get e x ( x 2 2 x + 2) + C . 3c) This is improper b/c it has an asymptote at x = 2. A graph is recom mended. So, it should be split in two (this also helps remove the absolute value signs). The first part isvalue signs)....
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This note was uploaded on 12/26/2011 for the course MAC 2312 taught by Professor Storfer during the Summer '08 term at FIU.
 Summer '08
 Storfer
 Calculus, Power Series, Polar Coordinates

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