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Unformatted text preview: MA115 Mathematical Analysis I Test 3 Answer Key Instructions The test consists of three sections, Part 1: Questions 1 – 6 Part 2: Questions 7 – 10 You are to do all of Part 1. For Part 2, select and complete 2 of the 4 problems given. Please clearly indicate which two problems you choose. The two problems you indicate will be the only problems from Part 2 graded. Show your work. This does not mean your answers should be long—we just need you to tell us how you obtained your answer in addition to what you obtained. Any response to a question consisting of just an answer will be marked incorrect. Please type your answers up as you do your homework, and submit your completed test via email by 9PM Tuesday, August 17, to the course instructor and the two teaching assistants. You are allowed to use your textbook and all of the notes from the course. You are not permitted to use a graphing calculator or computer graphing program. Good luck! 1 Part 1 1. (15 points) (a) We use the substitution u = 1 x du = 1 x 2 dx arriving at Z 6 e 1 /x x 2 dx = 6 Z e u du = 6 e u + C = 6 e 1 /x + C (b) We use the substitution u = cos x + sin x du = ( sin x + cos x ) dx = (cos x sin x ) dx getting an indefinite integral of Z cos x sin x √ cos x + sin x dx = Z 1 √ u du = 2 √ u + C = 2 √ cos x + sin x + C meaning that our definite integral is Z π/ 4 cos x sin x √ cos x + sin x dx = 2 √ cos x + sin x i π/ 4 = 2 r cos π 4 + sin π 4 2 √ cos 0 + sin 0 = 2 s √ 2 2 + √ 2 2 2 √ 1 + 0 = 2 q √ 2 2 √ 1 = 2 4 √ 2 2 2 (c) Since cot x = cos x sin x we have Z cot x dx = Z cos x sin x dx Using the substitution u = sin x du = cos x we arrive at Z cos x sin x dx = Z 1 u du = ln  u  + C = ln  sin x  + C 2. (10 points) The area involved looks like this: To find where the two func tions intersect we need to solve 2 √ x = 2 x 4 or √ x = x 2 Squaring both sides and a little algebra gives 0 = ( x 1)( x 4) 3 or x = 1 , 4. Of these two only x = 4 works as a point of intersection when...
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This note was uploaded on 04/07/2008 for the course MA 115 taught by Professor Mahalanobis during the Fall '08 term at Stevens.
 Fall '08
 Mahalanobis
 Math

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