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Midterm 1 solution

Midterm 1 solution - UNIVERSITY OF WATERLOO TEST 1...

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UNIVERSITY OF WATERLOO TEST # 1 Solutions FALL TERM 2011 Student Name (Print Legibly) ( family name ) ( given name ) Signature Student ID Number COURSE NUMBER MATH 128 COURSE TITLE Calculus 2 for the Sciences COURSE SECTION 001 DATE OF EXAM Monday, October 17, 2011 TIME PERIOD 17:30 - 18:50 DURATION OF EXAM 80 minutes NUMBER OF EXAM PAGES (Including this sheet) 10 INSTRUCTOR Koray Karabina EXAM TYPE Closed Book ADDITIONAL MATERIALS ALLOWED NONE (NO CALCULATORS) Notes: 1. Fill in your name, ID number, section, and sign the paper. Don’t write formulas on this page. 2. Answer all questions in the space provided. The last page is for rough work. 3. Check that there are 10 sheets. 4. Your grade will be in- fluenced by how clearly you express your ideas, and how well you organize your solutions. Marking Scheme: Question Mark Out of 1a 3 1b 3 1c 3 1d 3 2a 5 2b 5 3a 3 3b 5 4 10 5a 5 5b 5 6 10 Total 60
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MATH 128 - Test # 1 Solutions Fall Term 2011 Page 2 of 10 1. Evaluate the integrals in parts (a)–(d). (a) Z sin 3 x cos 5 x dx Solution 1: Z sin 3 x cos 5 x dx = Z sin 2 x cos 5 x sin x dx = Z (1 - cos 2 x ) cos 5 x sin x dx = - Z (1 - u 2 ) u 5 du ( u = cos x du = - sin x dx ) = u 8 8 - u 6 6 + C = cos 8 x 8 - cos 6 x 6 + C Solution 2: Z sin 3 x cos 5 x dx = Z (cos 2 x ) 2 sin 3 x cos x dx = Z (1 - sin 2 x ) 2 sin 3 x cos x dx = Z (1 - u 2 ) 2 u 3 du ( u = sin x du = cos x dx ) = Z u 7 - 2 u 5 + u 3 du = u 8 8 - u 6 3 + u 4 4 + C = sin 8 x 8 - sin 6 x 3 + sin 4 x 4 + C (b) Z 1 x 2 x 2 + 9 dx Solution: Let θ be such that - π/ 2 < θ < π/ 2, and let tan θ = x 3 . Then x = 3 tan θ dx = 3 sec 2 θ dθ x 2 + 9 = p 9 + 9 tan 2 θ = 3 sec 2 θ = 3 | sec θ | = 3 sec θ (sec θ > 0) Therefore, we can write Z 1 x 2 x 2 + 9 dx = Z 1 (9 tan 2 θ )(3 sec θ ) 3 sec 2 θ dθ = 1 9 Z cos θ sin 2 θ = 1 9 Z du u 2 ( u = sin θ du = cos θ dθ ) = - 1
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