MATH267_F17_midterm_1_solutions_good.pdf

# MATH267_F17_midterm_1_solutions_good.pdf - version 11 THE...

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version 11 THE UNIVERSITY OF CALGARY DEPARTMENT OF MATHEMATICS AND STATISTICS MIDTERM TEST 1 - SOLUTIONS MATH 267 L01 (Fall 2017) October 12, 2017 Time: 2 hours These solutions are based off of version 11 of midterm 1. The others are similar. I.D. NUMBER SURNAME OTHER NAMES NOTE: No calculators. No other aids. Closed-book. FOR TA USE ONLY Question Score MC /35 B1 /5 B2 /5 B3 /5 TOTAL /50 1

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Math 267 (L01) - M. Pawliuk - Midterm Test 1 Version 11 - Fall 2017 2 PART (A) Multiple Choice Questions (35 marks) Instructions: On the scantron, fill in one answer per question. Questions 1,3,5,7,9,11,13 are worth 3 marks each. Questions 2,4,6,8,10,12,14 are worth 2 marks each. 1. (3 marks) Evaluate Z 2 x cos( x 2 ) dx. (a) x 2 sin( x 3 3 ) + C , (b) x 2 sin( x 2 ) + C , (c) sin( x 2 ) + C , [***CORRECT***. This is a basic u -sub. Very similar to an example in Lecture 1-1 and 1-2.] (d) cos( x 2 ) + C , (e) 0. 2. (2 marks) Which u -substitution will turn Z dx x ln( x ) into Z du ln( u ) ? (a) u = x , [***CORRECT***. Remember ln( u 2 ) = 2 ln( u ).] (b) u = x , (c) u = ln( x ), (d) 1 x ln( x ) , (e) u = e x .
Math 267 (L01) - M. Pawliuk - Midterm Test 1 Version 11 - Fall 2017 3 3. (3 marks) Evaluate Z cos 2 ( x ) sin 3 ( x ) dx. (a) - sin 3 ( x ) 3 cos 4 ( x ) 4 + C , (b) - sin 3 ( x ) 3 + cos 4 ( x ) 4 + C , (c) - cos 3 ( x ) 3 cos 5 ( x ) 5 + C , (d) - cos 3 ( x ) 3 + cos 5 ( x ) 5 + C , [***CORRECT***. Use pythagoras on a sin 2 x , then it is a u -sub. This was an example in Lecture 2-1.] (e) - cos 3 ( x ) 3 . 4. (2 marks) Evaluate Z π 0 sin 2 ( θ ) dθ. (a) 0, (b) 1, (c) π 2 , [***CORRECT***. Use the half-angle formula for sine that reduces the power. The double-angle formula is not needed.] (d) π , (e) 2 π .

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Math 267 (L01) - M. Pawliuk - Midterm Test 1 Version 11 - Fall 2017 4 5. (3 marks) Evaluate Z dx 4 - 9 x 2 . (a) 1 3 arcsin( 1 2 x ) + C , (b) 2 3 arcsin( 1 2 x ) + C , (c) 2 3 arcsin( 3 2 x ) + C , (d) 1 3 arcsin( 3 2 x ) + C , [***CORRECT***. Use the trig sub x = 2 3 sin θ . This was an example in class.] (e) 1 3 arcsin( x ) + C . 6. (2 marks) By using an appropriate trig substitution, evaluate Z dx x ( x 2 + 1) . (a) arctan( x ) + C , (b) ln | x x 2 - 1 | + C , (c) ln | 1 1 + x 2 | + C , (d) ln | x 1 + x 2 | + C , [***CORRECT***. Use the trig sub x = tan θ . This is a sim- plified version of a Lyryx problem from assignment 1. This was also example 2 in lecture 3-1, where we solved it using partial fractions.] (e) x arctan( x ) + C .
Math 267 (L01) - M. Pawliuk - Midterm Test 1 Version 11 - Fall 2017 5 7. (3 marks) Evaluate Z x ln x dx. (a) x 2 2 (ln x - 1) + C , (b) x 2 4 (ln x - 1) + C , (c) x 2 2 (ln x - 1 2 ) + C , [***CORRECT***. By parts: u = ln( x ) , dv = xdx . This was the second example in lecture 2-2.] (d) x 2 4 (ln x - 1 2 ) + C , (e) x 2 2 ( x ln x - x ) + C .

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