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Prelim2Solutions

# Prelim2Solutions - Prelim 2 Math 1910 Calculus for...

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Prelim 2 – Math 1910 – Calculus for Engineers April 1, 2010 You have 90 minutes to complete this test. The test has 100 points. Calculators are not allowed. You should show all your work, and explain where calculations are coming from. Write clearly and legibly. Please write your name and the time you meet for recitation. 1 . Evaluate the following integrals: (a) [5 pts] Z 2 + x 1 - 9 x 2 dx (for | x | < 1 / 3). (b) [5 pts] Z tan - 1 ( θ ) . (c) [5 pts] Z 6 9 x 2 + 6 x + 2 dx . (d) [5 pts] Z x 3 - 2 x dx . Answer: (a) We split the integral in two: Z 2 + x 1 - 9 x 2 dx = 2 Z 1 1 - 9 x 2 dx + Z x 1 - 9 x 2 dx. The first one has a known form: 2 Z 1 1 - 9 x 2 dx = 2 3 Z 1 q 1 9 - x 2 dx = 2 3 sin - 1 (3 x ) . For the second one we use the substitution u = 9 x 2 (so du = 18 xdx ): Z x 1 - 9 x 2 dx = Z 1 18 1 - u du = - 1 9 1 - u = - 1 9 p 1 - 9 x 2 . Thus Z 2 + x 1 - 9 x 2 dx = 2 3 sin - 1 (3 x ) - 1 9 p 1 - 9 x 2 + C. (b) We use integration by parts with u = tan - 1 ( θ ) and dv = , so du = (1 + θ 2 ) - 1 and v = θ : Z tan - 1 ( θ ) = θ tan - 1 ( θ ) - Z θ 1 + θ 2 = θ tan - 1 ( θ ) - 1 2 ln(1 + θ 2 ) + C. (c) We need to complete squares: Z 6 9 x 2 + 6 x + 2 dx = 6 Z 1 (3 x + 1) 2 + 2 - 1 dx = 6 Z 1 (3 x + 1) 2 + 1 dx. Now we use the substitution u = 3 x + 1 (so du = 3 dx ) to get Z 6 9 x 2 + 6 x + 2 dx = 6 Z 1 3 u 2 + 1 dx = 2 tan - 1 ( u ) + C = 2 tan - 1 (3 x + 1) + C. 1

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(d) We use integration by parts. Set u = x and dv = 3 - 2 x dx , so du = dx and v = - 1 3 (3 - 2 x ) 3 / 2 . Then Z x 3 - 2 x dx = - 1 3 x (3 - 2 x ) 3 / 2 + 1 3 Z (3 - 2 x ) 3 / 2 dx = - 1 3 x (3 - 2 x ) 3 / 2 - 1 15 (3 - 2 x ) 5 / 2 + C = - 1 5 ( x + 1)(3 - 2 x ) 3 / 2 .
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