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Unformatted text preview: armington (kma786) – 5.5 – Stepp – (55860) 1 This printout should have 7 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the value of I = integraldisplay π/ 2 4 sin2 t dt . 1. I = 3 2. I = 5 3. I = 4 correct 4. I = 7 2 5. I = 9 2 Explanation: Set u = 2 t . Then du = 2 dt , while t = 0 = ⇒ u = 0 , and t = π 2 = ⇒ u = π . In this case, I = 1 2 integraldisplay π 4 sin u du = 1 2 bracketleftBig 4 cos u bracketrightBig π . Now cos 0 = 0 , cos π = 1 . Consequently, I = 2 parenleftBig ( 1) ( 1) parenrightBig = 4 . keywords: IntSubst, IntSubstExam, 002 10.0 points Evaluate the integral I = integraldisplay 1 4 x (1 x 2 ) 5 dx . 1. I = 4 5 2. I = 2 5 3. I = 1 3 correct 4. I = 2 3 5. I = 1 3 6. I = 2 3 Explanation: Set u = 1 x 2 . Then du = 2 x dx while x = 0 = ⇒ u = 1 , x = 1 = ⇒ u = 0 ....
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This note was uploaded on 02/11/2011 for the course M 408S taught by Professor Stepp during the Spring '11 term at University of Texas at Austin.
 Spring '11
 STEPP
 Calculus

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