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Exam1Review-solutions - Van Ligten (hlv63) Exam1Review...

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Unformatted text preview: Van Ligten (hlv63) Exam1Review Gilbert (56650) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 0.0 points Rewrite the sum 5 n parenleftBig 2 + 6 n parenrightBig 2 + 5 n parenleftBig 2 + 12 n parenrightBig 2 + . . . + 5 n parenleftBig 2 + 6 n n parenrightBig 2 using sigma notation. 1. n summationdisplay i = 1 5 n parenleftBig 2 i + 6 i n parenrightBig 2 2. n summationdisplay i = 1 6 i n parenleftBig 2 + 5 i n parenrightBig 2 3. n summationdisplay i = 1 6 n parenleftBig 2 + 5 i n parenrightBig 2 4. n summationdisplay i = 1 5 n parenleftBig 2 + 6 i n parenrightBig 2 correct 5. n summationdisplay i = 1 6 n parenleftBig 2 i + 5 i n parenrightBig 2 6. n summationdisplay i = 1 5 i n parenleftBig 2 + 6 i n parenrightBig 2 Explanation: The terms are of the form 5 n parenleftBig 2 + 6 i n parenrightBig 2 , with i = 1 , 2 , . . . , n . Consequently in sigma notation the sum becomes n summationdisplay i = 1 5 n parenleftBig 2 + 6 i n parenrightBig 2 . 002 0.0 points Find the value of the integral I = integraldisplay 9 1 f ( x ) dx when integraldisplay 12 1 f ( x ) dx = 5 , integraldisplay 12 9 f ( x ) dx = 7 . 1. I = 2 correct 2. I = 4 3. I = 0 4. I = 3 5. I = 1 Explanation: By the linearity of integration, integraldisplay 9 1 f ( x ) dx + integraldisplay 12 9 f ( x ) dx = integraldisplay 12 1 f ( x ) dx . Consequently, I = integraldisplay 12 1 f ( x ) dx integraldisplay 12 9 f ( x ) dx = 2 . 003 0.0 points Determine F ( x ) when F ( x ) = integraldisplay x 4 4 sin t t dt . 1. F ( x ) = 2 sin( x ) x 2. F ( x ) = 4 cos( x ) x 3. F ( x ) = 4 cos x x 4. F ( x ) = 2 sin( x ) x correct 5. F ( x ) = 2 sin x x 6. F ( x ) = 4 cos x x Van Ligten (hlv63) Exam1Review Gilbert (56650) 2 7. F ( x ) = 4 sin x x 8. F ( x ) = 2 cos( x ) x Explanation: By the Fundamental Theorem of Calculus and the Chain Rule, d dx parenleftBig integraldisplay g ( x ) a f ( t ) dt parenrightBig = f ( g ( x )) g ( x ) . When F ( x ) = integraldisplay x 4 4 sin t t dt , therefore, F ( x ) = 4 sin( x ) x parenleftBig d dx x parenrightBig . Consequently, F ( x ) = 2 sin( x ) x , since d dx x = 1 2 x . 004 0.0 points When F is defined for x 6 by F ( x ) = integraldisplay x 6 3 t 2 + 13 t 10 4 + 3 t 2 dt, determine the largest interval(s) on which F is decreasing. 1. bracketleftbigg 5 , 2 3 bracketrightbigg correct 2. [ 5 , ) 3. [ 6 , 5] uniondisplay bracketleftbigg 2 3 , parenrightbigg 4. bracketleftbigg 6 , 2 3 bracketrightbigg uniondisplay [5 , ) 5....
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This note was uploaded on 04/13/2010 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas at Austin.

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Exam1Review-solutions - Van Ligten (hlv63) Exam1Review...

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