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exam1_solnfall07

# exam1_solnfall07 - Math 116 First Exam Name Instructor 1 Do...

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Math 116 / Exam 1 (October 10, 2007) page 2 You may find the following partial table of integrals to be useful: Z e ax sin( bx ) dx = 1 a 2 + b 2 e ax ( a sin( bx ) - b cos( bx ) + C ) , Z e ax cos( bx ) dx = 1 a 2 + b 2 e ax ( a cos( bx ) + b sin( bx ) + C ) , Z sin( ax ) sin( bx ) dx = 1 b 2 - a 2 ( a cos( ax ) sin( bx ) - b sin( ax ) cos( bx )) + C, a 6 = b, Z cos( ax ) cos( bx ) dx = 1 b 2 - a 2 ( b cos( ax ) sin( bx ) - a sin( ax ) cos( bx )) + C, a 6 = b, Z sin( ax ) cos( bx ) dx = 1 b 2 - a 2 ( b sin( ax ) sin( bx ) + a cos( ax ) cos( bx )) + C, a 6 = b. Z 1 x 2 + a 2 dx = 1 a arctan( x a ) + C, a 6 = 0 Z bx + c x 2 + a 2 dx = b 2 ln( x 2 + a 2 ) + c a arctan( x a ) + C, a 6 = 0 Z 1 ( x - a )( x - b ) dx = 1 a - b (ln | x - a | - ln | x - b | ) + C, a 6 = b Z cx + d ( x - a )( x - b ) dx = 1 a - b (( ac + d ) ln | x - a | - ( bc + d ) ln | x - b | ) + C, a 6 = b Z 1 a 2 - x 2 dx = arcsin( x a ) + C Z 1 x 2 ± a 2 dx = ln | x + p x 2 ± a 2 | + C You may use the convergence properties of the following integrals: Z 1 1 x p dx converges for p > 1 and diverges for p 1 , Z 1 0 1 x p dx converges for p < 1 and diverges for p 1 , Z 0 e - ax dx converges for a > 0 .
Math 116 / Exam 1 (October 10, 2007) page 3 1. [16 points] For this problem, R 5 1 g ( x ) dx = 12, and f ( x ) = 2 x - 9. Values of g ( x ) are given in the table below.

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exam1_solnfall07 - Math 116 First Exam Name Instructor 1 Do...

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