sol16-8 - SOLUTIONS FOR HOMEWORK PROBLEMS 16.8 MATH 132 -...

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SOLUTIONS FOR HOMEWORK PROBLEMS 16.8 MATH 132 - WI00 4 When you are given some fixed x -s then they will definitely be limit values for your integral; hence for our problem we’ll have, being given x = 2 and x = 4: R 4 2 x + 5 dx = x 2 2 + 5 x | 4 2 = ( 4 2 2 + 5 * 4) - ( 2 2 2 + 5 * 2) = 28 - 12 = 16 8 Same trick here, x = - 2 and x = - 1, so: R - 1 - 2 2 x 2 - x dx = 2 x 3 3 - x 2 2 | - 1 - 2 = (2 ( - 1) 3 3 - ( - 1) 2 2 ) - (2 ( - 2) 3 3 - ( - 2) 2 2 ) = (2 - 1 3 - 1 2 ) - (2 - 8 3 - 4 2 ) = - 7 6 - ( - 44 6 ) = 37 6 The reason I went into that much depth into computing the result is that I want to make sure you notice and remember that: ” Negative numbers raised to odd powers remain negative , but the same negative numbers raised to even powers become positive ” !! So a negative number will give up its negative sign when it has a even power (2, 4, 6, etc). 20 Notice that e 2 is a constant . .. I know you know, but just wanted to make sure :). R e 2 1 1 x dx = ln x | e 2 1 = ln (
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This note was uploaded on 07/26/2011 for the course MATH 132 taught by Professor Staff during the Spring '08 term at Ohio State.

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