sol16-2

# sol16-2 - -1) + C 1 = 3-2 + C 1 = 1 + C 1 = 5 → C 1 = 4....

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SOLUTIONS CHAPTER 16.2 MATH 132 - WI01 2. y = R ( x 2 - x ) dx = x 3 3 - x 2 2 + C . Using y (3) = 4 3 3 3 - 3 2 2 + C = 4 C = - 1 2 . Thus y = x 3 3 - x 2 2 - 1 2 . 6. y 00 = x + 1 y 0 = R ( x + 1) dx = x 2 2 + x + C 1 . Using y 0 (0) = 0 0 + 0 + C 1 = 0 C 1 = 0. Thus y = R [ x 2 2 + x ] dx = 1 2 R x 2 dx + R xdx = 1 2 x 3 3 + x 2 2 + C 2 = x 3 6 + x 2 2 + C 2 . y (0) = 5 0 + 0 + C 2 = 5, so C 2 = 5. Thus y = x 3 6 + x 2 2 + 5. 12. r 0 = 10000 - 2(2 q + q 3 ), so r = R (10000 - 4 q - 2 q 3 ) dq = 10000 q - 4 q 2 2 - 2 q 4 4 + C = 10000 q - 2 q 2 - q 4 2 + C . When q = 0 r = 0, hence 0 - 0 - 0+ C = 0 C = 0 r = 10000 q - 2 q 2 - q 4 4 . Therefore, the demand function is p = r q = 10000 - 2 q - q 3 2 . 22. f 00 ( x ) = 6 x + 2 f 0 ( x ) = R (6 x + 2) dx = 6 x 2 2 + 2 x + C 1 = 3 x 2 + 2 x + C 1 . From f 0 ( - 1) = 5 3( - 1) 2 + 2(
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Unformatted text preview: -1) + C 1 = 3-2 + C 1 = 1 + C 1 = 5 → C 1 = 4. Thus f ( x ) = 3 x 2 + 2 x + 4. f ( x ) = R (3 x 2 + 2 x + 4) dx = 3 x 3 3 + 2 x 2 2 + 4 x + C 2 = x 3 + x 2 + 4 x + C 2 Thus f (1)-f (-1) = (1+1+4+ C 2 )-(-1+1-4+ C 2 ) = 6+ C 2-(-4)-C 2 = 6 + 4 = 10. • Date : 02/23/2001. 1...
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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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