# =\$800+∫_0^300▒〖(0.0003x^2-0.18x+22)dx〗

May 19th, 2015
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=\$800+∫_0^300▒〖(0.0003x^2-0.18x+22)dx〗 = \$800 + [0.0003x3/3 - 0.18x2/2 + 22x] NOTE: the integral=0 when x=0 so we don't need to include the x=0 case for part "A". = \$800+(0.0003)(300)(300)(300)/3 - 0.18(300)(300)/2 + 22(300) = 800+2700-8100+6600=\$2000

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= \$800 + [0.0003x3/3 - 0.18x2/2 + 22x]NOTE:the integral=0 when x=0 so we don't need to include the x=0 case for part "A".= \$800+(0.0003)(300)(300)(300)/3 - 0.18(300)(300)/2 + 22(300) = 800+2700-8100+6600=\$2000

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