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Find the area of the bounded region enclosed by the curve y = 3 - |x - 1/2| - |x + 1| and the x-axis.

1. 9/4
2. 45/16
3. 27/8
4. 63/16

Correct answer: 27/8

Solution

Critical points: x = -1 and x = 1/2. For x < -1: y = 3-(1/2-x)-(-(x+1)) = 3-1/2+x+x+1 = 3.5+2x. For -1<=x<=1/2: y = 3-(1/2-x)-(x+1) = 3-1/2+x-x-1 = 3/2. For x > 1/2: y = 3-(x-1/2)-(x+1) = 3-x+1/2-x-1 = 5/2-2x. Zeros: 3.5+2x=0 -> x=-7/4; 5/2-2x=0 -> x=5/4. Area = integral from -7/4 to -1 of (3.5+2x)dx + integral from -1 to 1/2 of (3/2)dx + integral from 1/2 to 5/4 of (5/2-2x)dx. Part1: [3.5x+x²] from -7/4 to -1 = (3.5(-1)+1)-{3.5(-7/4)+49/16} = (-2.5)-{-49/8+49/16} = -2.5-(-49/16) = -2.5+49/16 = -40/16+49/16=9/16. Part2: (3/2)*(3/2)=9/4. Part3: [5x/2-x²] from 1/2 to 5/4 = (25/8-25/16)-(5/4-1/4)=25/16-1=9/16. Total = 9/16+9/4+9/16 = 9/16+36/16+9/16=54/16=27/8.

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