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Question

Let f(x)=max{3,x2,1x2} for 12x2.
Then the value of the integral  21/2f(x)dx is 
  1. 113
  2. 133
  3. 143
  4. 163


Solution

The correct option is C 143

21/2f(x)dx=1/31/21x2 dx+31/33 dx+23x2dx
=[1x]1/31/2+[3x]31/3+[x33]23=143

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