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Evaluation of definite integrals

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Question

 Let f(x)= maximum {x+|x|,x[x]}, where [x] is the greatest integer less than or equal to  x. Then 22f(x)dx=

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Solution

22f(x)dx=20{x}dx+022xdx=2×12×1×1+(40)=1+4=5

 


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