the value of integrand limit [0 to [x]] {x-[x]}dx (where [.] denotes greatest integer function.

3 years ago

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Arun Kumar
IIT Delhi
Askiitians Faculty
8 months ago
                                        
x-[x] is periodic with period 1. therefore, ?0[x]x-[x] dx = [x]?01x-[x] dx =[x]/2
we can get integral of x-[x] within limits 0 and 1 using method of areas as ½.
Therefore the answer is [x]/2
8 months ago

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