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

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Arun Kumar · Answer

Arun Kumar IIT Delhi Askiitians Faculty

Praneeth · Answer

x-[x] is periodic with period 1. therefore, ? 0 [x] x-[x] dx = [x]? 0 1 x-[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

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

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

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

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

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