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

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```                                        $\int_{0}^{[x]}x-[x]dx \\=>[x]^2/2-\int_{1}^{2}1dx-\int_{2}^{3}2dx-\int_{3}^{4}3dx..............depends\, value \,of\, [x]$Arun KumarIIT DelhiAskiitians Faculty
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one year ago
```                                        x-[x] is periodic with period 1. therefore, ?0[x]x-[x] dx = [x]?01x-[x] dx =[x]/2we can get integral of x-[x] within limits 0 and 1 using method of areas as ½.Therefore the answer is [x]/2
```
one year ago

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