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The value of $\int\limits_0^{[x]}(-x-[x]) d x$, is |
[x] 2[x] (1/2)[x] none of these |
(1/2)[x] |
We have, $I=\int\limits_0^{[x]}(x-[x]) d x$ $\Rightarrow I=[x] \int\limits_0^1(x-[x]) d x$ [∵ x - [x] is periodic] $\Rightarrow I=\frac{1}{2}[x]$ |
