The value of $\int\limits_0^{[x]}(-x-[x]) d x$, is
Answer & explanation
Correct answer: option 3
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]
$\Rightarrow I=\frac{1}{2}[x]$
The value of $\int\limits_0^{[x]}(-x-[x]) d x$, is
Correct answer: option 3
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]
$\Rightarrow I=\frac{1}{2}[x]$