Explanation

(c)
$I = \int_{- 1}^{0}{\lbrack x + \lbrack x + \lbrack x}\rbrack\rbrack\rbrack dx + \int_{0}^{1}{\lbrack x + \lbrack x + \lbrack x}\rbrack\rbrack\rbrack\ dx$
$\Rightarrow I = I_{1} + I_{2}$
For $I_{1} \rightarrow \lbrack x\rbrack = - 1$ and $- 1 < x < 0$
$- 2 < x + \lbrack x\rbrack < - 1\ \Rightarrow \lbrack x + \lbrack x\rbrack\rbrack\ = - 2$
and now, $- 3 < x + \lbrack x + \lbrack x\rbrack\rbrack < - 2$
$\Rightarrow \ \lbrack x + \lbrack x + \lbrack x\rbrack\rbrack\rbrack\ = - 3$
$I_{1} = \int_{- 1}^{0}{( - 3)dx + - 3\lbrack x\rbrack_{- 1}^{0}} = - 3$
For $I_{2} \rightarrow \lbrack x\rbrack = 0$ and $0 < x < 1$
$\Rightarrow 0 < x + \lbrack x\rbrack < 1\ \Rightarrow \lbrack x + \lbrack x\rbrack\rbrack = 0$
Again, $0 \leq x + \lbrack x + \lbrack x\rbrack\rbrack < \ 1$
$\Rightarrow \ \lbrack x + \lbrack x + \lbrack x\rbrack\rbrack\rbrack\ = 0$
So, $I_{2} = 0$
$\therefore\ I_{1} + I_{2} = - 3$