Explanation

(a)
$\int_{- 1}^{1}{f(x)dx = \int_{- 1}^{1}{\left( x - \lbrack x\rbrack \right)dx}}$
$= \int_{- 1}^{1}{x\ dx - \int_{- 1}^{1}{\lbrack x\rbrack dx}}$
$= 0 - \int_{- 1}^{1}{\lbrack x\rbrack dx\ \ \ \ \ (\because x}$ is an odd function)
But $\lbrack x\rbrack = \left\{ \begin{array}{r}
- 1, \\
0, \\
1,
\end{array} \right.\ \ \ \begin{matrix}
if & - 1 \leq x < 0 \\
if & 0 \leq x < 1 \\
if & x = 1
\end{matrix}$
$\therefore\ \int_{- 1}^{1}{\lbrack x\rbrack\ dx = \int_{- 1}^{0}{\lbrack x\rbrack\ dx} + \int_{0}^{1}{\lbrack x\rbrack\ dx}}$
$= \int_{- 1}^{0}{\lbrack - 1\rbrack\ dx} + \int_{0}^{1}{0\ dx = - \lbrack x\rbrack_{- 1}^{0} + 0 = 1}$
Thus, $\int_{- 1}^{1}{f(x)dx = 1}$