Explanation

(d)
$\int_{- 1}^{1}{\lbrack x\sin{\pi x\rbrack\ dx = 2\int_{0}^{1}{\lbrack x\sin{\pi x\rbrack\ dx}}}}$ …(i)
(as $\lbrack x\sin{\pi x\rbrack}$ is a even function)
Now, $0 \leq x \leq 1$
$\Rightarrow \ 0 \leq \pi x \leq \pi\$
$\Rightarrow \ 0 \leq \sin{\pi x \leq 1}$ ….(ii)
$\Rightarrow \ 0 \leq x\sin{\pi x < 1}$
$\Rightarrow \ \lbrack x\sin{\pi x\rbrack = 0}$
From Eqs. (i) and (ii), we get
$\int_{0}^{1}{\lbrack x\sin{\pi x\rbrack dx = 0}}$