Explanation

(a)
$\because x = \lbrack x\rbrack + \left\{ x \right\} = n + f$ (say)
Where $\lbrack x\rbrack = n,\ \left\{ x \right\} = f(0 \leq f < 1)$
Now, $\int_{0}^{x}{\left\lbrack t^{2} \right\rbrack dt = 2(x - 1)}$
$\Rightarrow \ \int_{0}^{n + f}{\lbrack t\rbrack^{2}dt = 2(n + f - 1)}$
$\Rightarrow \ \sum_{r = 0}^{n - 1}{\int_{r}^{r + 1}{\lbrack t\rbrack^{2}dt +}\int_{n}^{n + f}{\lbrack t\rbrack^{2}dt = 2(n + f - 1)}}$
$\Rightarrow \ \sum_{r = 0}^{n - 1}{r^{2} + n^{2}f = 2(n + f - 1)}$
$\Rightarrow \ \frac{(n - 1)n(2n - 1)}{6} - 2(n - 1) = \left( 2 - n^{2} \right)f$
For $n = 1,\ 0 - 0 = f\ \Rightarrow f = 0$
Then, $x = n + f = 1 + 0 = 1$
For $n = 2,\ 1 - 2 = - 2f\ \Rightarrow \ f = \frac{1}{2}$
Then, $x = 2 + \frac{1}{2} = \frac{5}{2}$
For $n > 2,\ f < 0$ impossible
Hence, $x = 1,\ \frac{5}{2}$