Integrating Floor

Let $f(x) = x^2-x+1$ .

When $f(x) = 1\quad \Rightarrow x^2-x+1 = 1$

$\Rightarrow x^2-x = 0 \quad \Rightarrow x(x-1) = 0 \quad \Rightarrow \begin{cases} x = 0 \\ x = 1 \end{cases}$

When $f(x) = 2\quad \Rightarrow x^2-x+1 = 2$

$\Rightarrow x^2-x - 1= 0 \quad \Rightarrow x = \dfrac {1+\sqrt{5}}{2}$ , since $x > 0$

When $x=2\quad \Rightarrow f(2) = 3$ .

Therefore, $\begin{cases} \lfloor x^2-x+1 \rfloor = 0 \text{ for } x \in (0,1) \\ \lfloor x^2-x+1 \rfloor = 1 \text{ for } x \in \left[1,\frac {1+\sqrt{5}}{2}\right) \\ \lfloor x^2-x+1 \rfloor = 2 \text{ for } x \in \left[\frac {1+\sqrt{5}}{2},2 \right) \end{cases}$

$\Rightarrow \begin{aligned} \displaystyle \int_0^2 {\lfloor x^2-x+1 \rfloor dx} & = \int_1^{\frac {1+\sqrt{5}}{2}} { dx} + \int_{\frac {1+\sqrt{5}}{2}}^2 {2 dx} \\ & = \frac {1+\sqrt{5}}{2} - 1 + 2\left( 2 - \frac {1+\sqrt{5}}{2} \right) \\ & = \frac {1 + \sqrt{5} + 4 - 2\sqrt{5}}{2} = \frac {5 - \sqrt{5}}{2} \end{aligned}$

$\Rightarrow a + b = 5 + 2 = \boxed{7}$

did the same.