$$\int_0^2 \left|2x^2 - 3x + \left[x - \frac{1}{2}\right]\right| dx$$, where $$[t]$$ is the greatest integer function, is equal to

We need to evaluate $$\displaystyle\int_0^2 \left|2x^2 - 3x + \left[x - \dfrac{1}{2}\right]\right| dx$$ where $$[t]$$ denotes the greatest integer function.

We split the interval based on where $$\left[x - \dfrac{1}{2}\right]$$ changes value. Setting $$u = x - 1/2$$, the floor function $$[u]$$ changes at integer values of $$u$$, i.e., at $$x = 1/2, 3/2, 5/2, \ldots$$

On $$[0, 1/2)$$: $$x - 1/2 \in [-1/2, 0)$$, so $$[x - 1/2] = -1$$. The integrand becomes $$|2x^2 - 3x - 1|$$.

On $$[1/2, 3/2)$$: $$x - 1/2 \in [0, 1)$$, so $$[x - 1/2] = 0$$. The integrand becomes $$|2x^2 - 3x|$$.

On $$[3/2, 2]$$: $$x - 1/2 \in [1, 3/2]$$, so $$[x - 1/2] = 1$$. The integrand becomes $$|2x^2 - 3x + 1|$$.

Interval [0, 1/2]: $$g_1(x) = 2x^2 - 3x - 1$$. At $$x = 0$$: $$g_1 = -1 < 0$$. At $$x = 1/2$$: $$g_1 = 1/2 - 3/2 - 1 = -2 < 0$$. So $$|g_1| = -(2x^2 - 3x - 1) = -2x^2 + 3x + 1$$.

$$\displaystyle\int_0^{1/2} (-2x^2 + 3x + 1)\,dx = \left[-\dfrac{2x^3}{3} + \dfrac{3x^2}{2} + x\right]_0^{1/2} = -\dfrac{1}{12} + \dfrac{3}{8} + \dfrac{1}{2} = -\dfrac{1}{12} + \dfrac{3}{8} + \dfrac{1}{2}$$

$$= \dfrac{-2 + 9 + 12}{24} = \dfrac{19}{24}$$

Interval [1/2, 3/2]: $$g_2(x) = 2x^2 - 3x = x(2x - 3)$$. This is zero at $$x = 0$$ and $$x = 3/2$$. On $$(1/2, 3/2)$$: $$g_2 < 0$$.

$$\displaystyle\int_{1/2}^{3/2} |2x^2 - 3x|\,dx = \int_{1/2}^{3/2} (3x - 2x^2)\,dx = \left[\dfrac{3x^2}{2} - \dfrac{2x^3}{3}\right]_{1/2}^{3/2}$$

At $$x = 3/2$$: $$\dfrac{3 \cdot 9/4}{2} - \dfrac{2 \cdot 27/8}{3} = \dfrac{27}{8} - \dfrac{54}{24} = \dfrac{27}{8} - \dfrac{9}{4} = \dfrac{27 - 18}{8} = \dfrac{9}{8}$$

At $$x = 1/2$$: $$\dfrac{3 \cdot 1/4}{2} - \dfrac{2 \cdot 1/8}{3} = \dfrac{3}{8} - \dfrac{1}{12} = \dfrac{9 - 2}{24} = \dfrac{7}{24}$$

Result: $$\dfrac{9}{8} - \dfrac{7}{24} = \dfrac{27 - 7}{24} = \dfrac{20}{24} = \dfrac{5}{6}$$

Interval [3/2, 2]: $$g_3(x) = 2x^2 - 3x + 1 = (2x - 1)(x - 1)$$. At $$x = 3/2$$: $$g_3 = 2(9/4) - 9/2 + 1 = 9/2 - 9/2 + 1 = 1 > 0$$. At $$x = 2$$: $$g_3 = 8 - 6 + 1 = 3 > 0$$. So $$|g_3| = g_3$$.

$$\displaystyle\int_{3/2}^{2} (2x^2 - 3x + 1)\,dx = \left[\dfrac{2x^3}{3} - \dfrac{3x^2}{2} + x\right]_{3/2}^{2}$$

At $$x = 2$$: $$\dfrac{16}{3} - 6 + 2 = \dfrac{16}{3} - 4 = \dfrac{4}{3}$$

At $$x = 3/2$$: $$\dfrac{2 \cdot 27/8}{3} - \dfrac{3 \cdot 9/4}{2} + \dfrac{3}{2} = \dfrac{9}{4} - \dfrac{27}{8} + \dfrac{3}{2} = \dfrac{18 - 27 + 12}{8} = \dfrac{3}{8}$$

Result: $$\dfrac{4}{3} - \dfrac{3}{8} = \dfrac{32 - 9}{24} = \dfrac{23}{24}$$

Total: $$\dfrac{19}{24} + \dfrac{5}{6} + \dfrac{23}{24} = \dfrac{19}{24} + \dfrac{20}{24} + \dfrac{23}{24} = \dfrac{62}{24} = \dfrac{31}{12}$$

The correct answer is Option C: $$\dfrac{31}{12}$$.