If $$[t]$$ denotes the greatest integer $$\leq t$$, then the value of $$\int_0^1 [2x - |3x^2 - 5x + 2| + 1] dx$$ is

We need to evaluate $$\displaystyle\int_0^1 [2x - |3x^2 - 5x + 2| + 1]\, dx$$ where $$[t]$$ is the greatest integer function.

We first analyse $$g(x) = 2x - |3x^2 - 5x + 2| + 1$$. The expression inside the absolute value is $$3x^2 - 5x + 2 = (3x - 2)(x - 1)$$. The roots are $$x = \dfrac{2}{3}$$ and $$x = 1$$.

For $$x \in [0, 2/3]$$: $$3x^2 - 5x + 2 \geq 0$$ (both factors have consistent sign), so $$|3x^2 - 5x + 2| = 3x^2 - 5x + 2$$, and $$g(x) = 2x - (3x^2 - 5x + 2) + 1 = -3x^2 + 7x - 1$$.

For $$x \in [2/3, 1]$$: $$3x^2 - 5x + 2 \leq 0$$, so $$|3x^2 - 5x + 2| = -(3x^2 - 5x + 2)$$, and $$g(x) = 2x + 3x^2 - 5x + 2 + 1 = 3x^2 - 3x + 3$$.

Now for the floor function, we need to determine where $$g(x)$$ crosses integer values.

On $$[0, 2/3]$$: $$g(x) = -3x^2 + 7x - 1$$. At $$x = 0$$: $$g(0) = -1$$. At $$x = 2/3$$: $$g(2/3) = -3(4/9) + 7(2/3) - 1 = -4/3 + 14/3 - 1 = 10/3 - 1 = 7/3 \approx 2.33$$. The vertex is at $$x = 7/6 > 2/3$$, so on $$[0, 2/3]$$, $$g$$ is increasing.

We find where $$g(x) = 0$$: $$-3x^2 + 7x - 1 = 0$$, so $$3x^2 - 7x + 1 = 0$$, giving $$x = \dfrac{7 \pm \sqrt{49 - 12}}{6} = \dfrac{7 \pm \sqrt{37}}{6}$$. The smaller root is $$x_1 = \dfrac{7 - \sqrt{37}}{6}$$.

Where $$g(x) = 1$$: $$-3x^2 + 7x - 1 = 1$$, so $$3x^2 - 7x + 2 = 0$$, giving $$x = \dfrac{7 \pm \sqrt{49-24}}{6} = \dfrac{7 \pm 5}{6}$$. So $$x = \dfrac{1}{3}$$ or $$x = 2$$. On our interval, $$x_2 = \dfrac{1}{3}$$.

Where $$g(x) = 2$$: $$-3x^2 + 7x - 1 = 2$$, so $$3x^2 - 7x + 3 = 0$$, giving $$x = \dfrac{7 \pm \sqrt{49-36}}{6} = \dfrac{7 \pm \sqrt{13}}{6}$$. The smaller root is $$x_3 = \dfrac{7 - \sqrt{13}}{6}$$.

On $$[2/3, 1]$$: $$g(x) = 3x^2 - 3x + 3$$. At $$x = 2/3$$: $$g = 3(4/9) - 2 + 3 = 4/3 + 1 = 7/3$$. At $$x = 1$$: $$g = 3 - 3 + 3 = 3$$. The minimum is at $$x = 1/2$$ (outside this interval), so on $$[2/3, 1]$$, $$g$$ is increasing from $$7/3$$ to $$3$$.

Where $$g(x) = 3$$: $$3x^2 - 3x + 3 = 3$$, so $$3x^2 - 3x = 0$$, giving $$x = 0$$ or $$x = 1$$. So $$g(x) = 3$$ only at $$x = 1$$. On $$[2/3, 1)$$, $$g(x) \in [7/3, 3)$$, so $$[g(x)] = 2$$.

Now on $$[0, 2/3]$$, the floor values are:

• $$[0, x_1)$$: $$g(x) \in [-1, 0)$$, so $$[g(x)] = -1$$
• $$[x_1, x_2)$$: $$g(x) \in [0, 1)$$, so $$[g(x)] = 0$$
• $$[x_2, x_3)$$: $$g(x) \in [1, 2)$$, so $$[g(x)] = 1$$
• $$[x_3, 2/3]$$: $$g(x) \in [2, 7/3]$$, so $$[g(x)] = 2$$

The integral becomes: $$I = (-1)(x_1 - 0) + 0 \cdot (x_2 - x_1) + 1 \cdot (x_3 - x_2) + 2 \cdot (2/3 - x_3) + 2 \cdot (1 - 2/3)$$

$$= -x_1 + (x_3 - x_2) + 2(2/3 - x_3) + 2/3$$

$$= -x_1 + x_3 - x_2 + 4/3 - 2x_3 + 2/3$$

$$= -x_1 - x_2 - x_3 + 2$$

Substituting: $$x_1 = \dfrac{7 - \sqrt{37}}{6}$$, $$x_2 = \dfrac{1}{3} = \dfrac{2}{6}$$, $$x_3 = \dfrac{7 - \sqrt{13}}{6}$$.

$$x_1 + x_2 + x_3 = \dfrac{7 - \sqrt{37} + 2 + 7 - \sqrt{13}}{6} = \dfrac{16 - \sqrt{37} - \sqrt{13}}{6}$$

$$I = 2 - \dfrac{16 - \sqrt{37} - \sqrt{13}}{6} = \dfrac{12 - 16 + \sqrt{37} + \sqrt{13}}{6} = \dfrac{\sqrt{37} + \sqrt{13} - 4}{6}$$

Hence, the correct answer is Option A: $$\dfrac{\sqrt{37}+\sqrt{13}-4}{6}$$.