Let \( [t] \) denote the greatest integer less than or equal to \( t \). Then, the value of the integral \( \int_{0}^{1} \left[ -8x^2 + 6x - 1 \right] dx \) is equal to:
\( -1 \)
\( -\frac{5}{4} \)
\( \frac{\sqrt{17} - 13}{8} \)
\( \frac{\sqrt{17} - 16}{8} \)