Let $$[t]$$ denote the greatest integer less than or equal to $$t$$. Then the value of the integral $$\int_0^1 [-8x^2 + 6x - 1] dx$$ is equal to

Given,

$$I=\int_0^1[-8x^2+6x-1]\,dx$$

Let

$$f(x)=-8x^2+6x-1$$

Now,

$$f(x)=0$$

gives

$$8x^2-6x+1=0$$

$$x=\frac{6\pm\sqrt{36-32}}{16}$$

$$x=\frac14,\frac12$$

Also,

$$f(x)>0\quad \text{for}\quad \frac14<x<\frac12$$

and

$$f(x)<0\quad \text{elsewhere}$$

Now,

For

$$0\le x<\frac14,\qquad -1<f(x)\le0$$

so,

$$[f(x)]=-1$$

For

$$\frac14<x<\frac12,\qquad 0<f(x)<1$$

so,

$$[f(x)]=0$$

For

$$\frac12<x\le1,\qquad -3\le f(x)<0$$

Now,

$$f(x)=-1$$

gives

$$-8x^2+6x=0$$

$$x=0,\frac34$$

Hence,

$$-1\le f(x)<0\quad \text{for}\quad \frac12<x<\frac34$$

so,

$$[f(x)]=-1$$

Also,

$$f(x)=-2$$

gives

$$8x^2-6x-1=0$$

$$x=\frac{3+\sqrt{17}}{8}$$

Thus,

$$-2\le f(x)<-1\quad \text{for}\quad \frac34<x<\frac{3+\sqrt{17}}{8}$$

so,

$$[f(x)]=-2$$

and

$$-3\le f(x)<-2\quad \text{for}\quad \frac{3+\sqrt{17}}{8}<x\le1$$

so,

$$[f(x)]=-3$$

Therefore,

$$I=\int_0^{1/4}(-1)\,dx+\int_{1/4}^{1/2}0\,dx+\int_{1/2}^{3/4}(-1)\,dx$$

$$+\int_{3/4}^{(3+\sqrt{17})/8}(-2)\,dx+\int_{(3+\sqrt{17})/8}^{1}(-3)\,dx$$

Evaluating,

$$I=-\frac14-\frac14-2\left(\frac{\sqrt{17}-3}{8}\right)-3\left(1-\frac{3+\sqrt{17}}{8}\right)$$

$$I=\frac{\sqrt{17}-13}{8}$$

Hence, the correct answer is

$$\boxed{\text{Option C }\frac{\sqrt{17}-13}{8}}$$