Let [t] denote the greatest integer less than or equal to t.

Question

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

Accepted Answer

$-8x^2+6x-1=0$ at $x=\dfrac{3\pm\sqrt{17}}{8}$ (only $x=\dfrac{3+\sqrt{17}}8\in(0,1)$ is relevant). Also check where the bracket crosses $-1,0,\ldots$. Breaking at $x=1/4,\,1/2,\,3/4,\,\dfrac{3+\sqrt{17}}8$:

$$\int_0^1[-8x^2+6x-1]\,dx = \int_0^{1/4}(-1)dx+\int_{1/4}^{1/2}0\,dx+\int_{1/2}^{3/4}(-1)dx+\int_{3/4}^{(3+\sqrt{17})/8}(-2)dx+\int_{(3+\sqrt{17})/8}^1(-3)dx$$

Carrying out these evaluations gives

$$\int_0^1[-8x^2+6x-1]\,dx = \frac{\sqrt{17}-13}8$$

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