Why so serious?

Clearly the roots of quadratic function f'(x) are x=-1 and x=1/3 . So

Suppose $f(x) = ax^{3} + bx^{2} + cx + d.$ We then have that

$f(2) = 8a + 4b + 2c + d = 0,$ (i).

Now since $f'(x) = 3ax^{2} + 2bx + c,$ we have that

$f'(-1) = 3a - 2b + c = 0,$ (ii), and

$f'(\frac{1}{3}) = \frac{a}{3} + \frac{2b}{3} + c = 0 \Longrightarrow a + 2b + 3c = 0,$ (iii).

Finally, $\displaystyle\int_{-1}^{1} f(x) dx = \frac{ax^{4}}{4} + \frac{bx^{3}}{3} + \frac{cx^{2}}{2} + d*x,$

which when evaluated from $x = -1$ to $x = 1$ results in the equation

$\frac{2b}{3} + 2d = \frac{14}{3} \Longrightarrow b + 3d = 7,$ (iv).

Solving equations (i) through (iv) simultaneously gives us that

$a = -\dfrac{7}{29}, b = -\dfrac{7}{29}, c = \dfrac{7}{29}$ and $d = \dfrac{70}{29}.$

Thus $f(x) = -\dfrac{7}{29}(x^{3} + x^{2} - x - 10),$ and so

$\lfloor f(-3) \rfloor = \lfloor -\dfrac{7}{29} * (-25) \rfloor = \lfloor 6.0344... \rfloor = \boxed{6}.$