y=f(x)= $\left \{ {{6x^{2}} \atop {a-a^{2}x}} \right.$ $\frac{khi}{khi}$ $\frac{x\leq0}{x\geq0}$ và I= $\int\limits^4_{-1}f(x) \, dx$ có tất cả bao nhiêu số nguyên a để I+22 $\geq$ 0

\[\begin{array}{l} I = \int\limits_{ - 1}^4 {f\left( x \right)dx} .\\ = \int\limits_{ - 1}^0 {6{x^2}dx} + \int\limits_0^4 {\left( {a - {a^2}x} \right)dx} \\ = \left. {\frac{{6{x^3}}}{3}} \right|_{ - 1}^0 + \left. {\left( {ax - \frac{{{a^2}{x^2}}}{2}} \right)} \right|_0^4\\ = 2 + 4a - 8{a^2}.\\ \Rightarrow I + 22 \ge 0\\ \Leftrightarrow 2 + 4a - 8{a^2} + 22 \ge 0\\ \Leftrightarrow 8{a^2} - 4a - 24 \le 0\\ \Leftrightarrow - \frac{3}{2} \le a \le 2\\ a \in Z \Rightarrow a \in \left\{ { - 1;\,0;\,\,1;\,\,2} \right\}. \end{array}\]

I = $\int_{-1}^0 6x^2 dx + \int_0^4 (a-a^2x) dx = -8a^2 + 4a + 2$

De $I + 22 \geq 0$ thi $-8a^2 + 4a + 24 \geq 0$ hay $a> 2$ hoac $a< -3/2$