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