Answer:
[tex]\geq \sqrt{x} \sqrt{x} \left \{ {{y=2} \atop {x=2}} \right. \int\limits^a_b {x} \, dx \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \alpha \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \pi \frac{x}{y} \int\limits^a_b {x} \, dx \sqrt{x}[/tex]
Step-by-step explanation:
